Discussion about $\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)}$ May I have some further discussion on
$$\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)} = \frac{\pi^3}{96} - \frac{\pi}{8}\ln^2\!2 \tag1$$
where $H_{n}$ is harmonic-number and $\psi(x)$ is digamma function.
Formerly, I deduce the result by this elementary approach, also the only solution I have due to the asset for me is quite insufficient.
Start with famous series
$$\frac1{2} \arctan x \ln (1+x^2) = \sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} x^{2n+1}}$$
by which somehow I find (1) actually equal to
$$\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)} = \int_{0}^{1} {\frac{\arctan x \ln(1+x^2)}{x(1+x)} \mathrm{d}x}$$
and this integral is not impossible to solve.
Question 1: Any other possible access to summation as (1)?
For digamma function has strong relation with harmonic-number, here $\psi(n+1)-\psi(n+\tfrac1{2})$ can also be substitution for $H_{2n}-H_{n}$ with some other constants. So what if I want a part of (1) like
$$\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}^{2}}{2n+1}} \quad \text{or} \quad \sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}H_{n}}{2n+1}} \tag2$$
but, of course, I have no idea how to deal with such summation involving higher order of harmonic-number.
Question 2: How to find the closed form for summation in (2)?
 A: \begin{align}
S_1&=\sum_{n=0}^\infty\frac{(-1)^{n}H_{2n}^2}{2n+1}=\sum_{n=0}^\infty\frac{(-1)^{n}\left(H_{2n+1}-\frac1{2n+1}\right)^2}{2n+1}\\
&=\sum_{n=0}^\infty\frac{(-1)^{n}H_{2n+1}^2}{2n+1}-2\sum_{n=0}^\infty\frac{(-1)^{n}H_{2n+1}}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^{n}}{(2n+1)^3}
\end{align}
Using the fact that $$\sum_{n=0}^\infty (-1)^n a_{2n+1}=\Im\sum_{n=1}^\infty (i)^n a_n $$
we get 
$$S_1=\Im\left\{\sum_{n=1}^\infty\frac{(i)^{n}H_{n}^2}{n}-2\sum_{n=1}^\infty\frac{(i)^{n}H_{n}}{n^2}+\operatorname{Li}_3(i)\right\}$$
And by using the generating functions:
$$\sum_{n=1}^\infty\frac{x^{n}H_{n}^2}{n}=\operatorname{Li}_3(x)-\ln(1-x)\operatorname{Li}_2(x)-\frac13\ln^3(1-x)$$ 
$$\sum_{n=1}^\infty\frac{x^{n}H_{n}}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$ 
and setting $x=i$ and considering only the imaginary part, we get 

$$\small{S_1=\Im\left\{2\operatorname{Li}_3(1-i)-\ln(1-i)\operatorname{Li}_2(i)-2\ln(1-i)\operatorname{Li}_2(1-i)-\ln(i)\ln^2(1-i)-\frac13\ln^3(1-i)\right\}}$$


Using:
$$\ln(i)=\frac{\pi}{2}i$$
$$\ln(1-i)=\frac12\ln2-\frac{\pi}{4}i$$
$$\operatorname{Li_2}(i)-\frac{\pi^2}{48}+G\ i$$
$$\operatorname{Li_2}(1-i)=\frac{\pi^2}{16}-\left(\frac{\pi}{4}\ln2+G\right)i$$
which give us:
$$\ln(1-i)\operatorname{Li_2}(i) =\frac{\pi}{4}\ G-\frac{\pi^2}{96}\ln2+\left(\frac{\pi^3}{192}+\frac12\ln2\ G\right)i$$
$$\ln(1-i)\operatorname{Li_2}(1-i) =-\frac{\pi}{4}\ G-\frac{\pi^2}{32}\ln2-\left(\frac{\pi^3}{64}+\frac{\pi}{8}\ln^22+\frac12\ln2\ G\right)i$$
$$\ln(i)\ln^2(1-i)=\frac{\pi^2}{8}\ln2-\left(\frac{\pi^3}{32}-\frac{\pi}{8}\ln^22\right)i$$
$$\ln^3(1-i)=\frac18\ln^32-\frac{3\pi^2}{32}\ln2-\left(\frac{\pi^3}{64}-\frac{3\pi}{16}\ln^22\right)i$$
So by combining the imaginary parts of the results above , we get

$$S_1=\sum_{n=0}^\infty\frac{(-1)^{n}H_{2n}^2}{2n+1}=\frac{5}{96}\pi^3+\frac{3\pi}{16}\ln^22+\frac12\ln2\ G+2\Im\left\{\operatorname{Li_3}(1-i)\right\}$$

A: Solution to question 1:
Since you managed to write your sum in (1) as $\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx$, so lets evaluate the integral.

From here we have 
$$\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx=\frac{3\pi}{8}\ln^22-\frac{3\pi^3}{96}\tag{1}$$
and from here we have 
$$\int_0^1\frac{\arctan x}x\ln\left(\frac{(1+x^2)^3}{(1+x)^2}\right)dx=0\tag{2}$$
By combining (1) and (2) we get

$$\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^22$$

.

BONUS:
Since 
$$\psi(n+1)-\psi(n+\tfrac12)=H_n-H_{n-\small{\frac12}}=2H_n-2H_{2n}-2\ln2$$
Then the quality in (1) can be written as
$$\small{\sum_{n=0}^\infty\frac{(-1)^nH_{2n}^2}{2n+1}-\sum_{n=0}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}-\ln2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}=\frac12\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx}$$
Rearranging
$$\small{\sum_{n=0}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}=\sum_{n=0}^\infty\frac{(-1)^nH_{2n}^2}{2n+1}-\ln2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}-\frac12\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx}$$
Lets evaluate the second sum:
\begin{align}
\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}&=\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{2n+1}-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}\\
&=\Im\left\{\sum_{n=1}^\infty\frac{(i)^nH_{n}}{n}-\sum_{n=1}^\infty\frac{(i)^n}{n^2}\right\}\\
&=\Im\left\{\frac12\ln^2(1-i)+\operatorname{Li}_2(i)-\operatorname{Li}_2(i)\right\}\\
&=\Im\left\{\frac12\ln^2(1-i)\right\}\\
&=-\frac{\pi}{8}\ln2
\end{align}
So by plugging this result, along with 
$$\sum_{n=0}^\infty\frac{(-1)^{n}H_{2n}^2}{2n+1}=\frac{5}{96}\pi^3+\frac{3\pi}{16}\ln^22+\frac12\ln2\ G+2\Im\left\{\operatorname{Li_3}(1-i)\right\}$$
and 
$$\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^22$$
we finally get

$$\sum_{n=0}^\infty\frac{(-1)^{n}H_{2n}H_n}{2n+1}=\frac{3}{64}\pi^3+\frac{3\pi}{8}\ln^22+\frac12\ln2\ G+2\Im\left\{\operatorname{Li_3}(1-i)\right\}$$

