Two unequal trilog integral twins $\int_0^1 \log^2(x)\frac{ \log(1\pm \sigma x)}{1\mp x}\,dx$ Trying to generalize this problem [1] I came up with these two integrals
$$i_{\pm}(\sigma) = \int_0^1 \log^2(x)\frac{ \log(1\pm \sigma x)}{1\pm x}\,dx\tag{1}$$
Here $-1 \le \sigma \le 1$ is a parameter. For $\sigma = \pm 1$ we are led back to the problem [1]. 
Although these two integrals look like twins they are quite different. 
Notice that the essential difference is in the denominator; under the $\log$ the sign change can be accomodated in $\sigma$.
While I was able to solve $i_{-}(\sigma)$ with the methods generalizing the solution to [1] I find $i_{+}(\sigma)$ tough.
Can you find to the closed forms of both integrals?
What I did so far:
The solution is based of the concept of the unified harmonic number $U_n$ I have introduced earlier [2] as a generalization of the harmonic number and the alternating harmonic number.
It is defined as
$$U_n(\sigma) = \sum_{k=1}^n \frac{\sigma^k}{k} = \sigma \int_0^1 \frac{1-(\sigma z)^n}{1-\sigma z}\,dz\tag{2}$$
The generating function is 
$$g_{-}(x,\sigma) = \sum_{k=1}^\infty x^n U_n (\sigma) = -\frac{\log(1-\sigma x)}{1-x}\tag{3}$$
Here we recognize a part of the integrand of $i_{-}(\sigma)$.
Hence multiplying $(3)$ with $\log(x)^2$ and integrating from $x=0$ to $x=1$ we have
$$-i_{-}(\sigma) = \sum_{n=1}^\infty U_n(\sigma) \int_0^1  x^n \log(x)^2\,dx 
\\= \sum_{k=1}^\infty U_n(\sigma) \frac{2}{(1+n)^3}
\\=\sigma \int_0^1 \sum_{n=1}^\infty  \frac{1-(\sigma z)^n}{1-\sigma z}\frac{2}{(1+n)^3}\,dz\tag{4}$$
$$=2\sigma \int_0^1\frac{ \sigma  z \zeta (3)-\text{Li}_3(z \sigma )}{z (1-\sigma  z)}\,dz\tag{5}$$
$$-i_{-}(\sigma)= \operatorname{Li}_2(\sigma ){}^2+2 \operatorname{Li}_3(\sigma ) \log (1-\sigma )\\
-2 (\operatorname {Li}_4(\sigma )+\zeta (3) \log (1-\sigma )) \tag{6}$$
Repeating the same for $i_{+}$ we have now
$$g_{+}(x,\sigma) = \sum_{k=1}^\infty (-1)^n x^n U_n (\sigma) = -\frac{\log(1+\sigma x)}{1+x}\tag{3a}$$
Multiplying $(3a)$ with $\log(x)^2$ and integrating from $x=0$ to $x=1$ we have
$$-i_{+}(\sigma) = \sum_{n=1}^\infty (-1)^n U_n(\sigma) \int_0^1  x^n \log(x)^2\,dx 
\\= \sum_{k=1}^\infty (-1)^n U_n(\sigma) \frac{2}{(1+n)^3}
\\=\sigma \int_0^1 \sum_{n=1}^\infty (-1)^n \frac{1-(\sigma z)^n}{1-\sigma z}\frac{2}{(1+n)^3}\,dz\tag{4a}$$
$$=-2\sigma \int_0^1\frac{4 \operatorname{Li}_3(-z \sigma )+3 \sigma  z \zeta (3)}{2 z (1-\sigma  z)}\,dz\tag{5a}$$
and here I am stuck.
While $(5)$ could be easily solved $(5a)$ is resistant.
My blocking point is this integral
$$\int_0^s \frac{\operatorname{Li}_3(t)}{1+t} \, dt\tag{7}$$
Ali Shather kindly pointed out to me that the integral for $s=1$ is well known, and he provided the useful hint that expanding the denominator gives
$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x} \, dt=\sum_{n=1}^{\infty} (-1)^{n-1}\int_0^1 x^{n-1}\operatorname{Li}_3(x)\,dx
\\= \sum_{n=1}^{\infty} (-1)^{n-1}\left(\frac{ \zeta(3)}{ n} -\frac{\zeta(2)}{ n^2}+\frac{H_{n}}{n^3}\right)\tag{8}$$
so that, finally borrowing the result for $\sum_{k=1}^{\infty}(-1)^{k+1} \frac{H_{k}}{k^3}$ from Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$ I find
$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x} \, dx=
-2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{3}{4} \zeta (3) \log (2)+\frac{\pi ^4}{60}-\frac{1}{12} \log ^4(2)+\frac{1}{12} \pi ^2 \log ^2(2)\simeq 0.339545\tag{9}$$
But, unfortunately, this does not solve the problem in question here which requires the integral $(7)$ as a function of $s$.
References
[1] Nice pair of trilog integrals $\int_0^z \frac{\log ^2(x) \log (1\pm x)}{1\mp x} \, dx$
[2] How to calculate the generating function of the unified harmonic sum $U(\sigma,n,p) = \sum_{k=1}^n \frac{\sigma^k}{k^p}$?
 A: Start with writing 
$$\text{Li}_3(x)=\frac14\text{Li}_3(x^2)-\text{Li}_3(-x)$$
$$\int\frac{\text{Li}_3(x)}{1+x}\ dx=\frac14\int\frac{\text{Li}_3(x^2)}{1+x}\ dx-\int\frac{\text{Li}_3(-x)}{1+x}\ dx=\frac14I_1-I_2$$
For $I_2$, apply integration by parts twice we get 
$$I_2=\ln(1+x)\text{Li}_3(-x)+\frac12\text{Li}_2^2(-x)$$
For $I_1$
$$I_1=\int\frac{\text{Li}_3(x^2)}{1+x}\ dx=\int\frac{(1-x)\text{Li}_3(x^2)}{1-x^2}\ dx$$
$$=\int\frac{\text{Li}_3(x^2)}{1-x^2}\ dx-\int\frac{x\text{Li}_3(x^2)}{1-x^2}\ dx=I_1'-I_1''$$
where
$$I_1''=\int\frac{x\text{Li}_3(x^2)}{1-x^2}\ dx\overset{x^2=y}{=}\frac12\int\frac{\text{Li}_3(y)}{1-y}\ dy$$
$$\overset{IBP}{=}-\frac12\ln(1-y)\text{Li}_3(y)-\frac14\text{Li}_2^2(y)$$
$$=-\frac12\ln(1-x^2)\text{Li}_3(x^2)-\frac14\text{Li}_2^2(x^2)$$
For $I_1'$ we can simply use the genrating function 
$$\sum_{n=1}^\infty H_n^{(a)}x^n=\frac{\text{Li}_a(x)}{1-x}$$
Set $a=3$ and replace $x$ with $x^2$ we get that 
$$I_1'=\int\frac{\text{Li}_3(x^2)}{1-x^2}\ dx=\sum_{n=1}^\infty H_n^{(3)}\int x^{2n}\ dx=\sum_{n=1}^\infty \frac{H_n^{(3)}}{2n+1}x^{2n+1}$$
Unfortunately I was not able to find the closed form of the latter generating sum. I hope you find this approach helpful. 

I also found that 
$$2\int_0^s\frac{\text{Li}_3(t)}{1+t}\ dt=\ln(1+s)\text{Li}_3(s)+\ln(1-s)\text{Li}_3(-s)+\text{Li}_2(-s)\text{Li}_2(s)+\int_{-s}^s\frac{\text{Li}_3(t)}{1+t}\ dt$$
