The following is from Mathematical Analysis $-$ A collection of Problems by Tolaso J. Kos $($Page $27$, Problem $282$$)$
$$\mathfrak{I}=\int\limits_0^1 \left[\log(x)\log(1-x)+\operatorname{Li}_2(x)\right]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx=4\zeta(2)\zeta(3)-9\zeta(5)\tag1$$
Today I came across this question asking for the evaluation of the integral
$$\mathfrak{J}=\int\limits_0^{\pi/2}\frac{\log^2(\sin x)\log^2(\cos x)}{\sin x\cos x}\mathrm dx=\frac12\zeta(5)-\frac14\zeta(2)\zeta(3)\tag2$$
Which can done be "rather simple" by invoking the fourth derivative of the Beta Function. The final structure of the result reminded me of the logarithmic integral $(1)$ I was not able to evaluate. It might turn out that this relation is by pure chance but nevertheless it motivated me to look at $(1)$ again. It is hardly probable that $(1)$ can be done in a similar way like $(2)$ in xpaul's answer to the linked question due the involved Dilogarithms$-$but anyway you can prove me wrong.
I have not got that far with $(1)$ but, however, I noticed two, I would say quite interesting, facts about the integral. First, consider the following, well-known functional relation of the Dilogarithm
$$\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\zeta(2)-\log(x)\log(1-x)$$
which can be used in order to get rid of the $\log(x)\log(1-x)$-term within $(1)$ and leading to
$$\small\int\limits_0^1 \left[\log(x)\log(1-x)+\operatorname{Li}_2(x)\right]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx=\int\limits_0^1 \left[\zeta(2)-\operatorname{Li}_2(1-x)\right]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx$$
Second, applying the substitution $x=1-x$ after a minor reshape yields to
$$\small\begin{align}
\int\limits_0^1 \left[\zeta(2)-\operatorname{Li}_2(1-x)\right]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx&=\int\limits_0^1 \left[\frac{\zeta(2)}{1-x}-\frac{\operatorname{Li}_2(1-x)}{1-x}\right]\left[\frac{\operatorname{Li}_2(x)}x-\zeta(2)\right]\mathrm dx\\
&=\int\limits_0^1 \left[\zeta(2)-\operatorname{Li}_2(1-x)\right]\left[\frac{\operatorname{Li}_2(x)}x-\zeta(2)\right]\frac{\mathrm dx}{1-x}\\
&=\int\limits_0^1 \left[\zeta(2)-\operatorname{Li}_2(x)\right]\left[\frac{\operatorname{Li}_2(1-x)}{1-x}-\zeta(2)\right]\frac{\mathrm dx}x\\
&=\int\limits_0^1 \left[\frac{\zeta(2)}x-\frac{\operatorname{Li}_2(x)}x\right]\left[\frac{\operatorname{Li}_2(1-x)}{1-x}-\zeta(2)\right]\mathrm dx
\end{align}$$
I want to point out the quite interesting one could say "almost"-symmetry of the two integrals
$$\begin{align}
\mathfrak{I}_1&=\int\limits_0^1 \left[\frac{\zeta(2)}{1-x}-\frac{\operatorname{Li}_2(1-x)}{1-x}\right]\left[\frac{\operatorname{Li}_2(x)}x-\zeta(2)\right]\mathrm dx\\
\mathfrak{I}_2&=\int\limits_0^1 \left[\frac{\zeta(2)}x-\frac{\operatorname{Li}_2(x)}x\right]\left[\frac{\operatorname{Li}_2(1-x)}{1-x}-\zeta(2)\right]\mathrm dx
\end{align}$$
which might be helpful for the actual evaluation. But from hereon I have no clue how to continue.
Just expanding the brackets out does not seem like a good idea to me. Since one the one hand it is not elegant at all and on the other hand it would lead to to the term $\operatorname{Li}_2(x)\operatorname{Li}_2(1-x)$ for which I have no idea how to deal with (I am not that confident using double series). I also tried various ways of IBP but this seems to be pointless since all variations ended up in producing a divergent term$-$unless I have missed a special choice of $u$ and $\mathrm v$. I have not figured out a suitable substitution nor an appropriate newly introduced parameter (for the application of Feynman's Trick) and the I do not know whether a series expansion would be helpful or not (connected with this issue is the possibility of a double summation with which I cannot really deal).
Thus, I am asking for the closed-form evaluation of $(1)$ hopefully equal to the given value (which works out numerically according to WolframAlpha). Even though I have troubles with double series I would accept an answer invoking these but I would appreciate a solution without involving them. As this integral appeared within a collection of Analysis Problems I am quite sure that it has been already evaluated somewhere (maybe even here on MSE where I was not able to find it!).
Thanks in advance!
Best Answer
Cross-posting this integral on AoPS brings Y. Sharifi's solution here after a day. Quite amazing one!
I will copy here his entire solution:
Edit. This integral was proposed two years ago in RMM and it appeared as problem UP $089$.
See in this link, at the page $70$.