TITLE: How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$
QUESTION [131 upvotes]: 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!

REPLY [2 votes]: Here is a proof for $\left(4\right)$ since i couldn't find one:
$$\int _0^1\frac{\ln ^3\left(1+x\right)\ln \left(x\right)}{x^2}\:dx$$
$$\overset{\operatorname{IBP}}=-\ln ^3\left(2\right)+3\int _0^1\frac{\ln ^2\left(1+x\right)}{x\left(1+x\right)}\:dx+3\int _0^1\frac{\ln \left(x\right)\ln ^2\left(1+x\right)}{x\left(1+x\right)}\:dx$$

$$3\underbrace{\int _0^1\frac{\ln ^2\left(1+x\right)}{x\left(1+x\right)}\:dx}_{x=\frac{1}{1+x}}=3\int _0^1\frac{\ln ^2\left(x\right)}{1-x}\:dx-3\int _0^{\frac{1}{2}}\frac{\ln ^2\left(x\right)}{1-x}\:dx$$
$$=6\sum _{k=1}^{\infty }\frac{1}{k^3}-6\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}-6\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^2\:2^k}-3\ln ^3\left(2\right)$$
$$=6\zeta \left(3\right)-6\operatorname{Li}_3\left(\frac{1}{2}\right)-6\ln \left(2\right)\operatorname{Li}_2\left(\frac{1}{2}\right)-3\ln ^3\left(2\right)$$
$$=\frac{3}{4}\zeta \left(3\right)-\ln ^3\left(2\right)$$

$$3\underbrace{\int _0^1\frac{\ln \left(x\right)\ln ^2\left(1+x\right)}{x\left(1+x\right)}\:dx}_{x=\frac{1}{1+x}}$$
$$=3\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln ^2\left(1-x\right)}{x}\:dx-3\int _{\frac{1}{2}}^1\frac{\ln ^3\left(x\right)}{1-x}\:dx$$
$$=-6\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}-6\ln \left(2\right)\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+6\sum _{k=1}^{\infty }\frac{1}{k^4\:2^k}+6\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}+18\sum _{k=1}^{\infty }\frac{1}{k^4}$$
$$-18\sum _{k=1}^{\infty }\frac{1}{k^4\:2^k}-18\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}-9\ln ^2\left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^2\:2^k}-3\ln ^4\left(2\right)$$
$$=\frac{69}{4}\zeta \left(4\right)-18\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{63}{4}\ln \left(2\right)\zeta \left(3\right)+\frac{9}{2}\ln ^2\left(2\right)\zeta \left(2\right)-\frac{3}{4}\ln ^4\left(2\right)$$
Where $\ln ^2\left(1-x\right)=2\sum _{k=1}^{\infty }\left(\frac{H_k}{k}-\frac{1}{k^2}\right)x^k$ is used on the $2$nd line.
See here and here for the $1$st and $2$nd sum.

Collecting the results yields:
$$\int _0^1\frac{\ln ^3\left(1+x\right)\ln \left(x\right)}{x^2}\:dx=\frac{69}{4}\zeta \left(4\right)+\frac{3}{4}\zeta \left(3\right)-18\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{63}{4}\ln \left(2\right)\zeta \left(3\right)$$
$$+\frac{9}{2}\ln ^2\left(2\right)\zeta \left(2\right)-2\ln ^3\left(2\right)-\frac{3}{4}\ln ^4\left(2\right)$$