We have the following result ($\text{Li}_{n}$ being the polylogarithm):
$$\tag{*}\small{ \int_0^1 \log^2 (1-x) \log^2 x \log^3(1+x) \frac{dx}{x} = -168 \text{Li}_5(\frac{1}{2}) \zeta (3)+96 \text{Li}_4(\frac{1}{2}){}^2-\frac{19}{15} \pi ^4 \text{Li}_4(\frac{1}{2})+\\ 12 \pi ^2 \text{Li}_6(\frac{1}{2})+8 \text{Li}_4(\frac{1}{2}) \log ^4(2)-2 \pi ^2 \text{Li}_4(\frac{1}{2}) \log ^2(2)+12 \pi ^2 \text{Li}_5(\frac{1}{2}) \log (2)+\frac{87 \pi ^2 \zeta (3)^2}{16}+\\ \frac{447 \zeta (3) \zeta (5)}{16}+\frac{7}{5} \zeta (3) \log ^5(2)-\frac{7}{12} \pi ^2 \zeta (3) \log ^3(2)-\frac{133}{120} \pi ^4 \zeta (3) \log (2)-\frac{\pi ^8}{9600}+\frac{\log ^8(2)}{6}- \\ \frac{1}{6} \pi ^2 \log ^6(2)-\frac{1}{90} \pi ^4 \log ^4(2)+\frac{19}{360} \pi ^6 \log ^2(2) }$$
This is extremely amazing: almost all other similar integrals are not expressible via ordinary polylogarithm.
The solution is however non-trivial. There are two methods: first is to find enough linear relations between similar integrals, once the rank is high enough, solving the system gives $(*)$; second method is to convert the integral into multiple zeta values, then use known linear relations between them. None of these methods can explain the result's simplicity.
Question: Is there a simpler method to prove (*), or a conceptual explanation of its elegance?
Any thought is welcomed. Thank you very much.
I wrote a Mathematica package, it can calculate the integral in subject and many similar ones. The following command calculates $(*)$:
MZIntegrate[Log[1-x]^2*Log[x]^2*Log[1+x]^3/x, {x,0,1}]
It can also solve some other integrals.
The package can be obtained here. I hope it can benefit those interested in related integral/series.
Remarks on the question:
- It's known that $\zeta(\bar{3},1,\bar{3},1)$ is very reminiscent to the RHS of $(*)$. But both the simplicity of $\zeta(\bar{3},1,\bar{3},1)$ and its connection to the integral are elusive to me.
- (Added by Iridescent) This contains nearly all known general formulas of these log integrals. However it does not help much on solving OP's problem.