Nice pair of trilog integrals $\int_0^z \frac{\log ^2(x) \log (1\pm x)}{1\mp x} \, dx$ Recently, in the wake of the solution of Compute $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$, I stumbled on this symmetric pair of integrals
$$i_{\pm}(z) = \int_0^z \frac{\log ^2(x) \log (1\pm x)}{1\mp x} \, dx$$
I tried several integrations by part and a series expansion but I could not solve them. 
(a) Can you do better?
(b) A slightly easier version asks for the case $z=1$. 
Here we have numerically $i_{+}(1) = 0.345691, i_{-}(1) = -0.235752$
Remark: the integrals would be easy if the denominator were identical with the argument of the logarithm.
 A: I am going to solve $(b)$
Rgarding $i_+(1)$, use 
$$\frac{\ln(1+x)}{1-x}=\sum_{n=1}^\infty \overline{H}_nx^n\tag1$$
multiply both sides by $\frac{\ln^2x}{x}$ then $\int_0^1$ and use the fact that $\int_0^1 x^{n-1}\ln^2x\ dx=\frac{2}{n^3}$ we get
$$2\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}=\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\ dx+\int_0^1\frac{\ln^2x\ln(1+x)}{x}\ dx$$
where 
$$\int_0^1\frac{\ln^2x\ln(1+x)}{x}\ dx=-\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^1 x^{n-1}\ln^2(x)\ dx=-2\sum_{n=1}^\infty\frac{(-1)^n}{n^4}=\frac74\zeta(4)$$
and $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}$ can be found using the generalization 
$$\sum_{k = 1}^\infty \frac{\overline H_k}{k^m} = \zeta (m) \log 2 - \frac{1}{2} m \zeta (m + 1) + \eta (m + 1) + \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1).$$
set $m=3$
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}=\frac74\ln2\zeta(3)-\frac5{16}\zeta(4)$$
Also I managed to calculate it here in a different way ( check the bonus).
Combine the two results we get
$$i_+(1)=\frac72\ln2\zeta(3)-\frac{19}{8}\zeta(4)$$
To find $i_-(1)$, just replace $x$ with $-x$ in $(1)$ and follow the same process coming across $\sum_{n=1}^\infty\frac{(-1)^n\overline{H}_n}{n^3}$ which is manageable using parity, but there is a nicer way
$$i_+(1)-i_-(1)=\int_0^1\ln^2(x)\left(\frac{\ln(1+x)}{1-x}-\frac{\ln(1-x)}{1+x}\right)\ dx$$ 
$$\overset{IBP}{=}\underbrace{-\ln^2(x)\ln(1-x)\ln(1+x)\bigg|_0^1}_{0}+2\int_0^1\frac{\ln x\ln(1-x)\ln(1+x)}{x}\ dx$$
the last integral can be done using the algebraic identity $4ab=(a+b)^2-(a-b)^2$ where $a=\ln(1-x)$ and $b=\ln(1+x)$ and you can find different solutions here
$$\int_0^1\frac{\ln x\ln(1-x)\ln(1+x)}{x}=2\operatorname{Li}_4\left(\frac12\right)-\frac12\ln^22\zeta(2)+\frac74\ln2\zeta(3)-\frac{27}{16}\zeta(4)+\frac1{12}\ln^42$$
$$\Longrightarrow i_-(1)=\zeta(4)+\ln^22\zeta(2)-4\operatorname{Li}_4\left(\frac12\right)-\frac1{6}\ln^42$$
A: Partial answer (I will examine just one of the cases and applied for $z = 1$).
I will analyse the following:
$$\int_0^1 \frac{\log^2(x)\log(1+x)}{1-x}\ \text{d}x$$
Let's start with $x \to e^z$ which transforms the integral into
$$\int_{-\infty}^0 \frac{z^2 e^z}{1-e^z}\log(1+e^z)\ \text{d}z$$
Use the Geometric series for
$$\frac{1}{1-e^z} = \sum_{k\in\mathbb{W}} e^{zk}$$
Where $\mathbb{W} = \mathbb{N} + \{0\}$.
Whence
$$\sum_{k\in\mathbb{W}}\int_{-\infty}^0 z^2 e^{z(k+1)}\log(1 + e^z)\ \text{d}z$$
We can now use the integration by parts with the following choice
$$f'(z) = z^2 e^{z(k+1)}$$
$$g(z) = \log(1 + e^z)$$
From which 
$$f(z) = \frac{e^{(k+1) z} \left((k+1)^2 z^2-2 (k+1) z+2\right)}{(k+1)^3}$$
$$g'(z) = \frac{e^z}{1+e^z}$$
Hence
$$\sum_{k\in\mathbb{W}}\left[\left(\frac{e^{(k+1) z} \left((k+1)^2 z^2-2 (k+1) z+2\right)}{(k+1)^3}\right) \cdot \log(1 + e^z)\bigg|_{-\infty}^0  - \left(\frac{e^{(k+1) z} \left((k+1)^2 z^2-2 (k+1) z+2\right)}{(k+1)^3}\right)\cdot \frac{e^z}{1+e^z}\bigg|_{-\infty}^0 \right]$$
Computing the two limits is rather easy, and we ge
$$\sum_{k\in\mathbb{W}} \left(\frac{2\log(2)}{(1+k)^3}  - \frac{1}{(1+k)^3}\right)$$
$$(\log(4) - 1)\sum_{k\in\mathbb{W}} \frac{1}{(1+k)^3}$$
The last sum is very well known, it's the Riemann Zeta function of three.
$$ \to (\log(4) - 1)\zeta(3)$$
Warning
The numerical value in this case is $\approx 0.464348(...)$ which is different from the true numerical one. I suspect I have made some error somewhere, hence I just wrote this down in order to read it clearly, and I will check it better later!
A: Partial self answer for $z=1$
Starting with the idea of @Mycroft I have transformed the remaining problem into the calculation of the following Euler sums
$$\sum_{k=0}^{\infty} \left\{\frac{H_{\frac{k-1}{2}}^{(2)}-H_{\frac{k}{2}}^{(2)}}{2 (k+1)^2},\frac{H_{\frac{k-1}{2}}^{(3)}-H_{\frac{k}{2}}^{(3)}}{4 (k+1)},\frac{H_{\frac{k-1}{2}}-H_{\frac{k}{2}}}{(k+1)^3}\right\}$$
