I managed to evaluate the integral based on many integrals/ sums results and solution turned out really long but I did my best to make it shorter. If any step is not clear please let me know.
In the body, we reached
$$\int_0^1\frac{(\operatorname{Li}_2(x)-\zeta(2))\operatorname{Li}_2(x^2)}{1-x^2}\ dx=-\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}-\color{blue}{\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^3}}$$
Or
$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}=-\int_0^1\frac{(\operatorname{Li}_2(x)-\zeta(2))\operatorname{Li}_2(x^2)}{1-x^2}\ dx-\color{blue}{\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^3}}$$
.
Evaluation of the integral:
By applying integration by parts we have
$$I=\int_0^1\frac{(\operatorname{Li}_2(x)-\zeta(2))\operatorname{Li}_2(x^2)}{1-x^2}\ dx\\=\int_0^1\frac{\tanh^{-1}x\ln(1-x)\operatorname{Li}_2(x^2)}{x}\ dx+2\int_0^1\frac{\tanh^{-1}x\ln(1-x^2)}{x}(\operatorname{Li}_2(x)-\zeta(2))\ dx\\
=A+2B$$
The first integral $A$: Using $\tanh^{-1}x=\frac12\ln\left(\frac{1+x}{1-x}\right)$
$$A=\frac12\int_0^1\frac{\ln(1+x)\ln(1-x)\operatorname{Li}_2(x^2)}{x}\ dx-\frac12\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x^2)}{x}\ dx\\
=\frac12(A_1-A_2)$$
where $A_1$ is nicely calculated by Cornel here:
$$A_1=\frac{275}{32}\zeta (5)-\frac{5 }{8}\zeta (2) \zeta (3)+\frac{4}{3} \ln ^32\zeta (2)-\frac{7}{2} \ln ^22\zeta (3)-\frac{4}{15}\ln ^52\\-8 \ln 2\operatorname{Li}_4\left(\frac{1}{2}\right)-8 \operatorname{Li}_5\left(\frac{1}{2}\right).$$
For $A_2$,
\begin{align}
A_2&=\sum_{n=1}^\infty\frac1{n^2}\int_0^1 x^{2n-1} \ln^2(1-x)\ dx\\
&=\sum_{n=1}^\infty\frac1{n^2}\left(\frac{H_{2n}^2+H_{2n}^{(2)}}{2n}\right)\\
&=2\sum_{n=1}^\infty\frac{H_n^2+H_n^{(2)}}{n^3}(1+(-1)^n)
\end{align}
collecting these results we get
$$A_2=-\frac{1}{8}\zeta (5)+\frac{11 }{2}\zeta (2) \zeta (3)+\frac{4}{3} \ln ^32\zeta (2)-\frac{7}{2}\ln^22\zeta (3)-\frac{4}{15}\ln ^52\\-8 \ln 2\operatorname{Li}_4\left(\frac{1}{2}\right)-8 \operatorname{Li}_5\left(\frac{1}{2}\right).$$
Combining $A_1$ and $A_2$ we get
$$\boxed{A=\frac{279}{64}\zeta(5)-\frac{49}{16}\zeta(2)\zeta(3)}$$
The second integral $B$:
By using the identity
$$\tanh^{-1}x\ln(1-x^2)=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}x^{2n+1}$$
we can write
\begin{align}
B&=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}\int_0^1 x^{2n}(\operatorname{Li}_2(x)-\zeta(2))\ dx\\
&=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}\left(-\frac{H_{2n}}{(2n+1)^2}-\frac{1}{(2n+1)^3}\right)\\
&=2\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n+1)^3}+2\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^4}\\
&=\sum_{n=1}^\infty\frac{H_{n}^2}{(n+1)^3}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}^2}{(n+1)^3}+\sum_{n=1}^\infty\frac{H_{n}}{(n+1)^4}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{(n+1)^4}\\
&=\sum_{n=1}^\infty\frac{H_{n-1}^2}{n^3}-\sum_{n=1}^\infty\frac{(-1)^nH_{n-1}^2}{n^3}+\sum_{n=1}^\infty\frac{H_{n-1}}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_{n-1}}{n^4}\\
&=\sum_{n=1}^\infty\frac{H_n^2}{n^3}-\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}-\sum_{n=1}^\infty\frac{H_n}{n^4}
\end{align}
collecting these results we get
$$\boxed{\small{B=-\frac{31}{16}\zeta (5)-\frac{7 }{8}\zeta (2) \zeta (3)-\frac{2}{3} \ln ^32\zeta (2)+\frac{7}{4}\ln^22\zeta (3)+\frac{2}{15}\ln ^52+4 \ln 2\operatorname{Li}_4\left(\frac{1}{2}\right)+4 \operatorname{Li}_5\left(\frac{1}{2}\right)}}$$
Finally, combine the boxed results of $A$ and $B$ we get
$$I=\frac{31}{64}\zeta (5)-\frac{77 }{16}\zeta (2) \zeta (3)-\frac{4}{3} \ln ^32\zeta (2)+\frac{7}{2}\ln^22\zeta (3)+\frac{4}{15}\ln ^52\\+8 \ln 2\operatorname{Li}_4\left(\frac{1}{2}\right)+8 \operatorname{Li}_5\left(\frac{1}{2}\right).$$
Evaluation of the blue sum:
By Abel's summation we have
$$\sum_{n=1}^\infty\frac{H_n^{(2)}-\zeta(2)}{(2n-1)^3}=\frac18\sum_{n=1}^\infty\frac{H_n^{(3)}}{(n+1)^2}-\sum_{n=1}^\infty\frac{H_{2n}^{(3)}}{(n+1)^2}$$
where
$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{(n+1)^2}=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-\zeta(5)=\frac92\zeta(5)-2\zeta(2)\zeta(3)$$
and
$$\sum_{n=1}^\infty\frac{H_{2n}^{(3)}}{(n+1)^2}=\sum_{n=1}^\infty\frac{H_{2n-2}^{(3)}}{n^2}=\sum_{n=1}^\infty\frac{H_{2n}^{(3)}-\frac1{(2n)^3}-\frac1{(2n-1)^3}}{n^2}\\
=\sum_{n=1}^\infty\frac{H_{2n}^{(3)}}{n^2}-\frac18\zeta(5)-\sum_{n=1}^\infty\frac1{n^2(2n-1)^3}\\$$
combining the two sums we get
$$\sum_{n=1}^\infty\frac{H_n^{(2)}-\zeta(2)}{(2n-1)^3}=\frac{11}{16}\zeta(5)-\frac14\zeta(2)\zeta(3)-\sum_{n=1}^\infty\frac{H_{2n}^{(3)}}{n^2}+\sum_{n=1}^\infty\frac1{n^2(2n-1)^3}\tag1$$
On the other hand:
\begin{align}
\sum_{n=1}^\infty\frac{H_n^{(2)}-\zeta(2)}{(2n-1)^3}&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n-1)^3}-\sum_{n=1}^\infty\frac{\zeta(2)}{(2n-1)^3}\\
&=\sum_{n=0}^\infty\frac{H_{n+1}^{(2)}}{(2n+1)^3}-\frac78\zeta(2)\zeta(3)\\
&=\sum_{n=0}^\infty\frac{H_{n}^{(2)}}{(2n+1)^3}+\sum_{n=0}^\infty\frac{1}{(n+1)^2(2n+1)^3}-\frac78\zeta(2)\zeta(3)\\
&=\sum_{n=0}^\infty\frac{H_{n}^{(2)}}{(2n+1)^3}+\sum_{n=1}^\infty\frac{1}{n^2(2n-1)^3}-\frac78\zeta(2)\zeta(3)\tag{2}
\end{align}
From (1) and (2) we get
$$\sum_{n=0}^\infty\frac{H_{n}^{(2)}}{(2n+1)^3}=\frac{11}{16}\zeta(5)+\frac58\zeta(2)\zeta(3)-\sum_{n=1}^\infty\frac{H_{2n}^{(3)}}{n^2}\\
=\frac{11}{16}\zeta(5)+\frac58\zeta(2)\zeta(3)-2\sum_{n=1}^\infty\frac{H_{n}^{(3)}}{n^2}-2\sum_{n=1}^\infty\frac{(-1)^nH_{n}^{(3)}}{n^2}$$
Thus
$$\sum_{n=1}^\infty\frac{H_{n}^{(2)}}{(2n+1)^3}=\frac{49}{8}\zeta(2)\zeta(3)-\frac{93}{8}\zeta(5)$$
Finally, by collecting the results of $I$ and the blue sum we get our closed form.
References
$\sum_{n=1}^\infty\frac{(-1)^{n}H_n^{(2)}}{n^3}=\frac{11}{32}\zeta(5)-\frac58\zeta(2)\zeta(3)\\
\small{\sum_{n=1}^\infty\frac{(-1)^{n}H_n^2}{n^3}=-4\operatorname{Li}_5\left(\frac12\right)-4\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{19}{32}\zeta(5)+\frac{11}8\zeta(2)\zeta(3)-\frac74\ln^22\zeta(3)+\frac23\ln^32\zeta(2)-\frac2{15}\ln^52}
$
$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$
$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)\\\sum_{n=1}^\infty\frac{H_n^2}{n^3}=\frac72\zeta(5)-\zeta(2)\zeta(3)$
$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}2\zeta(5)-2\zeta(2)\zeta(3)$
$\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}=-\frac{59}{32}\zeta(5)+\frac12\zeta(2)\zeta(3)$
$\sum^\infty_{n=1}\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ ( Euler Identity)
