Challenging Integral:
\begin{align} I=\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx&=6\operatorname{Li}_5\left(\frac12\right)+6\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{81}{16}\zeta(5)-\frac{21}{8}\zeta(2)\zeta(3)\\&\quad+\frac{21}8\ln^22\zeta(3)-\ln^32\zeta(2)+\frac15\ln^52 \end{align}
I came across this integral while i was trying to calculate $\displaystyle\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}$, proposed by Cornel on his FB page here, but he has not revealed his solution yet.
The integral is related to the sum through the identity ( see here): $$\int_0^1 \frac{\ln^a(1-x)\ln(1+x)}{x}dx=(-1)^a a! \sum_{n=1}^\infty\frac{H_n^{(a+1)}}{n2^n}$$
With $a=3$, We get $\quad\displaystyle I=-6\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}\quad$.
The way I computed this integral is really long as it's based on values of tough alternating Euler sums which themselves long to calculate. I hope we can find other approaches that save us such tedious calculations. Any way, here is my approach:
Using the identity from this solution: $\displaystyle\int_0^1 x^{n-1}\ln^3(1-x)\ dx=-\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$
Multiplying both sides by $\frac{(-1)^{n-1}}{n}$ then summing both sides from $n=1$ to $n=\infty$, gives: \begin{align} I&=\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty-\frac{(-x)^{n}}{n}dx=\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx\\ &=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}+3\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2} \end{align}
We have: \begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}&=-6\operatorname{Li}_5\left(\frac12\right)-6\ln2\operatorname{Li}_4\left(\frac12\right)+\ln^32\zeta(2)-\frac{21}{8}\ln^22\zeta(3)\\&\quad+\frac{27}{16}\zeta(2)\zeta(3)+\frac94\zeta(5)-\frac15\ln^52 \end{align}
\begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}&=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac23\ln^32\zeta(2)+\frac74\ln^22\zeta(3)\\&\quad-\frac{15}{16}\zeta(2)\zeta(3)-\frac{23}8\zeta(5)+\frac2{15}\ln^52 \end{align}
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$$
The proof of the first and second sum can be found here and the third sum can be found here.
By substituting these three sums,we get the closed form of $I$.
Thanks.
