<s>[INST] How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$ Please help me to find a closed form for this integral:
$$I=\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx\tag1$$
I suspect it might exist because there are similar integrals having closed forms:
$$\begin{align}\int_0^1\frac{\ln^3(1-x)\ln x}x\mathrm dx&=12\zeta(5)-\pi^2\zeta(3)\tag2\\
\int_0^1\frac{\ln^2(1+x)\ln x}x\mathrm dx&=\frac{\pi^4}{24}-\frac16\ln^42+\frac{\pi^2}6\ln^22-\frac72\zeta(3)\ln2-4\operatorname{Li}_4\!\left(\tfrac12\right)\tag3\\
\int_0^1\frac{\ln^3(1+x)\ln x}{x^2}\mathrm dx&=\frac34\zeta(3)-\frac{63}4\zeta(3)\ln2+\frac{23\pi^4}{120}\\&-\frac34\ln^42-2\ln^32+\frac{3\pi^2}4\ln^22-18\operatorname{Li}_4\!\left(\tfrac12\right).\tag4\end{align}$$
Thanks!
 [/INST] An alternate form of the answers given by @Cleo and @Tunk-Fey as sum of $1$ and $1/2$ argumented polylogarithm-products with rational coefficients:
$$I = \frac{99}{16}\operatorname{Li}_5(1)-12\operatorname{Li}_5\left(\frac{1}{2}\right) + 15\operatorname{Li}_1\left( \frac{1}{2} \right)\operatorname{Li}_4(1) - 12\operatorname{Li}_1\left(\frac{1}{2}\right)\operatorname{Li}_4\left(\frac{1}{2}\right) - 15\operatorname{Li}_2\left( \frac{1}{2} \right)\operatorname{Li}_3(1)-\frac{51}{4}\operatorname{Li}_1^2\left( \frac{1}{2} \right)\operatorname{Li}_3(1)+12\operatorname{Li}_2(1)\operatorname{Li}_3\left( \frac{1}{2} \right) - \frac{2}{5}\operatorname{Li}_1^5\left(\frac{1}{2}\right),$$
where $\operatorname{Li}_n$ is the polylogarithm function, and specifically
$$\begin{align}
& \operatorname{Li}_5(1) \ \ \ =  \zeta(5) \\
& \operatorname{Li}_5\left(\textstyle\frac{1}{2}\right) = \textstyle \sum_{k=1}^\infty {2^{-k} \over k^5} \\
& \operatorname{Li}_4(1) \ \ \ = \zeta(4) = \frac{\pi^4}{90} \\
& \operatorname{Li}_4\left(\textstyle\frac{1}{2}\right) = \textstyle \sum_{k=1}^\infty {2^{-k} \over k^4} \\
& \operatorname{Li}_3(1) \ \ \ =  \zeta(3) \\
& \operatorname{Li}_3\left(\textstyle\frac{1}{2}\right) = \frac{7}{8} \zeta(3) - \frac{\pi^2}{12} \ln 2 + \frac{1}{6} \ln^3 2 \\
& \operatorname{Li}_2(1) \ \ \ =  \zeta(2) = \frac{\pi^2}{6} \\
& \operatorname{Li}_2\left(\textstyle\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{1}{2} \ln^2 2 \\
& \operatorname{Li}_1\left(\textstyle\frac{1}{2}\right) = \ln2,
\end{align}$$
where $\zeta$ is the Riemann zeta function.
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