What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n\ ?$ Recently, the generating function of order 2 for the alternating harmonic series was calculated (What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$).
I would like take the next step to the order 3. Is there a closed form function here as well?
Defining $\overline{H}_n = \sum_{k=1}^{n} (-1)^{k+1}/k$ and 
$$g_q(z) = \sum_{n=0}^{\infty}\frac{z^n}{n^q} \overline{H}_n\tag{1}$$ 
I ask the
Question
Can you calculate the function defined by the sum $g_3(z)$, i.e. express it through known functions?
My effort so far
The alternating harmonic series has an integral representation
$$\overline{H}_n= \int_0^1 \frac{1-(-1)^n x^n}{x+1} \, dx\tag{2}$$
Hence we can easily calculate its generating function
$$g_0(z) = \sum_{n=1}^{\infty} \overline{H}_n z^n = \int_0^1 \left(\sum _{n=1}^{\infty } \frac{\left(1-(-1)^n x^n\right) z^n}{x+1}\right) \, dx\\=\int_0^1 \frac{z}{(1-z) (x z+1)} \, dx=\frac{\log (z+1)}{1-z}\tag{3}$$ 
The next orders can of $g(z)$  be generated successively by dividing by $z$ and integrating, i.e.
$$g_{q+1}(z)=\int_0^z \frac{ g_{q}(z)}{z}\,dz, q=0,1,2,...\tag{4}$$
Because $g_2$ is known we could just plug it into $(4)$ and integrate. The problem is, however, that $g_2$ already consists of about 20 summands, and hence in the first place there are about 20 integrals to calculate. In view of the exploding number of possible transformations (substitution, partial integration, utilizing relations between the polylog functions involved etc.) it is highly desirabable to keep the number of integrals to be "cracked" as small as possible.
For the case of order 3 I have now boiled it down to just one (!) integral having derived (by partial integration) this formula
$$g_3(z) = g_2(z) \log(z) -\frac{1}{2} g_1(z) \log(z)^2 +\frac{1}{2}i(z)\tag{5}$$
where the remaining integral is
$$i(z) = \int_0^z \frac{\log(t)^2 \log(1+t)} {t(1-t)} \,dt\tag{6}$$
The integral is convergent in the range $0<z<1$. In fact, the integrand has the expansion
$$\frac{\log(t)^2 \log(1+t)} {t(1-t)} \underset{t\to0}\simeq \log ^2(t)\left(1+\frac{t}{2}+\frac{5 t^2}{6}+ O(t^3)\right) $$
and close to $z=1$ as well
$$\frac{\log(t)^2 \log(1+t)} {t(1-t)} \underset{t\to1}\simeq (1 - t)\log(2) + (1 - t)^2 \left(-\frac{1}{2} + 2 \log(2)\right)+O((1 - t)^3)$$
For $z<0$ the integral becomes complex as we can see from the example case
$$\int_0^{-\frac{1}{2}} \log ^2(t) \, dt=\frac{1}{2} \left(\pi ^2-2-\log ^2(2)+2 i \pi  (1+\log (2))-\log (4)\right)$$
This is surprising because the original power series for the generating function $(1)$ is convergent for $|z|<1$ and hence defines a real function. The problem is resolved considering that the complete expression containes other terms which (in some way) compensate the singularities.
Here I'm stuck and I was not able to solve the integral. But as I know that there are many experienced and skilled experts in this forum I'm confident they can solve $i(z)$.
Discussion
I have moved the text to https://math.stackexchange.com/a/3544006/198592
 A: Incomplete solution
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n=x+\sum_{n=2}^\infty\frac{\overline{H}_n}{n^3}x^n$$
By using  
$$\sum_{n=2}^\infty f(n)=\sum_{n=1}^\infty f(2n+1)+\sum_{n=1}^\infty f(2n)$$
we have 
$$\Longrightarrow \sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n=x+\sum_{n=1}^\infty\frac{\overline{H}_{2n+1}}{(2n+1)^3}x^{2n+1}+\sum_{n=1}^\infty\frac{\overline{H}_{2n}}{(2n)^3}x^{2n}$$
now use $$\overline{H}_{2n+1}=H_{2n+1}-H_n, \quad \overline{H}_{2n}=H_{2n}-H_n$$
We get that 
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n=\color{blue}{x+\sum_{n=1}^\infty\frac{H_{2n+1}}{(2n+1)^3}x^{2n+1}+\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}x^{2n}}-\frac18\sum_{n=1}^\infty\frac{H_n}{n^3}x^{2n}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}x^{2n+1}$$
$$=\color{blue}{\sum_{n=1}^\infty\frac{H_n}{n^3}x^n}-\frac18\sum_{n=1}^\infty\frac{H_n}{n^3}x^{2n}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}x^{2n+1}$$
The first sum is already calculated here and the second sum is the same as the the first one but just replace $x$ with $x^2$. The last sum seems annoying but I will give it a try.
