Computing $\sum\limits_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}$ 
How to evaluate the following challenging sum:
  $$S=\sum_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}=\frac74\zeta(3)+\frac3{32}\pi^3-\frac{\pi}{2}G-2\ln2G+\frac{\pi}{8}\ln^22-2\Im\operatorname{Li}_3(1+i)?$$

where $H_{n}$  is the harmonic number, $\operatorname{Li}_n$ is the polylogarithm function and $G$ is Catalan constant. 

This problem was proposed by a friend and here is my approach:
Using the identity 
$$\int_0^1\frac{x^n}{1+x}\ dx=H_{n/2}-H_n+\ln2, \quad x\mapsto x^2$$
$$2\int_0^1\frac{x^{2n+1}}{1+x^2}\ dx=H_{n/2}-H_n+\ln2$$
Multiply both sides by $\frac{(-1)^n}{(2n+1)^2}$ to get 
$$\sum_{n=0}^\infty (-1)^n\frac{H_{n/2}-H_n+\ln2}{(2n+1)^2}=2\int_0^1\frac{1}{1+x^2}\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2}\ dx\\=2\int_0^1\frac{1}{1+x^2}\Im\sum_{n=0}^\infty\frac{(i)^nx^n}{n^2}\ dx=2\Im\int_0^1\frac{\operatorname{Li}_2(ix)}{1+x^2}\ dx$$
Rearranging the terms to get
$$S=\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}-\ln2G+2\Im\int_0^1\frac{\operatorname{Li}_2(ix)}{1+x^2}\ dx$$
where 
$$\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}=-\int_0^1\ln x\sum_{n=0}^\infty(-1)^n x^{2n}H_n\ dx=\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx$$
which is calculated here:
$$\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx=\frac3{32}\pi^3+\frac{\pi}8\ln^22-\ln2~G-2\text{Im}\operatorname{Li_3}(1+i)$$

And the question here is how to evaluate the dilogarithm integral or a different way to compute $S$? Thank you.
 A: We have that:$$\Im\operatorname{Li}_2 (ix)=\operatorname{Ti}_2 (x)=\int_0^x \frac{\arctan t}{t}dt=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)^2}$$
So let's rewrite the integral in terms of the inverse tangent function:$$\int_0^1\frac{1}{1+x^2}\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2}dx=\int_0^1 \frac{\operatorname{Ti}_2(x)}{1+x^2}dx$$
$$\overset{IBP}=\arctan x\operatorname{Ti}_2(x)\bigg|_0^1-\int_0^1 \frac{\arctan^2 x}{x}dx=\frac{\pi}{4}G-\mathcal J$$
$$\mathcal J=\int_0^1\frac{\arctan^2 x}{x}dx\overset{x=\tan t}=2\int_0^\frac{\pi}{4} \frac{t^2}{\sin(2t)}dt\overset{2t=x}=\frac14\int_0^\frac{\pi}{2}\frac{x^2}{\sin x}dx=\frac{\pi}{2}G-\frac{7}{8}\zeta(3)$$
The last integral is nicely evaluated here.
$$\Rightarrow \boxed{\Im\int_0^1 \frac{\operatorname{Li}_2(ix)}{1+x^2}dx=\int_0^1 \frac{\operatorname{Ti}_2(x)}{1+x^2}dx=\frac{7}{8}\zeta(3)-\frac{\pi}{4}G}$$
A: From here we got
$$X=2\sum_{n=0}^\infty\frac{H_{n/2}-H_n+\ln2}{(2n+1)^3}=\frac{\pi^2}{4}G-\int_0^{\pi/2}\frac{x^3}{2\sin x}\ dx$$
Using the identity in (4) 
$$\sum_{n=0}^\infty(H_{n/2}-H_n+\ln2)\sin(x(2n+1))=\frac{\pi/2-x}{2\cos x}$$
multiply both sides by $x^2$ then integrate from $x=0$ to $\pi/2$ we get
\begin{align}
I&=\int_0^{\pi/2}\frac{(\pi/2-x)x^2}{2\cos x}\ dx=\sum_{n=0}^\infty(H_{n/2}-H_n+\ln2)\int_0^{\pi/2} x^2 \sin(x(2n+1))\ dx\\
&=\sum_{n=0}^\infty(H_{n/2}-H_n+\ln2)\left(\frac{\pi(-1)^n}{(2n+1)^2}-\frac{2}{(2n+1)^3}\right)\\
&=\pi\sum_{n=0}^\infty\frac{H_{n/2}}{(2n+1)^2}-\pi\sum_{n=0}^\infty\frac{H_{n}}{(2n+1)^2}+\pi\ln2G-X\\
&=\pi\sum_{n=0}^\infty\frac{H_{n/2}}{(2n+1)^2}-\pi\sum_{n=0}^\infty\frac{H_{n}}{(2n+1)^2}+\pi\ln2G-\frac{\pi^2}{4}G+\int_0^{\pi/2}\frac{x^3}{2\sin x}\ dx
\end{align}
But 
\begin{align}
I&=\int_0^{\pi/2}\frac{(\pi/2-x)x^2}{2\cos x}\ dx=\int_0^{\pi/2}\frac{x(\pi/2-x)^2}{2\sin x}\ dx\\
&=\frac{\pi^2}{4}\underbrace{\int_0^{\pi/2}\frac{x}{2\sin x}\ dx}_{G}-\pi\underbrace{\int_0^{\pi/2}\frac{x^2}{2\sin x}\ dx}_{\pi G-\frac74\zeta(3)}+\int_0^{\pi/2}\frac{x^3}{2\sin x}\ dx\\
&=\frac74\pi\zeta(3)-\frac34\pi^2G+\int_0^{\pi/2}\frac{x^3}{2\sin x}\ dx
\end{align}
Rearrange and divide by $\pi$ to get
$$\sum_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}=\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}+\frac74\zeta(3)-\frac{\pi}{2}G-\ln2G$$
Substitute the result of $\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}$ calculated in the question body, we get the desired closed form.
