Two hard integrals: $\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx$ and $\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx$ I found two integrals that seem hard to evaluate:
$$I_1=\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx$$
$$I_2=\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx$$
I am just a beginner in logarithmic integral. So, I searched to find substitution like $x=\frac{1}{1+x}$, $x=\frac{1}{1-x}$, or $x=\frac{1-x}{1+x}$, but they didn't work.
Can I ask some ideas from every one? Thank you.
EDIT: After using Mathematica, with MZIntegrate paclet gives closed-form:
$$\begin{align}I_1&=\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx\\&=G^2+\frac{5 \text{Li}_4\left(\frac{1}{2}\right)}{4}+\frac{35}{32} \zeta (3) \log (2)-\frac{119 \pi ^4}{5760}+\frac{5 \log ^4(2)}{96}-\frac{5}{96} \pi ^2 \log ^2(2)\\I_2&=\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx\\&=2 G^2+\frac{35}{16} \zeta (3) \log (2)-\frac{199 \pi ^4}{5760}\end{align}$$
where $G$ is Catalan's constant.

Express the first integral via integration by parts in terms of the second and a third integral
$$
I_1=\int_{0}^{1}\frac{\ln x\ln(1-x)\ln (1+x^2)}{x}dx=\frac12I_2 -I_3\tag1$$
where
\begin{align}
I_2=&\int_{0}^{1}\frac{\ln^2x\ln (1+x^2)}{1-x}dx
= 2G^2 +\frac{35}{16}\ln2\zeta(3)-\frac{199\pi^4}{5760}\\
I_3=& \int_{0}^{1}\frac{x\ln^2 x\ln(1-x)}{1+x^2}dx
=-\frac54\text{Li}_4\left(\frac{1}{2}\right)+\frac{13 \pi ^4}{3840}+\frac{5\pi^2}{96} \ln ^22-\frac{5 }{96}\ln ^42
\end{align}
Similarly to $I_2$, $I_3$ is relatively manageable and is evaluated in the Appendix. Substitute above into (1) to obtain
$$I_1=\frac54\text{Li}_4\left(\frac{1}{2}\right)+ G^2+\frac{35}{32} \ln2\zeta (3) -\frac{119 \pi ^4}{5760}-\frac{5\pi^2}{96} \ln ^22+\frac{5 }{96}\ln ^42
$$

Appendix: (to be completed)
\begin{align}
I_3=\frac18 \int_{0}^{1}\frac{\ln^2 x\ln (1-x)}{1+x}\ \overset{x\to x^2}{dx}
 - \int_{0}^{1}\frac{x\ln^2x\ln(1+x)}{1+x^2}dx\\
\end{align}
\begin{align}
\int_{0}^{1}\frac{\ln^2x\ln(1-x)}{1+x}dx=&\ 
-4\text{Li}_4\left(\frac{1}{2}\right) +\frac{\pi^4}{90}+\frac{\pi^2}6\ln^22-\frac16\ln^42\\
\int_{0}^{1}\frac{x\ln^2x\ln(1+x)}{1+x^2}dx=&\
\frac34 \text{Li}_4\left(\frac{1}{2}\right)-\frac{23\pi^4}{11520}-\frac{\pi^2}{32}\ln^22+\frac1{32}\ln^42
\end{align}
