Prove that
I encountered this integral while working on the sum $\displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$. Both of the integral and the sum were proposed by Cornel Valean:
$$I=\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx=\frac12\ln^32\zeta(2)-\frac78\ln^22\zeta(3)-\frac58\ln2\zeta(4)+\frac{27}{32}\zeta(5)+\frac78\zeta(2)\zeta(3)\\-\frac{7}{60}\ln^52-2\ln2\operatorname{Li}_4\left(\frac12\right)-2\operatorname{Li}_5\left(\frac12\right);$$
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-3\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{23}{64}\zeta(5)-\frac1{16}\ln2\zeta(4)+\frac{23}{16}\zeta(2)\zeta(3)\\-\frac{23}{16}\ln^22\zeta(3)+\frac7{12}\ln^32\zeta(2)-\frac{13}{120}\ln^52.$$
This sum itself is related, through the Cauchy product of $\ln(1-x)\text{Li}_3(x)$, to the following sum:
$$\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^22^n}=4\operatorname{Li}_5\left(\frac12\right)+3\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{81}{64}\zeta(5)+\frac5{16}\ln2\zeta(4)\nonumber\\ -\frac78\zeta(2)\zeta(3)+\frac{7}{8}\ln^22\zeta(3)-\frac5{12}\ln^32\zeta(2)+\frac{11}{120}\ln^52$$
The main integral is very related to the integral $\int_0^1 \frac{\ln^3(1-x)\ln(1+x)}{x}dx$ which I managed to solve using three tough results of alternating series, so again I am looking for a different approach that does not use these results ( mentioned in the link) to compute $I$.
Here is how the two integrals are related:
$$\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx\overset{IBP}{=}\operatorname{Li}_2\left(\frac12\right)\operatorname{Li}_3\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)-\operatorname{Li}_5\left(\frac12\right)+\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}$$
From this identity, we have $\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}=-\frac16\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx$
Then
$$\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx=\operatorname{Li}_2\left(\frac12\right)\operatorname{Li}_3\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)-\operatorname{Li}_5\left(\frac12\right)\\-\frac16\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx$$
So, any elegant way to solve any of these two integrals?