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 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!
