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).

The series involving $H_k$ and $H_{2k}$ can be studied in a similar way: since
$$ \frac{-\log(1-x)}{1-x} = \sum_{n\geq 1} H_n x^{n} $$
we have $ \frac{-\log(1+x^2)}{1+x^2} = \sum_{n\geq 1} H_n(-1)^n x^{2n} $ and
$$ \sum_{k\geq 1}\frac{(-1)^k}{(2k+1)^2}H_k = \int_{0}^{1}\frac{\log(1+x^2)\log(x)}{1+x^2}\,dx$$
boils down to
$$ \int_{0}^{\pi/4} -2\log(\cos\theta) \log(\tan\theta)\,d\theta $$
which is simple to tackle through well-known Fourier series. It equals 
$$ -\frac{\pi^3}{64}-K\log(2)-\frac{\pi}{16}\log^2(2)+2\,\text{Im}\,\text{Li}_3\left(\frac{1+i}{2}\right)\approx -0.07355395672853217. $$
