The posed question is very similar in some sense to the one treated in my previous post (and also the answers might look similar to some extent, in some sense).
A solution in (large steps) by Cornel Ioan Valean
In the following, the general line of a straightforward solution (it can be counted so if we have some key and extremely hard to derive series at hand, and we have them!) will be given. More details appearing on the way will be found in the sequel of (Almost) Impossible Integrals, Sums, and Series.
We may start directly from the main integral as posed and turn it into a sum of series by using that $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} x^{2n-1}=\frac{x}{1+x^2}, \ \int_0^1 x^{n-1} \log(x)\log^2(1-x)\textrm{d}x=2\zeta(3)\frac{1}{n}+2\zeta(2)\frac{H_n}{n}-\frac{H_n^{(2)}}{n^2}-2\frac{H_n^{(3)}}{n}-2\frac{H_n H_n^{(2)}}{n}-\frac{H_n^2}{n^2},$ where the latter is immediately obtained by differentiating once with respect to $n$ the well-known logarithmic integral, $\displaystyle \int_0^1 x^{n-1}\log^2(1-x)\textrm{d}x=\frac{(\psi(n+1)+\gamma)^2+\zeta(2)-\psi^{(1)}(n+1)}{n}$ (observe the Beta function structure). When $n$ is a positive integer, the previous logarithmic integral result immediately turns into $\displaystyle \int_0^1 x^{n-1}\log^2(1-x)\textrm{d}x=\frac{H_n^2+H_n^{(2)}}{n}.$ An elementary derivation of the latter special case may be found in (Almost) Impossible Integrals, Sums, and Series, pages $60-62$.
So, returning to the main integral, and using the results above, we have
$$\int_0^1\frac{x\log(x)\log^2(1-x)}{1+x^2}\textrm{d}x=\sum_{n=1}^{\infty}(-1)^{n-1}\int_0^1 x^{2n-1}\log{(x)}\log^2(1-x)\textrm{d}x$$
$$=\zeta(3)\underbrace{\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n}}_{\displaystyle \log(2)}+\zeta(2)\underbrace{\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}}_{\displaystyle 5/48 \pi^2-\log^2(2)/4}-\frac{1}{4}\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^2}{n^2}-\frac{1}{4}\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(2)}}{n^2}$$
$$-\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n} H_{2n}^{(2)}}{n}-\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(3)}}{n}$$
$$=\frac{167}{256}\zeta(4)+\frac{3}{16}\log^2(2)\zeta(2)-\frac{5}{64}\log^4(2)-\frac{15}{8}\operatorname{Li}_4\left(\frac{1}{2}\right).\tag1$$
Now, a first important point to mention is that during the extraction of some advanced harmonic series appearing above we need the well-known polylogarithmic relations on this page, and the special tetralogarithmic values,
$$\Re\biggr \{\operatorname{Li}_4\left(\frac{1\pm i}{2}\right)\biggr\}=\frac{343}{1024}\zeta(4)-\frac{5}{128}\log^2(2)\zeta(2)+\frac{1}{96}\log^4(2)+\frac{5}{16}\operatorname{Li}_4\left(\frac{1}{2}\right);$$
$$\Re\{\operatorname{Li}_4(1\pm i)\}=\frac{485}{512}\zeta(4)+\frac{1}{8}\log^2(2)\zeta(2)-\frac{5}{384}\log^4(2)-\frac{5}{16}\operatorname{Li}_4\left(\frac{1}{2}\right).$$
In order to extract the tetralogarithmic values we need the alternating harmonic series $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{n^3}$ you may find calculated in a very simple and elegant way in this answer. Other steps are trivial as explained in this 2014 answer, where we use the classical generating function $\displaystyle \sum_{n=1}^{\infty}x^n \frac{H_n}{n^3}$ and the tetralogarithmic inverse relation in order to make the connections between that series and the special tetralogarithmic values. Full details about the extraction process of these results (and other related ones) will be found in the sequel of (Almost) Impossible Integrals, Sums, and Series.
The fourth series in $(1)$ is calculated in this awesome answer. The third and the fifth series may be extracted immediately by exploiting their corresponding generating functions
$$\sum_{n=1}^{\infty} x^n \frac{H_n^2}{n^2}$$
$$=2\zeta(4)-\frac{1}{3}\log(x)\log^3(1-x)-\log^2(1-x)\operatorname{Li}_2(1-x)+\frac{1}{2}(\operatorname{Li}_2(x))^2$$
$$+2\log(1-x)\operatorname{Li}_3(1-x)+\operatorname{Li}_4(x)-2\operatorname{Li}_4(1-x), \ |x|\le1 \land \ x\neq1,$$
and
$$\sum_{n=1}^{\infty} x^n \frac{H_n H_n^{(2)}}{n}$$
$$=-\zeta(4)+\frac{1}{6}\log(x)\log^3(1-x)-\frac{1}{24}\log^4(1-x)+\frac{1}{2}\log^2(1-x)\operatorname{Li}_2(1-x)$$
$$-\log(1-x)\operatorname{Li}_3(1-x)+\operatorname{Li}_4(1-x)-\operatorname{Li}_4\left(\frac{x}{x-1}\right), \ |x|\le1 \land \ x\neq1.$$
leading to
$$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{2n}^2}{n^2}$$
$$=2 G^2-\log(2)\pi G+\frac{231}{32}\zeta(4)-\frac{35}{16}\log(2)\zeta(3)+\log^2(2)\zeta(2)-\frac{5}{48}\log^4(2)$$
$$-2 \pi \Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}-\frac{5}{2}\operatorname{Li}_4\left(\frac{1}{2}\right)$$
and
$$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{2n}H_{2n}^{(2)}}{n}$$
$$
=\frac{1}{4}\log(2)\pi G-\frac{137}{128}\zeta(4)+\frac{35}{64}\log(2)\zeta(3)-\frac{3}{8}\log^2(2)\zeta(2)+\frac{5}{96}\log^4(2)$$
$$+\frac{\pi}{2}\Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}+\frac{5}{4}\operatorname{Li}_4\left(\frac{1}{2}\right).$$
Both series above require tetralogarithmic values with a complex argument as previously mentioned.
And the final series in $(1)$ is a particular case of the more general result,
$$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x^2}{2}\right)}{1-x}\textrm{d}x$$
$$=m\zeta (m+1)- 2^{-m} \left(1-2^{-m+1}\right) \log(2 ) \zeta (m) -\sum_{k=0}^{m-1}\beta(k+1)\beta(m-k)$$
$$-\sum_{k=1}^{m-2}2^{- m-1}(1-2^{-k})(1-2^{-m+k+1}) \zeta (k+1)\zeta (m-k),$$
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
