Compute $\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}}{(2n+1)^3}$ and $\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}$ How to prove  

$$S_1=\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}}{(2n+1)^3}=1+\frac{35}{128}\pi\zeta(3)+\frac{1}{48}\zeta(4)-\frac1{384}\psi^{(3)}\left(\frac14\right)$$
$$S_2=\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}=1+\frac18G\zeta(2)-\frac{35}{64}\pi\zeta(3)-\frac{15}{16}\zeta(4)+\frac1{768}\psi^{(3)}\left(\frac14\right)$$
  where $H_n=\sum_{n=1}^\infty\frac1n$ is the $n$th harmonic number, $G$ denotes the Catalan's constant, $\zeta$ denotes the Riemman Zeta function and $\psi^{(n)}$ designates the Polygamma function.

These two sums were proposed by Cornel and can be found here and here . My solution to $S_1$ can be found in the first link but its long, so can we find a better way to find $S_1$ and $S_2$ ?
Thanks.

Note: Using the generating function of $\ \sum_{n=1}^\infty x^n\frac{H_n}{n^3}$ to evaluate $S_1$ is not allowed.

Different approach to evaluate $S_1$:
From here we have
$$I=\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\frac{\pi^3}{16}\ln2-\frac{7\pi}{64}\zeta(3)-\frac{\pi^4}{96}+\frac1{768}\psi^{(3)}\left(\frac14\right)\tag1$$
On the other hand 
$$I=\int_0^1 \frac{\ln^2x\arctan x}{x}\ dx-\int_0^1 \frac{x\ln^2x\arctan x}{1+x^2}\ dx$$
For the first integral, use $\arctan x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}$ and for the second integral, use  the identity $\frac{\arctan x}{1+x^2}=\frac12\sum_{n=0}^\infty(-1)^n\left(H_n-2H_{2n}\right)x^{2n-1}$ we have
$$I=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln^2x\ dx-\frac12\sum_{n=0}^\infty(-1)^n(H_n-2H_{2n})\int_0^1x^{2n}\ln^2x\ dx$$
$$=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}-\sum_{n=0}^\infty(-1)^n\frac{H_n-2H_{2n}}{(2n+1)^3}$$
$$=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}-\sum_{n=0}^\infty\frac{(-1)^nH_n}{(2n+1)^3}+2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^3},\quad H_{2n}=H_{2n+1}-\frac{1}{2n+1}$$
$$=\sum_{n=0}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}+2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^3}\tag2$$
Combine $(1)$ and $(2)$ and substitute
$$\sum_{n=0}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$
we obtain that 
$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac1{384}\psi^{(3)}\left(\frac14\right)-\frac{1}{48}\pi^4-\frac{35}{128}\pi\zeta(3)$$
