Closed-forms for several tough integrals

Question

These integrals came up in the process of finding solution to Vladimir Reshetnikov's problem. I wonder if there are closed-forms for the following integrals: \begin{array}{1,1} &[\text{1}] &\quad\int_0^1\frac{\operatorname{Li}_3(ax)}{1+2x}\ dx\\[12pt] &[\text{2}] &\quad\int_0^1\frac{\operatorname{Li}_2(ax)\ln x}{1+2x}\ dx\\[12pt] &[\text{3}] &\quad\int_0^1\frac{\ln(1-ax)\ln^2 x}{1+2x}\ dx \end{array} I have tried many substitutions, integration by parts, or differentiation under integral sign method, but without success so far. I do not need a complete or rigorous answer and your answer can be only Mathematica's or Maple's output since I don't have those software packages in my computer or links of related papers. I'd be grateful for any help you are able to provide.

Answer 1

Here are values of the integral $[3]$ for some specific values of the parameter $a$: $$\int_0^1\frac{\ln(1-x)\ln^2x}{1+2x}dx=2\operatorname{Li}_4\left(\frac12\right)-\operatorname{Li}_4\left(\frac13\right)-\operatorname{Li}_4\left(\frac23\right)-\frac14\operatorname{Li}_4\left(\frac14\right)\\-\frac{\ln^42}{12}-\frac{\ln^43}{12}+\frac16\ln2\cdot\ln^33+\frac{\pi^2}6\left(\ln2\cdot\ln3-\ln^22-\operatorname{Li}_2\left(\frac13\right)\right).\tag1$$

$$\int_0^1\frac{\ln(1+x)\ln^2x}{1+2x}dx=3\operatorname{Li}_4\left(\frac12\right)-\frac34\operatorname{Li}_4\left(\frac14\right)\\+\left(7\zeta(3)-\operatorname{Li}_3\left(\frac14\right)\right)\frac{\ln2}4-\frac{3\pi^4}{160}-\frac{\ln^42}{24}-\frac{\pi^2}6\ln^22.\tag2$$

$$\int_0^1\frac{\ln(1+2x)\ln^2x}{1+2x}dx=\operatorname{Li}_4\left(\frac12\right)+\operatorname{Li}_4\left(\frac13\right)+\operatorname{Li}_4\left(\frac23\right)-\frac18\operatorname{Li}_4\left(\frac14\right)\\+\left(\operatorname{Li}_3\left(\frac13\right)+\operatorname{Li}_3\left(\frac23\right)\right)\ln3-\frac{11\pi^4}{360}-\frac{\ln^42}{24}-\frac{\ln^43}4\\+\frac{\pi^2}{12}\left(\ln^23-\ln^22\right)+\frac13\ln2\cdot\ln^33.\tag3$$

Answer 2

I numerically found some more closed forms for the integrals. Perhaps you might find these helpful. (Sorry about the formatting.) I like how they all have closed forms at the same arguments.$\def\tfrac#1#2{{\textstyle\frac{#1}{#2}}}$

The third integral

$$\begin{eqnarray} \mathrm{III}(-8) &=& -6 \text{Li}_2(\tfrac{1}{3}) \zeta (2)-\tfrac{4}{3} \text{Li}_4(\tfrac{1}{2})+\tfrac{13}{4} \text{Li}_4(\tfrac{1}{4})+\tfrac{7}{2} \text{Li}_4(\tfrac{1}{3})-\tfrac{15}{4} \text{Li}_4(\tfrac{3}{4})-4\text{Li}_4(\tfrac{2}{3})+6 \text{Li}_2(\tfrac{1}{3}){}^2+6 \text{Li}_2(\tfrac{1}{3}) \log^23-6 \text{Li}_3(\tfrac{2}{3}) \log3-6 \text{Li}_3(\tfrac{1}{3}) \log3-12\text{Li}_2(\tfrac{1}{3}) \log2 \log3+13 \zeta (3) \log3+\tfrac{31}{6} \zeta (4)+\tfrac{25}{3} \zeta (2) \log^22-7 \zeta (2) \log^23+\tfrac{7}{18} (-\log^42)+\tfrac{167}{48} \log^43-\tfrac{25}{3} \log2 \log^3(3)+5 \log^32 \log3+\tfrac{9}{4} \log^22 \log^23 \\ \mathrm{III}(-3) &=& -\tfrac{1}{2} \text{Li}_4(\tfrac{3}{4})-\tfrac{17}{8} \text{Li}_4(\tfrac{1}{4})-2 \text{Li}_4(\tfrac{2}{3})+7\text{Li}_4(\tfrac{1}{2})+\text{Li}_4(\tfrac{1}{3})-\text{Li}_2(\tfrac{1}{3}){}^2-\text{Li}_2(\tfrac{1}{3}) \log^23+4 \text{Li}_3(\tfrac{2}{3}) \log2+2 \text{Li}_3(\tfrac{1}{3})\log2-2 \zeta (3) \log2-\tfrac{7}{2} \zeta (4)-\tfrac{5}{2} \zeta (2) \log^22+\zeta (2) \log^23+2 \zeta (2) \log2 \log3+\tfrac{7}{24} (-\log^43)-\tfrac{35}{24} \log^42-\tfrac{2}{3} \log2 \log^33+\tfrac{2}{3} \log^3(2) \log3+\tfrac{3}{2} \log^22 \log^23 \end{eqnarray}$$

$$\begin{eqnarray} \def\tfrac#1#2{{\textstyle\frac{#1}{#2}}} \mathrm{III}(\tfrac23) &=& 2 \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{1}{2} \text{Li}_4(\tfrac{3}{4})+\tfrac{7}{4} \text{Li}_4(\tfrac{1}{4})+2 \text{Li}_4(\tfrac{2}{3})-4\text{Li}_4(\tfrac{1}{2})-\text{Li}_4(\tfrac{1}{3})-\text{Li}_2(\tfrac{1}{3}){}^2-2 \text{Li}_2(\tfrac{1}{3}) \log^23-\text{Li}_3(\tfrac{2}{3}) \log3-3 \text{Li}_3(\tfrac{1}{3})\log3+2 \text{Li}_2(\tfrac{1}{3}) \log2 \log3+\tfrac{19}{6} \zeta (3) \log3-\tfrac{7}{4} \zeta (4)-\zeta (2) \log^23+\zeta (2) \log^22+\tfrac{1}{24} (-\log^43)+\tfrac{4}{3} \log^42+\tfrac{1}{2} \log2 \log^3(3)-\tfrac{2}{3} \log^32 \log3-\tfrac{1}{2} \log^22 \log^23 \\ \mathrm{III}(1) &=& -\text{Li}_2(\tfrac{1}{3}) \zeta (2)-\tfrac{1}{4} \text{Li}_4(\tfrac{1}{4})-\text{Li}_4(\tfrac{2}{3})+2 \text{Li}_4(\tfrac{1}{2})-\text{Li}_4(\tfrac{1}{3})-\zeta (2) \log^22+\zeta (2) \log2 \log(3)+\tfrac{1}{12} (-\log^43)-\tfrac{1}{12} \log^42+\tfrac{1}{6} \log2 \log^33 \\ \mathrm{III}(-1) &=& -\tfrac{3}{4} \text{Li}_4(\tfrac{1}{4})+3 \text{Li}_4(\tfrac{1}{2})+\text{Li}_3(\tfrac{2}{3}) \log2+\text{Li}_3(\tfrac{1}{3}) \log2+\tfrac{1}{8} \zeta (3) (-\log2)-\tfrac{27}{16} \zeta (4)-\tfrac{3}{2}\zeta (2) \log^22+\zeta (2) \log2 \log3-\tfrac{3}{8} \log^42-\tfrac{1}{3} \log2 \log^33+\tfrac{1}{2} \log^22 \log^23 \\ \mathrm{III}(-2) &=& -\tfrac{1}{8} \text{Li}_4(\tfrac{1}{4})+\text{Li}_4(\tfrac{2}{3})+\text{Li}_4(\tfrac{1}{2})+\text{Li}_4(\tfrac{1}{3})+\text{Li}_3(\tfrac{2}{3}) \log3+\text{Li}_3(\tfrac{1}{3}) \log 3-\tfrac{11}{4} \zeta (4)+\tfrac{1}{2} \zeta (2) \log^23-\tfrac{1}{2} \zeta (2) \log^22+\tfrac{1}{24} (-\log^42)-\tfrac{1}{4} \log^43+\tfrac{1}{3} \log2 \log^33 \\ \mathrm{III}(-\tfrac12) &=& \tfrac{1}{8} (-\text{Li}_4(\tfrac{1}{4}))-\tfrac{2}{3} \text{Li}_4(\tfrac{1}{2})+\tfrac{3}{2} \text{Li}_4(\tfrac{3}{4})+3\text{Li}_4(\tfrac{2}{3})+\text{Li}_2(\tfrac{1}{3}){}^2+\text{Li}_2(\tfrac{1}{3}) \log^23+\text{Li}_2(\tfrac{1}{3}) \log^22-3 \text{Li}_3(\tfrac{2}{3}) \log3+3\text{Li}_3(\tfrac{2}{3}) \log2-3 \text{Li}_3(\tfrac{1}{3}) \log3+3 \text{Li}_3(\tfrac{1}{3}) \log2-2 \text{Li}_2(\tfrac{1}{3}) \log2 \log3+\tfrac{13}{2} \zeta (3) \log3-\tfrac{13}{2} \zeta(3) \log2-\tfrac{79}{24} \zeta (4)-\tfrac{9}{2} \zeta (2) \log^23-\tfrac{19}{3} \zeta (2) \log^22+10 \zeta (2) \log2 \log3+\tfrac{8}{9} \log^42+\tfrac{11}{8} \log^43-\tfrac{17}{6} \log^32 \log3-4 \log2 \log^3(3)+\tfrac{9}{2} \log^22 \log^23 \end{eqnarray}$$

$$\begin{eqnarray} \mathrm{III}(\tfrac14) &=& -6 \text{Li}_2(\tfrac{1}{3}) \zeta (2)-\tfrac{7}{2} \text{Li}_4(\tfrac{1}{3})-\tfrac{29}{8} \text{Li}_4(\tfrac{1}{4})+\tfrac{15}{4} \text{Li}_4(\tfrac{3}{4})+\tfrac{13}{3} \text{Li}_4(\tfrac{1}{2})+4\text{Li}_4(\tfrac{2}{3})+6 \text{Li}_2(\tfrac{1}{3}){}^2+6 \text{Li}_2(\tfrac{1}{3}) \log^23+12 \text{Li}_2(\tfrac{1}{3}) \log^22-17 \text{Li}_3(\tfrac{2}{3}) \log3+34\text{Li}_3(\tfrac{2}{3}) \log2-17 \text{Li}_3(\tfrac{1}{3}) \log3+34 \text{Li}_3(\tfrac{1}{3}) \log2-12 \text{Li}_2(\tfrac{1}{3}) \log2 \log3+\tfrac{117}{4} \zeta (3) \log3-\tfrac{117}{2}\zeta (3) \log2-\tfrac{35}{12} \zeta (4)-\tfrac{203}{6} \zeta (2) \log^22-22 \zeta (2) \log^23+59 \zeta (2) \log2 \log3+\tfrac{19}{72} \log^42+\tfrac{115}{16} \log^43-\tfrac{43}{3} \log^32 \log3-\tfrac{53}{2} \log2 \log^3(3)+\tfrac{131}{4} \log^22 \log^23 \end{eqnarray}$$

The second integral

$$\begin{eqnarray} \mathrm{II}(-8) &=& \tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{1}{4} \text{Li}_4(\tfrac{1}{4})+\tfrac{2}{3} \text{Li}_4(\tfrac{1}{2})-3 \text{Li}_4(\tfrac{3}{4})-6\text{Li}_4(\tfrac{2}{3})-\text{Li}_2(\tfrac{1}{3}){}^2-\text{Li}_2(\tfrac{1}{3}) \log^23+6 \text{Li}_3(\tfrac{2}{3}) \log3+6 \text{Li}_3(\tfrac{1}{3}) \log3+2\text{Li}_2(\tfrac{1}{3}) \log2 \log3-13 \zeta (3) \log3+\tfrac{205}{24} \zeta (4)+\tfrac{19}{3} \zeta (2) \log^22+\tfrac{37}{4} \zeta (2) \log^23-\tfrac{25}{2} \zeta (2) \log2 \log3-\tfrac{65}{36} \log^42-\tfrac{5}{2}\log^43+5 \log2 \log^33+4 \log^32 \log3-4 \log^22 \log^23 \\ \mathrm{II}(-3) &=& -\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{1}{4} \text{Li}_2(\tfrac{1}{3}){}^2+\tfrac{7}{4} \text{Li}_4(\tfrac{1}{3})-\tfrac{17}{8} \text{Li}_4(\tfrac{3}{4})-\tfrac{8}{3}\text{Li}_4(\tfrac{1}{2})-\tfrac{7}{2} \text{Li}_4(\tfrac{2}{3})+\tfrac{1}{4} \text{Li}_2(\tfrac{1}{3}) \log^23-\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \log2 \log3-4 \text{Li}_3(\tfrac{2}{3})\log2-2 \text{Li}_3(\tfrac{1}{3}) \log2+2 \zeta (3) \log2+\tfrac{173}{24} \zeta (4)+\tfrac{3}{2} \zeta (2) \log^23+\tfrac{23}{3} \zeta (2) \log^22-\tfrac{17}{2} \zeta (2) \log2 \log3+\tfrac{1}{96} (-\log^4(3))-\tfrac{55}{36} \log^42+\tfrac{4}{3} \log2 \log^33+\tfrac{17}{6} \log^32 \log3-\tfrac{23}{8} \log^22 \log^23 \\ \mathrm{II}(-2) &=& -\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{1}{8} \text{Li}_4(\tfrac{1}{4})+\tfrac{1}{4}\text{Li}_2(\tfrac{1}{3}){}^2-\text{Li}_4(\tfrac{2}{3})-\text{Li}_4(\tfrac{1}{2})-\text{Li}_4(\tfrac{1}{3})+\tfrac{1}{4} \text{Li}_2(\tfrac{1}{3}) \log^23-\tfrac{1}{2}\text{Li}_2(\tfrac{1}{3}) \log2 \log3-\text{Li}_3(\tfrac{2}{3}) \log3-\text{Li}_3(\tfrac{1}{3}) \log3+\tfrac{27}{8} \zeta (4)+\tfrac{1}{2} \zeta (2) \log^22-\tfrac{3}{4} \zeta (2) \log^23+\tfrac{1}{2}\zeta (2) \log2 \log3+\tfrac{1}{24} \log^42+\tfrac{5}{16} \log^43-\tfrac{7}{12} \log2 \log^33+\tfrac{1}{4} \log^22 \log^23 \\ \mathrm{II}(-1) &=& -\tfrac{1}{4} \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{9}{16} \text{Li}_4(\tfrac{1}{4})-\tfrac{13}{6} \text{Li}_4(\tfrac{1}{2})-\text{Li}_3(\tfrac{2}{3}) \log2-\text{Li}_3(\tfrac{1}{3}) \log(2)+\tfrac{1}{8} \zeta (3) \log2+\tfrac{203}{96} \zeta (4)+\tfrac{1}{8} \zeta (2) (-\log^23)+\tfrac{11}{12} \zeta (2) \log^22-\tfrac{3}{4} \zeta (2) \log2 \log3+\tfrac{41}{144} \log^42+\tfrac{1}{3} \log2 \log^3(3)-\tfrac{1}{2} \log^22 \log^23 \\ \mathrm{II}(1) &=& \tfrac{3}{2} \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{3}{16} \text{Li}_4(\tfrac{1}{4})-\tfrac{1}{4} \text{Li}_2(\tfrac{1}{3}){}^2-\tfrac{3}{2} \text{Li}_4(\tfrac{1}{2})-\tfrac{1}{4}\text{Li}_2(\tfrac{1}{3}) \log^23+\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \log2 \log3-\tfrac{1}{2} \zeta (4)+\tfrac{3}{4} \zeta (2) \log^23+\tfrac{3}{4} \zeta (2) \log^22-\tfrac{3}{2} \zeta (2) \log2 \log(3)+\tfrac{1}{16} (-\log^43)+\tfrac{1}{16} \log^42+\tfrac{1}{4} \log2 \log^33-\tfrac{1}{4} \log^22 \log^23 \\ \mathrm{II}(-\tfrac12) &=& \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{3}{16} \text{Li}_4(\tfrac{1}{4})-\tfrac{5}{4} \text{Li}_2(\tfrac{1}{3}){}^2-\tfrac{3}{2} \text{Li}_4(\tfrac{3}{4})+\tfrac{13}{6} \text{Li}_4(\tfrac{1}{2})-3\text{Li}_4(\tfrac{2}{3})-\tfrac{3}{4} \text{Li}_2(\tfrac{1}{3}) \log^22-\tfrac{5}{4} \text{Li}_2(\tfrac{1}{3}) \log^23+\tfrac{5}{2} \text{Li}_2(\tfrac{1}{3}) \log2 \log3+3\text{Li}_3(\tfrac{2}{3}) \log3-3 \text{Li}_3(\tfrac{2}{3}) \log2+3 \text{Li}_3(\tfrac{1}{3}) \log3-3 \text{Li}_3(\tfrac{1}{3}) \log2-\tfrac{13}{2} \zeta (3) \log3+\tfrac{13}{2} \zeta (3) \log(2)+\tfrac{49}{24} \zeta (4)+\tfrac{61}{12} \zeta (2) \log^22+5 \zeta (2) \log^23-10 \zeta (2) \log2 \log3-\tfrac{113}{144} \log^42-\tfrac{23}{16} \log^43+\tfrac{11}{4} \log^32 \log3+\tfrac{17}{4} \log2 \log^33-\tfrac{37}{8}\log^22 \log^23 \\ \mathrm{II}(\tfrac23) &=& -\tfrac{3}{2} \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{3}{4} \text{Li}_2(\tfrac{1}{3}){}^2-\tfrac{31}{16} \text{Li}_4(\tfrac{1}{4})-\tfrac{21}{8} \text{Li}_4(\tfrac{3}{4})+\tfrac{11}{4}\text{Li}_4(\tfrac{1}{3})+\tfrac{17}{6} \text{Li}_4(\tfrac{1}{2})-\tfrac{11}{2} \text{Li}_4(\tfrac{2}{3})+\text{Li}_2(\tfrac{1}{3}) \log^23-\tfrac{3}{2} \text{Li}_2(\tfrac{1}{3}) \log2 \log(3)+\text{Li}_3(\tfrac{2}{3}) \log3+3 \text{Li}_3(\tfrac{1}{3}) \log3-\tfrac{19}{6} \zeta (3) \log3+\tfrac{61}{12} \zeta (4)+\tfrac{13}{4} \zeta (2) \log^23+\tfrac{71}{12} \zeta (2) \log^22-\tfrac{15}{2} \zeta (2)\log2 \log3+\tfrac{15}{32} (-\log^43)-\tfrac{421}{144} \log^42+\tfrac{11}{12} \log2 \log^33+\tfrac{7}{2} \log^32 \log3-\tfrac{15}{8} \log^22 \log^23 \end{eqnarray}$$

$$\begin{eqnarray} \mathrm{II}(\tfrac14) &=& \tfrac{13}{2} \text{Li}_2(\tfrac{1}{3}) \zeta (2)-\tfrac{13}{6} \text{Li}_4(\tfrac{1}{2})+\tfrac{7}{2} \text{Li}_4(\tfrac{1}{3})+\tfrac{59}{16} \text{Li}_4(\tfrac{1}{4})-\tfrac{27}{4}\text{Li}_4(\tfrac{3}{4})-10 \text{Li}_4(\tfrac{2}{3})-5 \text{Li}_2(\tfrac{1}{3}){}^2-5 \text{Li}_2(\tfrac{1}{3}) \log^23-7 \text{Li}_2(\tfrac{1}{3}) \log^22+17\text{Li}_3(\tfrac{2}{3}) \log3-34 \text{Li}_3(\tfrac{2}{3}) \log2+17 \text{Li}_3(\tfrac{1}{3}) \log3-34 \text{Li}_3(\tfrac{1}{3}) \log2+10 \text{Li}_2(\tfrac{1}{3}) \log2 \log(3)-\tfrac{117}{4} \zeta (3) \log3+\tfrac{117}{2} \zeta (3) \log2+\tfrac{91}{12} \zeta (4)+\tfrac{101}{4} \zeta (2) \log^23+\tfrac{449}{12} \zeta (2) \log^22-\tfrac{135}{2} \zeta (2) \log2 \log3+\tfrac{307}{144} (-\log^4(2))-\tfrac{115}{16} \log^43+\tfrac{53}{2} \log2 \log^33+16 \log^32 \log3-\tfrac{129}{4} \log^22 \log^23 \end{eqnarray}$$

The first integral

$$\begin{eqnarray} \def\tfrac#1#2{{\textstyle\frac{#1}{#2}}} \mathrm{I}(-8) &=& -\tfrac{3}{2} \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{3}{2} \text{Li}_2(\tfrac{1}{3}){}^2+\tfrac{7}{4} \text{Li}_4(\tfrac{1}{3})-\tfrac{33}{8} \text{Li}_4(\tfrac{3}{4})-\tfrac{69}{16}\text{Li}_4(\tfrac{1}{4})-\tfrac{13}{2} \text{Li}_4(\tfrac{2}{3})+\tfrac{17}{2} \text{Li}_4(\tfrac{1}{2})+\tfrac{3}{2} \text{Li}_2(\tfrac{1}{3}) \log^23+\text{Li}_3(\tfrac{2}{3}) \log(3)+\text{Li}_3(\tfrac{1}{3}) \log3-3 \text{Li}_2(\tfrac{1}{3}) \log2 \log3-\tfrac{11}{4} \zeta (3) \log3+\tfrac{45}{8} \zeta (4)+\tfrac{7}{2} \zeta (2) \log^23+\tfrac{35}{4} \zeta (2) \log^22-\tfrac{21}{2} \zeta(2) \log2 \log3-\tfrac{5}{32} \log^43-\tfrac{253}{48} \log^42+\tfrac{1}{12} \log2 \log^33+\tfrac{11}{2} \log^32 \log3-\tfrac{21}{8} \log^22 \log^23 \\ \mathrm{I}(-3) &=& -\tfrac{1}{4} \text{Li}_4(\tfrac{1}{4})-\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}){}^2-\tfrac{7}{8} \text{Li}_4(\tfrac{3}{4})+\tfrac{3}{2} \text{Li}_4(\tfrac{2}{3})-\tfrac{9}{4}\text{Li}_4(\tfrac{1}{3})+\tfrac{7}{3} \text{Li}_4(\tfrac{1}{2})-\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \log^23+\tfrac{1}{2} \text{Li}_3(\tfrac{2}{3}) \log3-2 \text{Li}_3(\tfrac{2}{3}) \log(2)-\text{Li}_3(\tfrac{1}{3}) \log2+\zeta (3) \log2-\tfrac{1}{6} \zeta (4)+\tfrac{1}{4} \zeta (2) \log^23+\tfrac{5}{3} \zeta (2) \log^22-2 \zeta (2) \log2 \log3+\tfrac{23}{96} (-\log^43)-\tfrac{47}{72} \log^4(2)+\tfrac{1}{2} \log2 \log^33+\tfrac{7}{6} \log^32 \log3-\tfrac{7}{8} \log^22 \log^23 \\ \mathrm{I}(-2) &=& -\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{1}{4} \text{Li}_2(\tfrac{1}{3}){}^2+\tfrac{1}{4} \text{Li}_2(\tfrac{1}{3}) \log^23-\tfrac{1}{2} \text{Li}_3(\tfrac{2}{3}) \log3-\tfrac{1}{2}\text{Li}_3(\tfrac{1}{3}) \log3-\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \log2 \log3+\tfrac{1}{2} \zeta (3) \log3+\tfrac{5}{8} \zeta (4)-\tfrac{3}{4} \zeta (2) \log^23+\tfrac{1}{2} \zeta (2) \log2 \log(3)+\tfrac{11}{48} \log^43-\tfrac{1}{2} \log2 \log^33+\tfrac{1}{4} \log^22 \log^23 \\ \mathrm{I}(-1) &=& -\tfrac{1}{4} \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{3}{16} \text{Li}_4(\tfrac{3}{4})+\tfrac{11}{32} \text{Li}_4(\tfrac{1}{4})+\tfrac{3}{8} \text{Li}_4(\tfrac{1}{3})+\tfrac{3}{4}\text{Li}_4(\tfrac{2}{3})-\tfrac{11}{12} \text{Li}_4(\tfrac{1}{2})-\tfrac{1}{2} \text{Li}_3(\tfrac{2}{3}) \log2-\tfrac{1}{2} \text{Li}_3(\tfrac{1}{3}) \log2+\tfrac{1}{16} \zeta (3) \log2+\tfrac{7}{16}\zeta (3) \log3-\tfrac{11}{48} \zeta (4)-\tfrac{5}{24} \zeta (2) \log^22-\tfrac{1}{2} \zeta (2) \log^23+\tfrac{1}{2} \zeta (2) \log2 \log3+\tfrac{3}{64} \log^43+\tfrac{91}{288} \log^42+\tfrac{1}{24} \log2 \log^33-\tfrac{1}{4}\log^32 \log3-\tfrac{1}{16} \log^22 \log^23 \\ \mathrm{I}(1) &=& \tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{1}{8} \text{Li}_4(\tfrac{1}{4})-\tfrac{1}{2} \text{Li}_4(\tfrac{2}{3})-\tfrac{1}{2} \text{Li}_4(\tfrac{1}{3})-\text{Li}_4(\tfrac{1}{2})-\tfrac{1}{2}\text{Li}_3(\tfrac{2}{3}) \log3-\tfrac{1}{2} \text{Li}_3(\tfrac{1}{3}) \log3+\tfrac{1}{2} \zeta (3) \log3+\zeta (4)+\tfrac{1}{2} \zeta (2) \log^22-\tfrac{1}{2} \zeta (2) \log2 \log3+\tfrac{1}{24} \log^4(2)+\tfrac{1}{8} \log^43-\tfrac{1}{6} \log2 \log^33 \\ \mathrm{I}(-\tfrac12) &=& -\tfrac{1}{4} \text{Li}_2(\tfrac{1}{3}){}^2-\tfrac{1}{2} \text{Li}_4(\tfrac{1}{2})+\tfrac{13}{16} \text{Li}_4(\tfrac{1}{4})-\tfrac{1}{4} \text{Li}_2(\tfrac{1}{3}) \log^23-\tfrac{1}{4}\text{Li}_2(\tfrac{1}{3}) \log^22-\tfrac{1}{2} \text{Li}_3(\tfrac{2}{3}) \log3-\tfrac{1}{2} \text{Li}_3(\tfrac{1}{3}) \log3+\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \log2 \log3-\tfrac{3}{2}\text{Li}_3(\tfrac{2}{3}) \log2-\tfrac{3}{2} \text{Li}_3(\tfrac{1}{3}) \log2+\tfrac{1}{2} \zeta (3) \log3+\tfrac{13}{4} \zeta (3) \log2-\tfrac{1}{2} \zeta (2) \log^23+\tfrac{1}{2} \zeta (2) \log^22-\zeta (2) \log(2) \log3+\tfrac{5}{48} \log^43+\tfrac{25}{48} \log^42+\tfrac{1}{3} \log^32 \log3+\tfrac{1}{2} \log2 \log^33-\tfrac{9}{8} \log^22 \log^23 \\ \mathrm{I}(\tfrac23) &=& \tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \zeta (2)+\tfrac{1}{4} (-\text{Li}_2(\tfrac{1}{3}){}^2)-\tfrac{3}{4} \text{Li}_4(\tfrac{1}{4})-\tfrac{3}{2} \text{Li}_4(\tfrac{3}{4})+\tfrac{9}{2}\text{Li}_4(\tfrac{1}{3})-6 \text{Li}_4(\tfrac{2}{3})-2 \text{Li}_4(\tfrac{1}{2})-\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \log^23+\tfrac{1}{2} \text{Li}_3(\tfrac{2}{3}) \log3+\tfrac{1}{2}\text{Li}_2(\tfrac{1}{3}) \log2 \log3+\tfrac{5}{2} \text{Li}_3(\tfrac{1}{3}) \log3-\tfrac{8}{3} \zeta (3) \log3+\tfrac{21}{4} \zeta (4)+3 \zeta (2) \log^23+5 \zeta (2) \log^22-\tfrac{13}{2} \zeta (2) \log2 \log(3)-\tfrac{2}{3} \log^43-\tfrac{19}{12} \log^42+\tfrac{3}{2} \log2 \log^33+2 \log^32 \log3-\tfrac{7}{4} \log^22 \log^23 \end{eqnarray} $$

$$\begin{eqnarray} \mathrm{I}(\tfrac14) &=& \tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \zeta (2)-\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}){}^2-\tfrac{3}{4} \text{Li}_4(\tfrac{3}{4})-\tfrac{3}{2} \text{Li}_4(\tfrac{2}{3})+\tfrac{25}{4}\text{Li}_4(\tfrac{1}{4})-9 \text{Li}_4(\tfrac{1}{2})-\tfrac{1}{2} \text{Li}_2(\tfrac{1}{3}) \log^23-\text{Li}_2(\tfrac{1}{3}) \log^22-\tfrac{1}{2} \text{Li}_3(\tfrac{2}{3}) \log(3)-\tfrac{1}{2} \text{Li}_3(\tfrac{1}{3}) \log3-17 \text{Li}_3(\tfrac{2}{3}) \log2-17 \text{Li}_3(\tfrac{1}{3}) \log2+\text{Li}_2(\tfrac{1}{3}) \log2 \log3+\tfrac{1}{2} \zeta (3) \log(3)+\tfrac{117}{4} \zeta (3) \log2+\tfrac{41}{8} \zeta (4)+\tfrac{1}{2} \zeta (2) \log^23+12 \zeta (2) \log^22-\tfrac{39}{2} \zeta (2) \log2 \log3+\tfrac{1}{48} (-\log^43)+\tfrac{79}{24} \log^42+\tfrac{8}{3} \log^32\log3+\tfrac{37}{6} \log2 \log^33-\tfrac{41}{4} \log^22 \log^23 \end{eqnarray}$$

Answer 3

$\def\Li{{\rm Li}}$I think I know how Mr. Tunk-Fey's approach to find solution to Mr. Vladimir Reshetnikov's problem: \begin{equation} I=\int_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx \end{equation} He used integration by parts method. Here might be the way he evaluated $I$:

Let \begin{equation} u=\ln(1+2x)\quad\Rightarrow\quad du=\dfrac{2\ dx}{1+2x} \end{equation} and \begin{equation} v=\int\frac{\ln(1-x)\ln(1+x)}{1+2x}\,dx \end{equation} In order to evaluate $v$, he might be using his technique in Mr. Vladimir Reshetnikov's another problem: \begin{equation} \int_0^1\frac{\ln^2x}{\sqrt{x^2-x+1}}dx \end{equation} See also David H's answer. I will just use that technique to evaluate $I$ without providing the complete steps. After several substitutions, we will arrive at the following equation \begin{align} v&=\frac{1}{6}\int\frac{\ln^2x}{1-\frac{2}{3}x}\ dx-\frac{1}{2}\int\frac{\ln^2x}{1-2x}\,dx-\frac{1}{2}\int\frac{\ln^2x}{1+x}\ dx-\frac{1}{6}\int\frac{\ln^2x}{1-\frac{1}{3}x}\,dx\\ &=\frac{1}{6}J_1-\frac{1}{2}J_2-\frac{1}{2}J_3-\frac{1}{6}J_4 \end{align} where \begin{align} J_n=\int\frac{\ln^2x}{1-ax}\ dx=\frac{2}{a}\Li_3(ax)-\frac{2}{a}\Li_2(ax)\ln x-\frac{1}{a}\ln(1-ax)\ln^2x+C \end{align} Using $J_n$ we get \begin{align} J_1&=3\Li_3\left(\frac{2}{3}x\right)-3\Li_2\left(\frac{2}{3}x\right)\ln x-\frac{3}{2}\ln\left(1-\frac{2}{3}x\right)\ln^2x\\ J_2&=\Li_3\left(2x\right)-\Li_2\left(2x\right)\ln x-\frac12\ln\left(1-2x\right)\ln^2x\\ J_3&=-2\Li_3(-x)+2\Li_2(-x)\ln x+\ln(1+x)\ln^2x\\ J_4&=6\Li_3\left(\frac{1}{3}x\right)-6\Li_2\left(\frac{1}{3}x\right)\ln x-3\ln\left(1-\frac{1}{3}x\right)\ln^2x\\ \end{align} Hence, we have \begin{align} uv&=\left.\left[\frac{1}{6}J_1-\frac{1}{2}J_2-\frac{1}{2}J_3-\frac{1}{6}J_4\right]\ln(1+2x)\right|_{x=0}^1\\ &=\left[\frac{1}{2}\Li_3\left(\frac23\right)-\frac{1}{2}\Li_3\left(2\right)+\Li_3\left(-1\right)-\Li_3\left(\frac{1}{3}\right)\right]\ln3\\ &=\left[\frac{1}{2}\Li_3\left(\frac{2}{3}\right)-\frac{1}{2}\Li_3\left(2\right)-\frac{3}{4}\zeta(3)-\Li_3\left(\frac{1}{3}\right)\right]\ln3 \end{align} The next step is to evaluate $\displaystyle\int v\,du$ of which consists of three general form of integrals: \begin{align} \frac{\Li_3(ax)}{1+2x}\,dx\tag1\\[10pt] \int_0^1\frac{\Li_2(ax)\ln x}{1+2x}\,dx\tag2\\[10pt] \int_0^1\frac{\ln(1-ax)\ln^2 x}{1+2x}\,dx\tag3 \end{align} from which the OP follows. To evaluate $I$, the corresponding values of $a$ are $\displaystyle\frac{2}{3},\,2,\,-1,\,$and $\displaystyle\frac{1}{3}$.

I wish I could evaluate $(1),\,(2),\,$and $(3)$ or, at least, evaluate the integrals with the corresponding values of $a$, but I couldn't (Sorry...). Luckily, Mr. Kirill has provided the results for $a=\displaystyle\frac{2}{3}$ and $a=-1$. I hope he will be generous to provide the results for $a=\displaystyle\frac{1}{3}$ and $a=2$.

Answer 4

For $(3)$ you can have the form

$$ I = - \rm Li_3( -2 ) \ln\left( {\frac {a+2}{2}} \right) - {\rm Li_4}( a )+ \int _{0}^{a}\,{\frac {{\rm Li_3}(t) }{t+2}}{dt},$$

where $\rm Li_s(z)$ is the polylogarithm function. For instance when $a=1/2$ we have

$$ I \sim - 0.08900930960. $$