On the alternating quadratic Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$ My question is:

Can a closed-form expression for the following alternating quadratic Euler sum be found? Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$.
  $$S = \sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$$

What I have managed to do so far is to convert $S$ to two rather difficult integrals as follows.
Starting with the result
$$\frac{H_{2n}}{2n} = -\int_0^1 x^{2n - 1} \ln (1 - x) \, dx \tag1$$
Multiplying (1) by $(-1)^n H_n/n$ then summing the result from $n = 1$ to $\infty$ gives
$$S = -2 \int_0^1 \frac{\ln (1 - x)}{x} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n}. \tag2$$
From the following generating function for the harmonic numbers
$$\sum_{n = 1}^\infty \frac{H_n x^n}{n} = \frac{1}{2} \ln^2 (1 - x) + \operatorname{Li}_2 (x),$$
replacing $x$ with $-x^2$ leads to
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n} = \frac{1}{2} \ln^2 (1 + x^2) + \operatorname{Li}_2 (-x^2).$$
Substituting this result into (2) yields 
$$S = -2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx,$$
or, after integrating the first of the integrals by parts twice
$$S = -\frac{5}{2} \zeta (4) + 4 \zeta (3) \ln 2 - 8 \int_0^1 \frac{x \operatorname{Li}_3 (x)}{1 + x^2} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx. \tag3$$
I have a slim hope the first of these integrals can be found (I cannot find it). As for the second of the integrals, it is proving to be a little difficult. 
Can someone find each of the integrals appearing in (3)? Or perhaps an alternative approach to the sum will deliver the closed-form I seek, I am fine either way. 

Update
Thanks to Ali Shather, the first of the integrals can be found. Here
\begin{align}
\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \ dx &=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{2n-1}\ln(1-x)\ dx\\
&= -\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n^3}\\
&=-4\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^3}\\
&=-4 \operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3}.
\end{align}
And using the result I calculated here, namely
$$\operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3} = \frac{5}{8} \operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{195}{256} \zeta (4) + \frac{5}{192} \ln^4 2 - \frac{5}{32} \zeta (2) \ln^2 2 + \frac{35}{64} \zeta (3) \ln 2,$$
gives
\begin{align}
\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx &= -\frac{5}{2} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{195}{64} \zeta (4) - \frac{5}{48} \ln^4 2\\
& \qquad + \frac{5}{8} \zeta (2) \ln^2 2 - \frac{35}{16} \zeta (3) \ln 2.
\end{align}
 A: Using your integral representation, the sum equals to: $$\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}= -2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx$$ $$\small=-2 C^2+2 \pi  C \log (2)-4 \pi  \Im(\text{Li}_3(1+i))+3 \text{Li}_4\left(\frac{1}{2}\right)+\frac{21}{8} \zeta (3) \log (2)+\frac{487 \pi ^4}{5760}+\frac{\log ^4(2)}{8}+\frac{1}{8} \pi ^2 \log ^2(2)$$
For the second integral and its derivation, see here.
A: Remark: I have noticed too late that this integral has already been solved (in the update of omegadot).
Nevertheless I don't delete the contribution because, together with this information, it shows that the hypergeometric functions appearing here can be simplified appreciably which gives hope for other cases.
Original post
A closed expression of the integral
$$i = \int_0^1 \frac{x \operatorname{Li}_3(x)}{x^2+1}\tag{1}$$
can be found in terms of (sorry Ali) hypergeometric function as follows.
Partial integration gives
$$i=s_{0}-\int_0^1 \frac{\text{Li}_2(x) \log \left(x^2+1\right)}{2 x} \, dx\tag{2a}$$
where 
$$s_0 = \frac{1}{2} \zeta (3) \log (2)\tag{2b}$$
Expanding the denominator of the integrand we find that $i=s_{0}+\sum a_{k}$ where
$$a_{k} =-\frac{1}{2} \int_0^1 \frac{(-1)^{k+1} x^{2 k-1} \text{Li}_2(x)}{k} \, dx=-\frac{(-1)^{k+1} \left(\pi ^2 k-3 H_{2 k}\right)}{24 k^3}\tag{3}$$
The two sums are
$$s_{1}=\frac{1}{24} \left(-\pi ^2\right) \sum _{k=1}^{\infty } \frac{(-1)^{k+1}}{k^2}=-\frac{\pi ^4}{288}\tag{4}$$
$$s_{2} = +\frac{1}{8} \sum _{k=1}^{\infty } \frac{(-1)^{k+1} H_{2 k}}{k^3}=\frac{1}{32} \left(-2 \,_P\tilde{F}_Q^{(\{0,0,0,0\},\{0,0,1\},0)}(\{1,1,1,1\},\{2,2,2\},-1)\\-\sqrt{\pi } \,_P\tilde{F}_Q^{(\{0,0,0,0,0\},\{0,0,0,1\},0)}\left(\left\{1,1,1,1,\frac{3}{2}\right\},\left\{2,2,2,\frac{3}{2}\right\},-1\right)\\+3 \zeta (3) (\gamma +\log (2))\right)\tag{5}$$
Where $\,_P\tilde{F}_Q$ is the regularized hypergeometric function. For more details see https://math.stackexchange.com/a/3544006/198592.
Two terms appear in $s_{2}$ due to the relation
$$H_{2 k}=\frac{1}{2} \left( H_{k-\frac{1}{2}}+ H_k \right)+\log (2)$$
The complete integral is then given by
$$i = s_{0}+s_{1}+s_{2}$$
The numeric check shows good agreement.
Discussion
I'm almost sure that the sum
$$\sum _{k=1}^{\infty } \frac{(-1)^{k+1} H_k}{k^3}$$
has a simpler expression, and so might
$$\sum _{k=1}^{\infty } \frac{(-1)^{k+1} H_{k-\frac{1}{2}}}{k^3}$$
and I'd be happy to replace the hypergeometric constructs.
No need to conjecture: omegadot has done it, see https://math.stackexchange.com/a/3290607/198592
A: A Second (Magical) Solution by Cornel Ioan Valean
To get a different solution, we start from Dan Fulea's integral in this post where Cornel provided with an extremely simple solution, and by simply rearranging it, we have that
$$\frac{\pi^4}{64}$$
$$=\frac{\pi^2}{4}\underbrace{\int_0^1\frac{\operatorname{arctanh}(t)}{t}\textrm{d}t}_{\displaystyle \pi^2/8}-\pi \int_0^1\frac{\arctan(t)\operatorname{arctanh}(t)}{t}\textrm{d}t+\int_0^1\frac{\arctan^2(t)\operatorname{arctanh}(t)}{t}\textrm{d}t,$$
and since the middle integral is provided in the first Cornel's solution and the transformation from the last integral to series is again shown in the first Cornel's solution together with the values of the resulting auxiliary series, we arrive at the desired value of the series,
$$\color{red}{\sum _{n=1}^{\infty} (-1)^{n-1}\frac{ H_n H_{2 n}}{n^2}}$$
$$\color{red}{=2 G^2-2\log(2)\pi G-\frac{1}{8}\log^4(2)-\frac{21}{8}\log(2)\zeta(3)+\frac{1}{4}\log^2(2)\pi ^2+\frac{773}{5760}\pi ^4}$$
$$\color{red}{-4 \pi  \Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}-3 \operatorname{Li}_4\left(\frac{1}{2}\right)},$$
and the solution is complete.
End of story
A note: The connection with Dan Fulea's integral was observed later, and this latter way is definitely more wonderful to consider.
A: A Third (Magical) Solution by Cornel Ioan Valean
This time we assume we have in hand that $\displaystyle \int_0^1 \frac{\arctan^2(x)}{x}\log\left(\frac{x}{(1-x)^2}\right)=G^2$, which is calculated extremely easily at this link. Furthermore, we also assume we have  at our disposal the value of the powerful harmonic series $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{n^3}$ which is derived in a simple manner in this answer. Now, as seen in this post, the integral is reducible to the previously mentioned harmonic series and $\color{red}{\text{the main harmonic series}}$.
Using the  simple facts stated above, we conclude that
$$\color{red}{\sum _{n=1}^{\infty} (-1)^{n-1}\frac{ H_n H_{2 n}}{n^2}}$$
$$\color{red}{=2 G^2-2\log(2)\pi G-\frac{1}{8}\log^4(2)-\frac{21}{8}\log(2)\zeta(3)+\frac{1}{4}\log^2(2)\pi ^2+\frac{773}{5760}\pi ^4}$$
$$\color{red}{-4 \pi  \Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}-3 \operatorname{Li}_4\left(\frac{1}{2}\right)},$$
and we are done!
