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.
 A: Solutions by Cornel Ioan Valean
It's straightforward to show that $\displaystyle \int_0^1 x^{n-1}\log(x)\log(1-x)\textrm{d}x=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\zeta(2)\frac{1}{n}$, and this is easily  extracted by differentiating once with respect to $n$ the well-known result {with a Beta function (involving derivates) structure}, $\displaystyle\int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-\frac{\psi(n+1)+\gamma}{n}$. Observe the latter result, when $n$ is a positive integer, may be put in the form with harmonic numbers, that is $\displaystyle\int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-\frac{H_n}{n}$. The last integral form  can be found elementary calculated in (Almost) Impossible Integrals, Sums, and Series, page $59$.
Returning to the first main integral and using the previous results, we get
$$\int_0^1\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}\textrm{d}x=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}\int_0^1x^{2n-1}\log(x)\log(1-x)\textrm{d}x 
$$
$$=-\frac{1}{2}\zeta(2)\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{n^2}+\frac{1}{4}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{n^3}+\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}^{(2)}}{n^2}$$
$$=G^2-\frac{5}{16}\log^2(2)\zeta(2)+\frac{35}{32}\log(2)\zeta(3)-\frac{119}{64}\zeta(4)+\frac{5}{96}\log^4(2)+\frac{5}{4}\operatorname{Li}_4\left(\frac{1}{2}\right),$$
and  the last equality follows by using the fact that the last two series are computed here, and they are some of the most difficult to calculate harmonic series (of low weight class,  $\le7$) in the mathematical literature.
For the second main integral, we have the more general result,
\begin{equation*}
\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x^2}{2}\right)}{1-x}\textrm{d}x
\end{equation*}
\begin{equation*}
=m\zeta (m+1)- 2^{-m} \left(1-2^{-m+1}\right) \log(2 ) \zeta (m) -\sum_{k=0}^{m-1}\beta(k+1)\beta(m-k)
\end{equation*}
\begin{equation*}
-\sum_{k=1}^{m-2}2^{- m-1}(1-2^{-k})(1-2^{-m+k+1}) \zeta (k+1)\zeta (m-k),
\end{equation*}
where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$ represents the $n$th generalized harmonic number of order $m$,
$\zeta$ denotes the Riemann zeta function and $\beta$ designates the Dirichlet beta function.
A very simple solution may be found in this answer that exploits the symmetry in double integrals.
End of story
A: 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}
A: 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}
