Prove that: $$\displaystyle \int_{0}^{1}\frac{1-x}{1-x^{6}}{\ln^4{x}} \ {dx} = \frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54} $$
I was able to simplify it a bit by substituting ${y = -\ln{x}}$ and some further mathematical manipulation but was not able to get the correct form.
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{1 - x \over 1 - x^{6}}\,\ln^{4}\pars{x}\,\dd x= {16 \over 243\root{3}}\,\pi^{5} + {605 \over 54}\,\zeta\pars{5}}$
Lets $\ds{\quad x_{n} = \expo{n\pi\ic/3}\,,\quad n = 0,1,2,3,4,5\quad}$ such that \begin{align} {1 - x \over 1 - x^{6}}& =\pars{x - 1}\sum_{n = 0}^{5}{x_{n}/6 \over x - x_{n}} ={1 \over 6}\sum_{n = 0}^{5}x_{n} \pars{{x - x_{n} \over x - x_{n}} + {x_{n} - 1 \over x - x_{n}}} \\[3mm]&={1 \over 6}\,\underbrace{\sum_{n = 0}^{5}x_{n}}_{\ds{=\ 0}} +{1 \over 6}\sum_{n = 0}^{5}{x_{n}\pars{x_{n} - 1} \over x - x_{n}} ={1 \over 6}\sum_{n = 1}^{5}{x_{n}\pars{x_{n} - 1} \over x - x_{n}} ={1 \over 6}\sum_{n = -2}^{2}{x_{n + 3}\pars{x_{n + 3} - 1} \over x - x_{n + 3}} \\[3mm]&={1 \over 6}\sum_{n = -2}^{2}{x_{n}\pars{x_{n} + 1} \over x + x_{n}} \end{align}
Then, $$ \color{#c00000}{\int_{0}^{1}{1 - x \over 1 - x^{6}}\,\ln^{4}\pars{x}\,\dd x} ={1 \over 6}\sum_{n = -2}^{2}x_{n}\pars{x_{n} + 1} \color{#00f}{\int_{0}^{1}{\ln^{4}\pars{x} \over x + x_{n}}\,\dd x}\tag{1} $$
Lets evaluate the integral: \begin{align} &\color{#00f}{\int_{0}^{1}{\ln^{k}\pars{x} \over x - a}\,\dd x} =-\int_{0}^{1}{\ln^{k}\pars{a\bracks{x/a}} \over 1 - x/a} \,{\dd x \over a} =-\int_{0}^{1/a}{\ln^{k}\pars{ax} \over 1 - x}\,\dd x \\[3mm]&=-\int_{0}^{1/a}\ln\pars{1 - x}k\ln^{k - 1}\pars{ax}\,{1 \over x}\,\dd x =k\int_{0}^{1/a}{{\rm Li}_{1}\pars{x} \over x}\,\ln^{k - 1}\pars{ax}\,\dd x \\[3mm]&=-k\pars{k - 1}\int_{0}^{1/a} {{\rm Li}_{2}\pars{x} \over x}\,\ln^{k - 2}\pars{ax}\,\dd x =\cdots \\[3mm]&=\pars{-1}^{r}\,{k! \over \pars{k - r - 1}!} \int_{0}^{1/a} {{\rm Li}_{1 + r}\pars{x} \over x}\,\ln^{k - r - 1}\pars{ax}\,\dd x =\cdots \\[3mm]&=\pars{-1}^{k - 1}k! \int_{0}^{1/a}{{\rm Li}_{k}\pars{x} \over x}\,\dd x =\pars{-1}^{k + 1}k!\,{\rm Li}_{k + 1}\pars{1 \over a} \end{align}
such that $$ \color{#00f}{\int_{0}^{1}{\ln^{4}\pars{x} \over x - x_{n}}\,\dd x} =-24\,{\rm Li}_{5}\pars{-\,{1 \over x_{n}}} =-24\,{\rm Li}_{5}\pars{-x_{-n}} $$
With expression $\pars{1}$: \begin{align} &\color{#c00000}{\int_{0}^{1}{1 - x \over 1 - x^{6}}\,\ln^{4}\pars{x}\,\dd x} =-4\sum_{n = -2}^{2}x_{n}\pars{x_{n} + 1}{\rm Li}_{5}\pars{-x_{-n}} \\[3mm]&=-8\,{\rm Li}_{5}\pars{-1} -8\,\Re\sum_{n = 1}^{2}x_{n}\pars{x_{n} + 1}{\rm Li}_{5}\pars{-x_{-n}} \\[3mm]&=-8\,{\rm Li}_{5}\pars{-1} -8\,\Re\sum_{n = 1}^{2}\expo{n\pi\ic/3}\pars{\expo{n\pi\ic/3} + 1} {\rm Li}_{5}\pars{\expo{\bracks{3 - n}\pi\ic/3}} \\[3mm]&=-8\,{\rm Li}_{5}\pars{-1} -8\,\Re\sum_{n = 1}^{2}\expo{n\pi\ic/2}\pars{\expo{n\pi\ic/6} + \expo{-n\pi\ic/6}} {\rm Li}_{5}\pars{\expo{\bracks{3 - n}\pi\ic/3}} \\[3mm]&=-8\,{\rm Li}_{5}\pars{-1} -16\,\Re\sum_{n = 1}^{2}\expo{n\pi\ic/2}\cos\pars{n\pi \over 6} {\rm Li}_{5}\pars{\expo{\bracks{3 - n}\pi\ic/3}} \\[3mm]&=-8\ \underbrace{{\rm Li}_{5}\pars{-1}} _{\ds{\color{#c00000}{-\,{15 \over 16}\,\zeta\pars{5}}}}\ +\ 8\root{3}\ \underbrace{\Im{\rm Li}_{5}\pars{\expo{2\pi\ic/3}}} _{\ds{\color{#c00000}{2\pi^{5} \over 729}}}\ +\ 8\ \underbrace{\Re{\rm Li}_{5}\pars{\expo{\pi\ic/3}}} _{\ds{\color{#c00000}{{25 \over 54}\,\zeta\pars{5}}}} \end{align}
So, $$\color{#66f}{\large% \int_{0}^{1}{1 - x \over 1 - x^{6}}\,\ln^{4}\pars{x}\,\dd x = {16 \over 243\root{3}}\,\pi^{5} + {605 \over 54}\,\zeta\pars{5}} $$
Note that $\frac{1-x}{1-x^6}=\sum_{k=0}^\infty (x^{6k}-x^{6k+1})$. And the integration $\int_0^1 x^n \ln{x}^4=\partial_n^4 \int_0^1 x^n dx=\frac{24}{(n+1)^5}$. We have $$LHS = 24\sum_{k=0}^\infty \left(\frac{1}{(6k+1)^5}-\frac{1}{(6k+2)^5}\right)$$ Use the discrete Fourier, and denote $\xi=\exp(i\frac{\pi}{3})$, $\xi_i =\xi^i$. Then $$LHS=24\sum_{i=1}^6a_i\sum_{k=1}^\infty \frac{\xi_i^k}{k^5}=24\sum_{i=1}^6 a_i Li_5(\xi_i).$$ where $a_1=a_5=\frac{1}{6}$,$a_2=-\frac{i}{2\sqrt{3}}=-a_4$,$a_6=0$,$a_3=-\frac{1}{3}$.Thus we use the summation formula for polylogarithm $$Li_5(\xi_2)+(-1)^5 Li_5(\xi_4)=-\frac{(2\pi i)^5}{5!}B_5(\frac{1}{3})=\frac{4i\pi^5}{729}.$$ $B$ is the Bernoulli polynomial. Also, $$Li_5(\xi_1)+Li_5(\xi_5)=(1-\frac{2}{2^5}-\frac{3}{3^5}+\frac{6}{6^5})\sum_{k=1}^\infty\frac{1}{k^5}=\frac{25}{27}\sum_{k=1}^\infty\frac{1}{k^5}.$$ Also note that $L_5(-1)=-(1-\frac{2}{2^5})\zeta(5).$ We conclude that $$24\sum_{i=1}^6 a_i Li_5(\xi_i)=24\frac{-i}{2\sqrt{3}}\cdot\frac{4i\pi^5}{729}+24(\frac{1}{6}\frac{25}{27}+\frac{1}{3}\cdot\frac{15}{16})\zeta(5)=\frac{16\pi^5}{243\sqrt{3}}+\frac{605}{54}\zeta(5)$$
The series $$ 24\sum_{k=0}^\infty \left(\frac{1}{(6k+1)^5}-\frac{1}{(6k+2)^5}\right)$$ $$ =\frac{24}{6^5}\sum_{k=0}^\infty \left(\frac{1}{(k+1/6)^5}-\frac{1}{(k+2/6)^5}\right)$$ Can be evaluated by the polygamma function $$ =\frac{24}{6^5} \left(\frac{-\psi^4(1/6)}{24} - \frac{-\psi^4(1/3)}{24} \right) $$ $$=\frac{1}{6^5} \left({\psi^4(1/3)} - {\psi^4(1/6)} \right) $$ $$= {\frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54}} $$
