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.

The routine integrals $\int_0^1 \frac{\ln^2t}{1-t}dt = 2\zeta(3)$,
$\int_0^1 \frac{\ln^2t}{1+t}dt = \frac32\zeta(3)$,
$\int_0^1 \frac{\ln^3t}{1-t}dt = -\frac{\pi^4}{15}$,
$\int_0^1 \frac{\ln^2t}{1+t^2}dt =\frac{\pi^3}{16}$,
$\int_0^1 \frac{\ln t}{1+t^2}dt =-G$, and
$\int_0^1 \frac{\ln t}{1+t}dt =-\frac{\pi^2}{12}$ are used below without elaboration.
\begin{align}
I_2=&\int_0^1\frac{\ln^2x\ln(1+x^2)}{1-x}dx\\
=& \int_0^1\ln(1+x^2)\ d\left( \int_0^x \frac{\ln^2t}{1-t}dt\right)\\
=& \ \ln2 \int_0^1 \frac{\ln^2t}{1-t}dt
-\int_0^1\frac{2x}{1+x^2}\int_0^x\frac{\ln^2t}{1-t} \overset{t=xy}{dt}\\
=&\ 2\ln2\zeta(3) +2\int_0^1\int_0^1\frac{\ln^2xy}{1+y^2}\left(\frac{1+xy}{1+x^2}-\frac1{1-xy}\right)dxdy\tag1
\end{align}
Note that, with $\ln^2xy =\ln^2x +2\ln x\ln y +\ln^2 y$
\begin{align}
&\int_0^1\int_0^1\frac{(1+xy)\ln^2xy}{(1+x^2)(1+y^2)}dxdy\\
=& \int_0^1 \int_0^1\frac{\ln^2x +2\ln x\ln y +\ln^2 y}{(1+x^2)(1+y^2)}+\frac1{16}\frac{\ln^2x +2\ln x\ln y +\ln^2 y}{(1+x)(1+y)}\ dxdy\\
=&\ 2G^2 +\frac3{16}\ln2\zeta(3)+\frac{37\pi^4}{1152}\\
\\ & \int_0^1\int_0^1\frac{\ln^2xy}{(1+y^2)(1-xy)}\overset{x=t/y}{dx}dy\\
=& \ \frac12\int_0^1 d\left(\ln\frac{y^2}{1+y^2}\right)\int_0^y \frac{\ln^2t}{1-t}dt
\overset{ibp}=-\ln2\zeta(3)+\frac{\pi^4}{15}+\frac12I_2
\end{align}
Plug into (1) to obtain
\begin{align}
I_2 = &\ 2\ln2\zeta(3)+2 \left(2G^2 +\frac3{16}\ln2\zeta(3)+\frac{37\pi^4}{1152} \right)+2\ln2\zeta(3)-\frac{2\pi^4}{15}-I_2\\
=&\ 2G^2 +\frac{35}{16}\ln2\zeta(3)-\frac{199\pi^4}{5760}
\end{align}
