Logarithmic integral $ \int_0^1 \frac{x}{x^2+1} \, \log(x)\log(x+1) \, {\rm d}x $ At various places e.g.
Calculate $\int_0^1\frac{\log^2(1+x)\log(x)\log(1-x)}{1-x}dx$
and
How to prove $\int_0^1x\ln^2(1+x)\ln(\frac{x^2}{1+x})\frac{dx}{1+x^2}$
logarithmic integrals are connected to Euler-sums. In view of the last link I'm wondering about the following integral
$$
\int_0^1 \frac{x}{x^2+1} \, \log(x)\log(x+1) \, {\rm d}x \, .
$$
I see I can throw it into Wolfram Alpha and get some disgusting anti-derivative with Li's up to ${\rm Li}_3$. Anyway is there some manually more tractable way to solve this?
I have tried two things of which both don't seem to lead anywhere so far.
For the first one:
I expressed $\frac{x}{x^2+1}$ by it's Mellin transform $\frac{\pi/2}{\cos\left(\frac{\pi s}{2}\right)}$ and interchanged the integral order
$$
\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} {\rm d}s \, \frac{\pi/2}{\cos\left(\frac{\pi s}{2}\right)} \left( -\frac{{\rm d}}{{\rm d}s} \right)\int_0^1 {\rm d}x \, x^{-s} \log(x+1)
$$
where the constant $c>-1$ is right of the first pole of the cosine at $s=-1$ and the contour can be closed in a circle on the left hand side of the plane to use the residue theorem. The $x$-integral is equal to
$$
\int_0^1 {\rm d}x \, x^{-s} \log(x+1) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1-s)} = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{1-s} \left( \frac{1}{n} - \frac{1}{n+1-s} \right)\\
= {\frac {\Psi \left( 1-s/2 \right) - \Psi \left( 3/2-s/2 \right) }{2(1-s)}} + {\frac {\log  \left( 2 \right) }{1-s}}
$$
where $\Psi$ is the Digamma function, related to the harmonic numbers $H_n$. Deriving with respect to $s$ and picking up the residue $(-1)^k$ of the Mellin transform at $s=-2k-1$ ($k=0,1,2,3,...$) one obtains
$$
\sum_{k=0}^\infty (-1)^{k+1} \Bigg\{ {\frac {\Psi \left( 3/2+k \right) - \Psi \left( 2+k \right) }{ 8\left( 1+k \right) ^{2}}} - {\frac {\Psi' \left( 3/2+k \right) - \Psi' \left( 2+k \right) }{8(1+k)}} + {\frac {\log  \left( 2 \right) }{ 4\left( 1+k \right) ^{2}}} \Bigg\}
$$
where $\Psi'$ is the derivative of the Digamma function related to $H_{n,2}$. The terms with integral argument presumably can be evaluated in closed form, but I'm wondering if the half-integer argument terms can be also evaluated just by algebraic manipulations?
Second:
I tried to find closed form for the integral by partial integration
\begin{align}
I(a) &=\int_0^1 \frac{\log(x) \log(x+1)}{x+a} \, {\rm d}a \\
&=-\frac{\log(2)}{a+1} - \int_0^1 \frac{x\left(\log(x)-1\right)}{(x+1)(x+a)} \, {\rm d}x + \int_0^1 \frac{x\left(\log(x)-1\right) \log(x+1)}{(x+a)^2} \, {\rm d}x \\
&=-\frac{\log(2)}{a+1} - \int_0^1 \frac{x\left(\log(x)-1\right)}{(x+1)(x+a)} \, {\rm d}x - \int_0^1 \left( \frac{\log(x+1)}{x+a} - \frac{a\log(x+1)}{(x+a)^2} \right) + I(a) + a I'(a)
\end{align}
and thus
$$
I(a) = \int_\infty^a \frac{{\rm d}a'}{a'} \Bigg\{ \frac{\log(2)}{a'+1} + \int_0^1 \frac{x\left(\log(x)-1\right)}{(x+1)(x+a')} \, {\rm d}x + \int_0^1 \left( \frac{\log(x+1)}{x+a'} - \frac{a'\log(x+1)}{(x+a')^2} \right) {\rm d}x \Bigg\}
$$
of which many terms are easy to integrate, but there is one combination which seems very difficult, namely something like
$$
\int \frac{{\rm Li}_2(a')}{a'+1} \, {\rm d}a' \, .
$$
$a=\pm i$ at the end. Any insights?
 A: Integrate as follows
\begin{align}
& \hspace{5mm}\int_0^1 \frac{x \ln x\ln(1+x) }{1+x^2}dx\\
&= \frac12 \int_0^1 \frac{x \ln x\ln(1-x^2) }{1+x^2}\overset{x^2\to x}{dx}
-\frac12\int_0^1 \frac{x \ln x\ln\frac{1-x}{1+x}}{1+x^2}\overset{\to x}{dx}\\
&= \frac18 \int_0^1 \frac{\ln x\ln(1-x) }{1+x}dx
-\frac14\int_0^1 \frac{\ln x\ln\frac{1-x}{1+x}}{1+x}{dx}\\
&\overset{ibp} = -\frac18\int_0^1 \frac{\ln x\ln(1+x) }{1-x}dx
+\frac18\int_0^1 \frac{\ln (1-x)\ln(1+x)-\ln^2(1+x) }{x}dx\tag1
\end{align}
Evaluate
\begin{align}
& \int_0^1 \frac{\ln (1-x)\ln(1+x)-\ln^2(1+x) }{x}dx\\
=& \frac12\int_0^1 \frac{\ln^2(1-x^2)}xdx
 - \frac12\int_0^1 \frac{\ln^2(1-x)}xdx
 -  \frac32\int_0^1 \frac{\ln^2(1+x)}xdx\\
=& \frac12\cdot \zeta(3)-\frac12\cdot 2\zeta(3)-\frac32\cdot \frac14\zeta(3)=-\frac78\zeta(3)
\end{align}
and
\begin{align}
& \int_0^1 \frac{\ln x\ln(1+x) }{1-x}dx\\
\overset{ibp}=& \>\ln(1+x)\left(\int_0^x \frac{\ln t}{1-t}dt\right)\bigg|_0^1-\int_0^1\frac1{1+x} \left(\int_0^x \frac{\ln t}{1-t}\overset{t= xy}{dt}\right)dx\\
=& -\ln(2)\zeta(2)-\int_0^1 \int_0^1 \frac{x\ln x}{(1+x)(1-xy)}dy dx
 - \int_0^1 \int_0^1 \frac{x\ln y}{(1+x)(1-xy)}\overset{x\leftrightarrows y}{dy dx}\\
 =& -\ln(2)\zeta(2)+ \int_0^1 \int_0^1 \left(\frac{\ln x}{(1+x)(1+y)}- \frac{\ln x}{1-xy} \right) dy dx\\
 =& -\ln(2)\zeta(2)- \frac12\ln2 \zeta(2) + \int_0^1 \frac{\ln x\ln(1-x)}{x} \overset{ibp}{dx}\\
=& -\frac32\ln2\zeta(2)+\frac12\int_0^1 \frac{\ln^2x}{1-x}dx
= -\frac32\ln2\zeta(2)+\zeta(3)
\end{align}
Substitute the two integral results evaluated above into (1) to arrive at
$$\int_0^1 \frac{x \ln x\ln(1+x) }{1+x^2}dx= \frac3{16}\ln2\zeta(2)-\frac{15}{64}\zeta(3)$$
A: This is not an answer but it is too long for a comment.
Concerning the disgusting antiderivative, I tried (using another CAS) to undertand how it was computed. If I am not mistaken, they seem to start using
$$\frac x{x^2+1}=\frac 12 \left(\frac 1 {x+i}+\frac 1 {x-i} \right)$$ and from here starts the nightmare.
The problem is that we face almost the same kind of problem considering
$$I(a)=\int \frac{{\rm Li}_2(a)}{a+1} \, da$$
The result given by a CAS is again awful but simplifies a lot when making $a=\pm i$.
$$I(+i)=\frac{1}{4} C (\pi +2 i \log (2))-\text{Li}_3(1-i)-\text{Li}_3(1+i)+\frac{1}{192}
   \pi ^2 (\log (1024)-i \pi )$$
$$I(-i)=\frac{1}{4} C (\pi -2 i \log (2))-\text{Li}_3(1-i)-\text{Li}_3(1+i)+\frac{1}{192}
   \pi ^2 (\log (1024)+i \pi )$$
