Question: how to evaluate this logarithm integral? $$ I=\int_0^1 \frac{x \log ^2(x) \log (1-x)}{1+x^2} d x $$
My attempt: $$ \begin{aligned} I=&\int_0^1 \frac{x \log ^2(x) \log (1-x)}{1+x^2} d x\\ =&\frac{1}{2} \int_0^1 \log ^2(x) \log (1-x) d\left(\log \left(x^2+1\right)\right) \xrightarrow{I B P} \\ \stackrel{\text{IBP}}{=}&\underbrace{\left.\frac{1}{2} \log \left(x^2+1\right) \log ^2(x) \log (1-x)\right|_0 ^1}_{=0} \\ &-\int_0^1 \frac{\log (x) \log (1-x) \log \left(x^2+1\right)}{x} d x+\frac{1}{2} \int_0^1 \frac{\log ^2(x) \log \left(x^2+1\right)}{1-x} d x \\ \end{aligned} $$ Let $$J = \int_0^1 \frac{x \log ^2(x) \log (1-x)}{1+x^2} dX$$ $$K = \int_0^1 \frac{\log ^2(x) \log \left(x^2+1\right)}{1-x} dX$$
How to evaluate $J$ and $K$?
To continue the evaluation of the integral by using your work so far, you might first notice that $$\int_0^1 \frac{\log ^2(x) \log \left(1+x^2\right)}{1-x} \textrm{d}x=\int_0^1 \frac{\log ^2(x) \log((1-x^4)/(1-x^2))}{1-x} \textrm{d}x$$ $$=\int_0^1 \frac{\log^2(x) \log(1-x^4)}{1-x^4} (1+x)(1+x^2) \textrm{d}x-\int_0^1 \frac{\log^2(x) \log(1-x^2)}{1-x^2} (1+x) \textrm{d}x,$$
and here we observe we can exploit the Beta function, or we can use the generalization
$$ \int_0^1\frac{\displaystyle \log^m(x)\log\left(\frac{1+x^2}{2}\right)}{1-x}\textrm{d}x$$ $$=(-1)^{m-1}m! \biggr((m+1) \zeta(m+2)-\frac{1}{2^{m+2}}\sum_{k=0}^m \eta(k+1) \eta(m-k+1)$$ $$-\sum_{k=0}^m \beta(k+1) \beta(m-k+1)\biggr),$$
whose derivation is shown in this answer (and also found in the sequel mentioned below, p. $38$).
As regards the other integral, you can exploit $\displaystyle \int_{0}^{1}x^{n-1}\log(x)\log(1-x)\textrm{d}x=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\pi^2}{6}\frac{1}{n}$, which is given in a generalized form in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), pp. $32$-$33$, the sequel of (Almost) Impossible Integrals, Sums, and Series (2019), and combine it with $\displaystyle \log(1+x^2)=\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{2n}}{n}$ to arrive at $$\int_0^1 \frac{\log (x) \log (1-x) \log(1+x^2)}{x} \textrm{d}x$$ $$=\frac{1}{4}\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{2n}}{n^3}+\frac{1}{2}\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(2)}}{n^2}-\frac{\pi^2}{12}\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2},$$ and the desired values of the two very challenging harmonic series may be found in this post. They may also be found in the sequel, pp. $765$-$769$.
End of story