<s>[INST] 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}
 [/INST] 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
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