Closed form of $\int_{0}^{1} \frac{\log(1+x)\log(2+x) \log(3+x)}{1+x}\,dx$ In https://math.stackexchange.com/a/3414337/198592 I have mentioned some integrals which I could not solve.
One of them is
$$i_{1}=\int_{0}^{1} \frac{\log(1+x)\log(2+x)\log(3+x)}{1+x}\,dx \simeq 0.295123\tag{1}$$
I have tried several approaches like partial integration and series expansion but with no avail.
The substiution
$$\log(x) = \int_0^{\infty } \frac{e^{-t}-e^{-t\; x}}{t} \, dt\tag{2}$$
did allow doing the $x$-integral but already the first integration of the triple integral failed.
Interestingly, with "one $\log$ less" or without the denominator $1+x$ the integration leads to a closed expression.
Problem: find a closed expression for $i_1$ or, equivalently, for
$$i_{1s}=\int_{0}^{1} \frac{\log(1+x)\log(1+\frac{x}{2})\log(1+\frac{x}{3})}{1+x}\,dx \simeq 0.0130713\tag{3}$$
 A: The process flows elegantly if we simply make use of the variable change $x\mapsto (1-x)/(1+x)$. I'll develop it.
So, we have that 
$$\int_0^1\frac{\log(1+x)\log(2+x)\log(3+x)}{1+x}\textrm{d}x$$
$$=\frac{1}{2}\log^2(2)\int_0^1 \frac{\log(3+x)}{1+x}\textrm{d}x-\frac{1}{2}\log^2(2)\int_0^1 \frac{\log(1+x)}{1+x}\textrm{d}x+\log(2)\int_0^1 \frac{\log^2(1+x)}{1+x}\textrm{d}x$$
$$-\frac{1}{2}\int_0^1\frac{\log^3(1+x)}{1+x}\textrm{d}x-\frac{\log(2)}{2}\int_0^1 \frac{\log(1+x)\log(2+x)}{1+x}\textrm{d}x$$
$$+\frac{\log(2)}{2}\int_0^1 \frac{\log(2+x)\log(3+x)}{1+x}\textrm{d}x-\log(2)\int_0^1 \frac{\log(1+x)\log(3+x)}{1+x}\textrm{d}x$$
$$+\frac{1}{2}\int_0^1 \frac{\log^2(1+x)\log(2+x)}{1+x}\textrm{d}x+\frac{1}{2}\int_0^1 \frac{\log^2(1+x)\log(3+x)}{1+x}\textrm{d}x.$$
What's next? We look at these integral with a trained eye and see all are reducible immediately to known, trivial integrals (mainly with integration by parts and usual knowledge of polylogarithms).
End of story.
A: Not an answer but too long for a comment.
For the fun of it, I used Taylor expansion to $O(x^{n+1})$. Below are some numbers wich show a very slow convergence
$$\left(
\begin{array}{cc}
 n & \text{result} \\
 100 & \color{red} {0.295}088992683718 \\
 200 & \color{red} {0.2951}14319823039 \\
 300 & \color{red} {0.2951}19066043094 \\
 400 & \color{red} {0.29512}0734361895 \\
 500 & \color{red} {0.29512}1508301534 \\
 600 & \color{red} {0.29512}1929303681 \\
 700 & \color{red} {0.295122}183398346 \\
 800 & \color{red} {0.295122}348430736 \\
 900 & \color{red} {0.295122}461636524
\end{array}
\right)$$
