Integral $\int_0^1\frac{\log(x)\log^2(1-x)\log^2(1+x)}{x}\mathrm dx$

Question

I decided to follow a recent trend and ask a question about logarithmic integrals :)

Is there a closed form for this integral? $$\int_0^1\frac{\log(x)\log^2(1-x)\log^2(1+x)}{x}\mathrm dx$$

Answer 1

This integral is equal to $$ -4\big( \zeta(-3,-1,-1,-1) +\zeta(-3,-1,1,-1) +\zeta(-3,1,-1,1) +\zeta(3,-1,-1,-1) +\zeta(3,-1,1,-1) +\zeta(3,1,-1,1) \big) $$ in terms of the multiple zeta function, which can also be simplified to $$ 2\zeta(-5,-1)-2\zeta(-5,1)+2\zeta(5,-1)+{\textstyle\frac32}\zeta(5,1)+4\zeta(-3,1,1,1), $$ of which only $$ \begin{aligned} \zeta(5,1) &= {\textstyle\frac34}\zeta(6)-{\textstyle\frac12}\zeta(3)^2 \\ \zeta(5,-1) &= {\textstyle\frac{111}{64}} \zeta (6)-{\textstyle\frac{9}{32}} \zeta (3)^2-{\textstyle\frac{31}{16}} \zeta (5) \log (2) \end{aligned} $$ have a known closed form (see also this article about Euler sums, and also Euler Sums and Contour Integral Representations by Philippe Flajolet and Bruno Salvy).

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Update (by editor): Based on MZV reduction of weight $6$, expression above is furtherly simplified to: $$-2 \zeta(\bar5,1)+8 \text{Li}_6\left(\frac{1}{2}\right)+4 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+8 \text{Li}_5\left(\frac{1}{2}\right) \log (2)-\frac{13 \zeta (3)^2}{16}+\frac{7}{6} \zeta (3) \log ^3(2)-\frac{221 \pi ^6}{30240}+\frac{\log ^6(2)}{9}-\frac{1}{12} \pi ^2 \log ^4(2)$$