TITLE: Closed form of $\int_0^1(\ln(1-x)\ln(1+x)\ln(x))^2\,dx$
QUESTION [18 upvotes]: I remember that some time ago I was asking this question Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$ ,
and now, while I was making a review, I asked myself if we can get the closed form of
$$
\int_{0}^{1}\left[\vphantom{\Large A}\ln\left(1 - x\right)\ln\left(1 + x\right)\ln\left(x\right)\right]^{\,2}\,\mathrm{d}x
$$
by using the similar tools as in that proof. The problem is that we have to cope with some
series that seem far more complicated. In case you have some fruitful ideas here ...

REPLY [4 votes]: See this article. The integral is recorded as the most complicated result:

$\scriptsize \int_0^1 \log ^2(1-x) \log ^2(x) \log ^2(x+1) \, dx=4 \zeta(\bar5,1)-\frac{16}{3} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)-32 \text{Li}_4\left(\frac{1}{2}\right)+24 \text{Li}_5\left(\frac{1}{2}\right)-16 \text{Li}_6\left(\frac{1}{2}\right)-8 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+24 \text{Li}_4\left(\frac{1}{2}\right) \log (2)-16 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+\frac{275 \zeta (3)^2}{8}+\frac{34 \pi ^2 \zeta (3)}{3}-237 \zeta (3)-\frac{341 \zeta (5)}{2}+7 \zeta (3) \log ^3(2)-\frac{91}{2} \zeta (3) \log ^2(2)-\frac{77}{6} \pi ^2 \zeta (3) \log (2)+156 \zeta (3) \log (2)+\frac{217}{2} \zeta (5) \log (2)+\frac{101 \pi ^6}{5040}-40 \pi ^2-\frac{29 \pi ^4}{45}+720-\frac{2}{9}  \log ^6(2)+\frac{4 \log ^5(2)}{5}-\frac{1}{18} \pi ^2 \log ^4(2)+\frac{2 \log ^4(2)}{3}+\frac{2}{3} \pi ^2 \log ^3(2)-24 \log ^3(2)-\frac{1}{36} \pi ^4 \log ^2(2)-8 \pi ^2 \log ^2(2)+144 \log ^2(2)+\frac{2}{3} \pi ^4 \log (2)+32 \pi ^2 \log (2)-480 \log (2)$

A bonus:

$\scriptsize \int_0^1 \frac{\log ^2(1-x) \log ^2(x) \log ^2(x+1)}{(1-x)^2 x^2 (x+1)^2} \, dx=8 \zeta(\bar5,1)+15\zeta(\bar5,1,1)-\frac{15}{2}\zeta(5,\bar1,1)+\frac{15}{2} \log (2) \zeta(\bar5,1)+15 \text{Li}_4\left(\frac{1}{2}\right) \zeta (3)-\frac{17}{12} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)+14 \text{Li}_4\left(\frac{1}{2}\right)+\frac{9}{2} \pi ^2 \text{Li}_5\left(\frac{1}{2}\right)+30 \text{Li}_5\left(\frac{1}{2}\right)+16 \text{Li}_6\left(\frac{1}{2}\right)+8 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+30 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+16 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+\frac{239 \pi ^2 \zeta (3)}{48}-\frac{5807 \pi ^2 \zeta (5)}{256}+\frac{9087 \zeta (7)}{32}-\frac{1457 \zeta (5)}{16}-\frac{5417 \zeta (3)^2}{128}-\frac{1259 \pi ^4 \zeta (3)}{1280}+\frac{13}{4} \zeta (3) \log ^4(2)-\frac{49}{4} \zeta (3) \log ^3(2)-\frac{61}{8} \pi ^2 \zeta (3) \log ^2(2)-\frac{203}{8} \zeta (3) \log ^2(2)+\frac{4743}{64} \zeta (5) \log ^2(2)+\frac{123}{2} \zeta (3)^2 \log (2)+\frac{31}{3} \pi ^2 \zeta (3) \log (2)-\frac{57}{4} \zeta (3) \log (2)-\frac{4867}{32} \zeta (5) \log (2)+\frac{167 \pi ^6}{2520}+\frac{19 \pi ^4}{1440}+\frac{2 \log ^6(2)}{9}-\frac{3}{80} \pi ^2 \log ^5(2)+\log ^5(2)-\frac{29}{288} \pi ^2 \log ^4(2)-\frac{17 \log ^4(2)}{12}-\frac{1}{32} \pi ^4 \log ^3(2)+\frac{5}{6} \pi ^2 \log ^3(2)+\frac{235}{576} \pi ^4 \log ^2(2)+\frac{5}{4} \pi ^2 \log ^2(2)+\frac{13}{48} \pi ^4 \log (2)-\frac{521 \pi ^6 \log (2)}{13440}$

This is not easy via elemenatry means. For reference, see here.