Prove $\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)=\frac{\pi^3}{32}-2G\ln2$ How to prove

$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)\stackrel ?=\frac{\pi^3}{32}-2G\ln2,$$
  where $G$ is the Catalan's constant.

Attempt
For the first sum,
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_{2k}=\Re\left\{\sum_{k=1}^{\infty}\frac{i^k}{(k+1)^2}H_{k}\right\},$$
which can be evaluated by using the formula in this post:
$$\sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n=\zeta(3)+\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x),$$
but we cannot apply the similar approach to the second sum 
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_k.$$
Then, I tried to write the sum as
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}\int_0^1\frac{2x^{2k}+x^k-3}{x-1}~\mathrm dx$$
and it become more complicated.
Edit: 
Are we able to evaluate the sum directly (avoid calculating integrals and  polylogs as much as possible)? The integral given by @Jack D'Aurizio is a bit complicated (see this post).

\begin{align}
S&=2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}\\
&=2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}-2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\
&=2\Im\sum_{n=1}^\infty\frac{(i)^nH_{n}}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}-\frac{\pi^2}{16}\\
&=S_1+S_2-\frac{\pi^3}{16}\tag{1}
\end{align}
Using the generating function: $$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$
then 
\begin{align}
S_1&=2\Im\left(\operatorname{Li}_3(i)-\operatorname{Li}_3(1-i)+\ln(1-i)\operatorname{Li}_2(1-i)+\frac12\ln(i)\ln^2(1-i)\right)\\
&\boxed{S_1=-2\Im\operatorname{Li}_3(1-i)-G\ln2-\frac{\pi}{8}\ln^22}
\end{align}
$$S_2=\sum_{n=1}^\infty\frac{(-1)^nH_n}{(2n+1)^2}=\int_0^1\frac{\ln(1+x^2)\ln x}{1+x^2}\ dx$$
The last integral is evaluated here
$$\boxed{\int_0^1\frac{\ln(1+x^2)\ln x}{1+x^2}\ dx=\frac3{32}\pi^3+\frac{\pi}8\ln^22-G\ln2+2\text{Im}\operatorname{Li_3}(1-i)=S_2}$$
Plugging $S_1$ and $S_2$ in $(1)$, we get $$\color{blue}{S=\frac{\pi^3}{32}-2G\ln2}$$
note that we used:
$$\ln(i)=\frac{\pi}{2}i$$
$$\ln(1-i)=\frac12\ln2-\frac{\pi}{4}i$$
$$\operatorname{Li_2}(1-i)=\frac{\pi^2}{16}-\left(\frac{\pi}{4}\ln2+G\right)i$$
which give us:
$$\ln(i)\ln^2(1-i)=\frac{\pi^2}{8}\ln2-\left(\frac{\pi^3}{32}-\frac{\pi}{8}\ln^22\right)i$$
$$\ln(1-i)\operatorname{Li_2}(1-i) =-\frac{\pi}{4}G-\frac{\pi^2}{32}\ln2-\left(\frac12\ln2G+\frac{\pi^3}{64}+\frac{\pi}{8}\ln^22\right)i$$
