The log integrals $\int_{0}^{1/2} \frac{\log(1+2x) \log(x)}{1+x} \, dx $ and $ \int_{0}^{1/2} \frac{\log(1+2x) \log(1-x)}{1+x} \, dx$

In attempting to evaluate $ \int_{0}^{\infty} [\text{Ei}(-x)]^{4} \, dx$ (which can be evaluated in terms of polylogarithm values), I determined that $$ \begin{align} \int_{0}^{\infty} [\text{Ei}(-x)]^{4} \, dx &= 8 \int_{2}^{\infty} \frac{\log(2+x) \log(x-1)}{x(1+x)} \, dx \\ &=8 \Bigg(\int_{0}^{1/2} \frac{\log(1+2x) \log(1-x)}{1+x} \, dx - \int_{0}^{1/2} \frac{\log(1+2x) \log(x)}{1+x} \, dx \\ &- \int_{0}^{1/2} \frac{\log (x) \log(1-x)}{1+x} \, dx + \int_{0}^{1/2} \frac{\log^{2}(x)}{1+x} \, dx \Bigg). \end{align}$$

I will show this at the end of my post.

I can evaluate the 3rd and 4th integrals, but I'm having difficulties with the first two integrals.

$$ \begin{align} \int_{0}^{1/2} \frac{\log(1+2x)\log(x)}{1+x} \, dx &= \int_{0}^{1/2} \log(x) \sum_{n=1}^{\infty} (-1)^{n+1} x^{n} \sum_{k=1}^{n} \frac{2^{k}}{k} \, dx \\ &= \sum_{n=1}^{\infty} (-1)^{n+1} \sum_{k=1}^{n}\frac{2^{k}}{k} \int_{0}^{1/2} x^{n} \log(x) \, dx \\ &=-\sum_{n=1}^{\infty} \left( -\frac{1}{2} \right)^{n+1} \left[\frac{\log 2}{n+1} + \frac{1}{(n+1)^{2}} \right] \sum_{k=1}^{n}\frac{2^{k}}{k} \end{align}$$

This approach doesn't seem particularly useful.

Approaching the other integral in the same manner leads to even more of a mess:

$$ \begin{align} \int_{0}^{1/2} \frac{\log(1+2x)\log(1-x)}{1+x} \, dx &= \sum_{n=1}^{\infty} (-1)^{n+1} \sum_{k=1}^{n}\frac{2^{k}}{k} \int_{0}^{1/2} x^{n} \log(1-x) \, dx \\& = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1} \left[\log (2) -\left(\frac{1}{2}\right)^{n+1} \log(2) - \sum_{m=1}^{n-1} \frac{(\frac{1}{2})^{m}}{m}\right]\sum_{k=1}^{n}\frac{2^{k}}{k} \end{align}$$

$$ \begin{align} \int_{0}^{\infty} [\text{Ei}(-x)]^{4} \, dx &= -4 \int_{0}^{\infty} [\text{Ei}(-x)]^{3} e^{-x} \, dx \tag{1} \\ &= 4 \int_{0}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{wyz} e^{-(w+y+z+1)x} \, dw \, dy \, dz \,dx \\& =4 \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{wyz} \int_{0}^{\infty} e^{-(w+y+z+1)x} \, dx \, dw \, dy \, dz \\ &= 4 \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{wyz} \frac{1}{w+y+z+1} \, dw \, dy \, dz \\ &= 4 \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \frac{1}{y+z+1} \left(\frac{1}{w} - \frac{1}{w+y+z+1} \right) \, dw \, dy \, dz \\ &= 4 \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \frac{1}{y+z+1} \log(2+y+z) \, dy \, dz \\ &= 8 \int_{1}^{\infty} \int_{1}^{z} \frac{1}{yz} \frac{1}{y+z+1} \log(2+y+z) \, dy \, dz \\ &= 8 \int_{2}^{\infty} \int_{u-1}^{u^{2}/4} \frac{1}{v} \frac{1}{u+1} \log(2+u) \frac{dv \, du}{\sqrt{u^{2}-4v}} \tag{2} \\& =16 \int_{2}^{\infty} \frac{\log(2+u)}{u+1} \int_{0}^{u-2} \frac{1}{u^{2}-t^{2}} \, dt \, du \tag{3} \\& =16 \int_{2}^{\infty} \frac{\log(2+u)}{u+1} \frac{1}{u} \text{arctanh} \left( \frac{u-2}{2}\right) \, du \\ &= 8 \int_{2}^{\infty} \frac{\log(2+u) \log(u-1)}{u(1+u)} \, du \end{align}$$

$(1)$ Integrate by parts.

$(2)$ Make the change of variables $u=y+z$, $v=yz$.

$(3)$ Make the substitution $t^{2} = u^{2}-4v$.

EDIT:

Using M.N.C.E.'s suggestion in the comments, both integrals can be expressed in terms of integrals that can be evaluated using integration by parts.

I posted an answer below.

At Vladimir Reshetnikov's request I'm going to show how to find an antiderivative for $$ \frac{\log(1+2x) \log(1-x)}{1+x} $$ using the identity $$2 \log(x) \log(y) = \log^{2}(x) + \log^{2}(y) - \log^{2} \left(\frac{x}{y} \right) $$ where $x$ and $y$ are positive real values.

$$ \begin{align} &\int \frac{\log(1+2x) \log(1-x)}{1+x} \, dx \\ &= \frac{1}{2} \int \frac{\log^{2}(1+2x)}{1+x} \, dx + \frac{1}{2} \int \frac{\log^{2}(1-x)}{1+x} \, dx - \frac{1}{2} \int \frac{\log^{2} \left(\frac{1+2x}{1-x} \right)}{1+x} \, dx \\ &= \frac{1}{2} \int \frac{\log^{2}(1+2x)}{1+x} \, dx + \frac{1}{2} \int \frac{\log^{2}(1-x)}{1+x} \, dx - \frac{3}{2} \int \frac{\log^{2}(t)}{(2+t)(1+2t)} \ dt \\ &= \frac{1}{2} \int \frac{\log^{2}(1+2x)}{1+x} \, dx + \frac{1}{2} \int \frac{\log^{2}(1-x)}{1+x} \, dx + \frac{1}{2} \int \frac{\log^{2}t}{2+t} \, dt - \int \frac{\log^{2}(t)}{1+ 2t} \ dt \end{align}$$

All 4 indefinite integrals can be evaluated by integrating by parts twice.

For the first one let $u = \log^{2}(1+2x)$ and $dv= \frac{dx}{1+x}$. Then for $v$ choose $\log(2+2x)$.

$$ \begin{align} \int \frac{\log^{2}(1+2x)}{1+x} \, dx &= \log^{2}(1+2x) \log(2+2x) -4 \int \frac{\log(1+2x) \log(2+2x)}{1+2x} \\ &= \log^{2}(1+2x) \log(2+2x) - 2 \int \frac{ \log(1+w) \log(w) }{w} \ dw \\ &= \log^{2}(1+2x) \log(2+2x) -2 \left(-\log(w) \text{Li}_{2} (-w) + \int \frac{\text{Li}_{2}(-w)}{w} \ dt\right) \\ &=\log^{2}(1+2x) \log(2+2x) + 2 \log(1+2x) \text{Li}_{2} (-1-2x) - 2 \text{Li}_{3}(-1-2x) + C \end{align}$$

For the second one let $ u = \log^{2}(1-x)$ and $dv = \frac{dx}{1+x}$. This time choose $\log \left(\frac{1+x}{2} \right)$ for $v$.

$$ \begin{align} \int \frac{\log^{2}(1-x)}{1+x} \, dt &= \log^{2}(1-x) \log \left(\frac{x+1}{2} \right) + 2 \int \frac{\log(1-x) \log \left(\frac{1+x}{2} \right)}{1-x} \ dx \\ &= \log^{2}(1-x) \log \left(\frac{1+x}{2} \right) - 2 \int \frac{\log(w) \log \left(1- \frac{w}{2} \right)}{w} \, dw \\ &= \log^{2}(1-x) \log \left(\frac{1+x}{2} \right) - 2 \left(- \log (w) \text{Li}_{2} \left( \frac{w}{2} \right)+ \int \frac{\text{Li}_{2}(\frac{w}{2})}{w} \right) \ dw \\ &= \log^{2}(1-x) \log \left(\frac{1+x}{2} \right) + 2 \log(1-x) \text{Li}_{2} \left(\frac{1-x}{2} \right) - 2 \text{Li}_{3} \left(\frac{1-x}{2} \right) + C \end{align}$$

The third one can be found at the end of this post.

The fourth one is more straightforward.

$$ \begin{align} \int \frac{\log^{2}(t)}{1+2t} \, dt &= \frac{1}{2} \log^{2}(t) \log(1+2t) - \int \frac{\log(t) \log(1+2t)}{t} \, dt \\ &= \frac{1}{2} \log^{2}(t) \log(1+2t) - \left(- \log(t) \text{Li}_{2}(-2t) + \int \frac{\text{Li}_{2}(-2t)}{t} \, dt \right) \\ &=\frac{1}{2} \log^{2}(t) \log(1+2t) + \log(t) \text{Li}_{2}(-2t) - \text{Li}_{3}(-2t) + C . \end{align}$$

And don't forget that for the 3rd and 4th integrals, $t = \frac{1+2x}{1-x}$.

All 4 antiderivatives above agree with Wolfram Alpha.

If we combine everything, the antiderivative for $\frac{\log(1+2x) \log(1-x)}{1+x}$ appears to be a bit different from the one provided by Wolfram Alpha. This might be because of the identity I used at the beginning.

Plugging in the limits I get

$$ \begin{align} &\int_{0}^{1/2} \frac{\log(1+2x) \log(1-x)}{1+x} \, dx \\ &= -3 \text{Li}_{3}(-2) + 3 \text{Li}_{2}(-2) \log(2) + \frac{\zeta(3)}{8} - \text{Li}_{3} \left(\frac{1}{4} \right) - \text{Li}_{2} \left(\frac{1}{4} \right) \log(2) - \frac{5}{6} \log^{3}(2) \\ &- \frac{\pi^{2}}{12} \log(2) + \text{Li}_{3} \left(- \frac{1}{2} \right) + \text{Li}_{3} (-8) + 2 \text{Li}_{2}(-8) \log(2) - \log(3) \log^{2}(2) \\ &\approx -0.05738655697. \end{align}$$

I'll leave it to others to figure out how to express the result in the nicest form possible.

A slightly nicer form of the result is

$$ \begin{align} &\int_{0}^{1/2} \frac{\log(1+2x) \log(1-x)}{1+x} \, dx \\ &= \text{Li}_{3} \left( - \frac{1}{8}\right) - \frac{3}{2} \text{Li}_{3} \left(\frac{1}{4} \right) + \frac{15}{8} \zeta(3) + \frac{3}{2} \log^{3}(2) - \log(3) \log^{2}(2)- \frac{\pi^{2}}{6} \log(2) \\ &+ 2 \text{Li}_{2} \left(-\frac{1}{8} \right) \log(2) - \frac{5}{2} \text{Li}_{2} \left( \frac{1}{4}\right) \log(2). \end{align}$$