The development of a solution to the proposed integral
I asked Cornel for a reduction to known integrals and series and here is a possible way to go
First Step
Note that
$$\int_0^1\frac{\operatorname{Li}_2(x^2)}{1-x^2}\left(\frac{\ln(1+x)}{x}-\log(2)\right)\textrm{d}x$$
$$=\int_0^1\frac{\operatorname{Li}_2(x^2)}{1-x^2}\left(\frac{\log((1+x)/2)}{x}+\log(2)\frac{1-x}{x}\right)\textrm{d}x$$
$$=\log(2)\int_0^1 \frac{\operatorname{Li}_2\left(x^2\right)}{x(1+x)} \textrm{d}x-\int _0^1 \left(\int _x^1\frac{\text{Li}_2\left(x^2\right)}{x(1-x^2)(1+y)}\textrm{d}y\right)\textrm{d}x$$
$$=\log(2)\underbrace{\int_0^1 \frac{\operatorname{Li}_2\left(x^2\right)}{x(1+x)} \textrm{d}x}_{\text{Reducible to known integrals and series}}-\underbrace{\int _0^1 \left(\int _0^y\frac{\text{Li}_2\left(x^2\right)}{x(1-x^2)(1+y)}\textrm{d}x\right)\textrm{d}y}_{\displaystyle \mathcal{I}}.$$
Second Step
Since we have that $\displaystyle \sum_{n=1}^{\infty} x^n H_n^{(2)} =\frac{\operatorname{Li}_2(x)}{1-x}$ and $\displaystyle \int_0^1\frac{x^{2n}}{1+x}\textrm{d}x=H_{n}-H_{2n}+\log(2),$ then our integral $\mathcal{I}$ reduces to the harmonic series
$$\mathcal{I}=\frac{1}{2}\sum _{n=1}^{\infty } \frac{H_n^{(2)} }{n}\left(H_n-H_{2 n}+\log (2)\right).$$
Third step
We want to apply Abel's summation to the last series and then we get that
$$\mathcal{I}=\zeta(2)-\frac{1}{2}\zeta (3)-\frac{1}{2}\log (2)\zeta (3)-\frac{23}{16}\zeta(4)-2 \log ^2(2)$$
$$+\frac{1}{2} \sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(n+1)^2}+\frac{1}{2} \underbrace{\sum _{n=1}^{\infty } \frac{H_n H_n^{(2)}}{(2 n+1) (2 n+2)}}_{\displaystyle \mathcal{S}}.$$
The first series with $n$ instead of $n+1$ in denominator is calculated here.
Fourth step
We may recall that
$$\sum_{n=1}^{\infty} x^n H_n H_n^{(2)}$$
$$=\frac{1}{1-x}\biggr(\frac{1}{2}\log(x) \log^2(1-x)+\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)-\zeta(2)\log(1-x)-\zeta(3)\biggr),$$
which is also found and proved in the book (Almost) Impossible Integrals, Sums, and Series.
Thus, using this result where we replace $x$ by $x^2$, throwing a double integral over both sides in the form $\displaystyle \int_0^1 \int_0^y f(x)\textrm{d}x\textrm{d}y$, and then rearranging and calculating the elementary resulting integrals, everything reduces to
$$\mathcal{S}=\sum _{n=1}^{\infty } \frac{H_n H_n^{(2)}}{(2 n+1) (2 n+2)}=\frac{5 }{4}\zeta (4)- \log (2)\zeta (3)-\log ^2(2)\zeta (2)$$
$$+\underbrace{\int_0^1 \frac{\operatorname{Li}_3\left(x^2\right)}{1+x} \textrm{d}x}_{\text{Reducible to known integrals and series}}+ \underbrace{\int_0^1 \frac{\operatorname{Li}_3\left(1-x^2\right)}{1+x} \textrm{d}x}_{\text{Reducible to known integrals and series}}$$
$$+\underbrace{ \int_0^1 \frac{\log (x) \log ^2\left(1-x^2\right)}{1+x} \textrm{d}x}_{\text{Reducible to known integrals and series}}.$$
A first short note: The case $\displaystyle \sum_{n=0}^\infty(-1)^n(\overline{H}_n-\log(2))^5$ may be reduced to a similar integral which can be approached in a similar style, except that we'll need to deal with weight $5$ harmonic series, and some of them might be pretty challenging like you may see in this paper On the calculation of two essential harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$.
A second short note: The last integral may also be viewed in the form of a Beta function sum since
$$ \int_0^1 \frac{\log (x) \log ^2\left(1-x^2\right)}{1+x} \textrm{d}x=\int_0^1 \frac{\log (x) \log ^2\left(1-x^2\right)}{1-x^2}(1-x) \textrm{d}x$$
$$=\frac{1}{4} \underbrace{\int_0^1 \frac{\log (x) \log ^2(1-x)}{\sqrt{x} (1-x)} \textrm{d}x}_{\text{Beta function}}-\frac{1}{4} \underbrace{\int_0^1 \frac{\log (x) \log ^2(1-x)}{1-x} \textrm{d}x}_{\text{Beta function}},$$
and it's also good to see the second integral works nicely by letting $x\mapsto 1-x$ and using geometric series. Good to know the first integral can be finished very nicely with Cornel's Master Theorem of Series, and thus we avoid tedious calculations with Beta function.
A third short note: A good point for the second integral from the Fourth Step (see the final part).
Now, if you start with integration by parts and then use the Dilogarithm reflection formula, $\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\zeta(2)-\log(x)\log(1-x)$, you arrive at a bunch of integrals where you might also find very useful to know that
$$ \int_0^1 \frac{\displaystyle \log(1-x)\operatorname{Li}_2\left(\frac{x}{x-1}\right)}{1+x} \textrm{d}x=\frac{29}{16} \zeta (4)+\frac{1}{4}\log ^2(2) \zeta (2) -\frac{1}{8} \log ^4(2),$$ which is presented and calculated in (Almost) Impossible Integrals, Sums, and Series, see page $17$ (also you may see a different approach here). Also good to know this is a very important integral in the extraction process of some harmonic series with skew-harmonic numbers!
So, it's clear now we may extract the value of $\displaystyle \int_0^1 \frac{\displaystyle \log(1-x)\operatorname{Li}_2(x)}{1+x} \textrm{d}x$ by also using Landen's Identity.
Alternatively, if you have a taste for Fourier series, we may easily arrive at the form
$$\int_0^1 \frac{\operatorname{Li}_3\left(1-x^2\right)}{1+x} \textrm{d}x= \frac{1}{4} \int_0^1 \frac{\log (1-x) \log ^2(x)}{x} \textrm{d}x+4\int_0^{\pi/2} x \log ^2\left(\sin(x)\right) \textrm{d}x,$$
where in the calculations I used the integral representation of the Trilogarithm you may found on page $4$ in the mentioned book. The second integral in the right-hand side is easily manageable by Fourier series and it has also been calculated here tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$.
So, there are at least two elegant ways to deal with
$$\int_0^1 \frac{\operatorname{Li}_3\left(1-x^2\right)}{1+x} \textrm{d}x.$$
The closed-form of a challenging auxiliary integral
$$\int_0^1 \frac{\operatorname{Li}_3\left(1-x^2\right)}{1+x} \textrm{d}x=2 \log ^2(2)\zeta (2)-\frac{23 }{8}\zeta (4)+\frac{1}{6}\log ^4(2)+4 \operatorname{Li}_4\left(\frac{1}{2}\right).$$
The closed-form of the series $\mathcal{S}$
$$\sum _{n=1}^{\infty } \frac{H_n H_n^{(2)}}{(2 n+1) (2 n+2)}=\frac{19 }{8}\zeta (4)-\frac{1}{6} \log ^4(2)-4 \operatorname{Li}_4\left(\frac{1}{2}\right).$$
The final closed-form
$$\int_0^1\frac{\operatorname{Li}_2(x^2)}{1-x^2}\left(\frac{\ln(1+x)}{x}-\log(2)\right)\textrm{d}x$$
$$=\frac{1}{6}\log ^4(2)-\frac{7 }{2}\zeta (4)+\frac{7}{2}\log (2)\zeta (3)-\frac{3}{2}\log ^2(2)\zeta (2)+4 \operatorname{Li}_4\left(\frac{1}{2}\right).$$
