How to compute $\int_0^1\left(\operatorname{Li}_2(x)-\zeta(2)\right)\frac{\ln^2(1-x^2)}{1-x^2}\ dx$ How to compute 

$$\int_0^1\left(\operatorname{Li}_2(x)-\zeta(2)\right)\frac{\ln^2(1-x^2)}{1-x^2}\ dx$$

where $\operatorname{Li}_2(x)=\sum_{n=1}^\infty \frac{x^n}{n^2}$ is the dilogarithm function.

I came across this integral while trying to calculate $\sum_{n=1}^\infty \frac{H_{2n}H_n^2}{(2n+1)^2}$. If we apply integration by parts, one of the integral is manageable and the second one is $\int_0^1\left(\operatorname{Li}_2(x)-\zeta(2)\right)\frac{\tanh^{-1}x\ln(1-x^2)x}{1-x^2}\ dx$ which I could not finalize. So any idea how to solve any of these two integrals? 
In case you are curious about the sum, Cornel provided the closed form here.
 A: Here is the Cornel's reduction of the $\operatorname{arctanh}$ integral to manageable integrals by using clever integrations by parts and exploiting Dilogarithm reflection formula in $(5)$ here http://mathworld.wolfram.com/Dilogarithm.html.
$$I=\frac{1}{16} \underbrace{\int_0^1 \frac{\text{Li}_2(x) \log ^2(x)}{x} \textrm{d}x}_{\text{known}}-\frac{1}{16} \underbrace{\int_0^1 \frac{\text{Li}_2(x) \log ^2(2-x)}{x} \textrm{d}x}_{\text{may use series with 
powers of 2 in denominator}}\\-\frac{1}{4} \underbrace{\int_0^1 \frac{\text{Li}_2(x) \log ^2(x)}{2-x} \textrm{d}x}_{\text{may use series with 
powers of 2 in denominator}}-\frac{3}{8} \underbrace{\int_0^1 \frac{\text{Li}_2(x) \log (x) \log (2-x)}{2-x} \textrm{d}x}_{\text{may use series with 
powers of 2 in denominator}}\\+\frac{5}{64} \underbrace{\int_0^1 \frac{\log ^4(1-x)}{x} \textrm{d}x}_{\text{known}}-\frac{1}{12} \underbrace{\int_0^1 \frac{\log (1-x) \log ^3(1+x)}{x} \textrm{d}x}_{\text{known}}\\-\frac{7}{16} \underbrace{\int_0^1 \frac{\log^2(1-x) \log (x) \log (1+x)}{1+x} \textrm{d}x}_{\text{Calculated in (Almost) Impossible Integrals, Sums, and Series}}-\frac{1}{4}\underbrace{\int_0^1\frac{\log^3(x)\log(1-x)}{2-x}\textrm{d}x}_{\text{may use series with 
powers of 2 in denominator}}\\-\frac{7}{32} \underbrace{\int_0^1 \frac{\log ^2(1-x) \log ^2(1+x)}{x} \textrm{d}x}_{\text{known}}.$$
Some of these integrals are already known. The harder ones can be easily reduced to advanced alternating harmonic series of weight $5$ or to advanced harmonic series of weight $5$ with powers of $2$ in denominator (and these ones are all already known). Also, playing with these integrals it's possible to get even more shortcuts. With this form of the main integral to calculate you're done.
A NOTE: The integral $\displaystyle \int_0^1 \frac{\log^2(1-x) \log (x) \log (1+x)}{1+x} \textrm{d}x$ may be found calculated in the book, (Almost) Impossible Integrals, Sums, and Series, pages $527$-$528$. In general, the integrals above where it is suggested a reduction to the series with powers of $2$ in denominator may also be rearranged and reduced to alternating harmonic series of weight $5$. End of story.
