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This integral to the value

\begin{align} \int_0^1\frac{\ln^2(1+x)\ln^2 x}{1-x}\ dx=&\ \color{blue}{-\frac{13\pi^2}{24}\zeta(3)+\frac{47}{2}\zeta(5)-\frac15\ln^52+\frac{\pi^2}9\ln^32-\frac{49\pi^4}{360}\ln2+\frac{7}2\zeta(3)\ln^22}\\&\color{blue}{-8\operatorname{Li}_4\left(\frac12\right)\ln2-16\operatorname{Li}_5\left(\frac12\right)}, \end{align} How to find this result? In fact,I find that $$\int_0^1\frac{\ln(1+x)\ln{x}}{1-x}dx=\zeta(3)-\frac{\pi^2}{4}\ln2$$ $$\int_0^1\frac{\ln^2(1+x)\ln{x}}{1-x}dx=\frac{21}{4}\zeta(3)\ln{2}-\frac{5\pi^2}{12}\ln^2{2}+\frac{1}{6}\ln^4{2}-\frac{7\pi^4}{144}+4\operatorname{Li}_4\left(\frac12\right)$$

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