TITLE: Closed form of $\int_{0}^{1} \frac{\log(1+x)\log(2+x) \log(3+x)}{1+x}\,dx$
QUESTION [11 upvotes]: 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
$$i_{1s}=\int_{0}^{1} \frac{\log(1+x)\log(1+\frac{x}{2})\log(1+\frac{x}{3})}{1+x}\,dx \simeq 0.0130713\tag{3}$$

REPLY [3 votes]: We have seen a great elegantly flowing answer of user97357329 (https://math.stackexchange.com/a/3522251/198592), the main step of which consisted in the substitution $x\to\frac{1-t}{1+t}$.
The next steps have been sketched very briefly in the answer, so that I take the freedom to elaborate a bit more and, in the end, provide the explit closed form  of our integral.
The transformation has the nice property (which can't be seen in the exposition of that user) that after expanding the resulting logs in the integrand we get one term that is the exact negative of the original term, and all other terms are simpler.
The transformation gives (with $dx\to- 2dt/(t+1)^2$)
$$i_{1}=\int_0^1 \frac{\log (x+1) \log (x+2) \log (x+3))}{x+1}\,dx\to\int_0^1 \frac{\log \left(\frac{2}{t+1}\right) \log \left(\frac{2 (t+2)}{t+1}\right) \log \left(\frac{t+3}{t+1}\right)}{t+1}\,dt$$
Expanding the logs in the integrand yields this list of terms
$$\left\{-\frac{\log ^2(2) \log (t+1)}{t+1},\frac{2 \log (2) \log ^2(t+1)}{t+1},-\frac{\log ^3(t+1)}{t+1},-\frac{\log (2) \log (t+1) \log (t+2)}{t+1},\frac{\log ^2(t+1) \log (t+2)}{t+1},\frac{\log ^2(2) \log (t+3)}{t+1},-\frac{2 \log (2) \log (t+1) \log (t+3)}{t+1},\frac{\log ^2(t+1) \log (t+3)}{t+1},\frac{\log (2) \log (t+2) \log (t+3)}{t+1},-\frac{\log (t+1) \log (t+2) \log (t+3)}{t+1}\right\}$$
We see that the last term is exactly the opposite of the original integrand, so that the complicated part drops out (move it to the l.h.s. and get twice the original expression).
The remaining integrals are in fact easy to solve. Using Mathematica we get some terms in the complex domain which, however compensate when we use the relations
$$\operatorname{Li}_2\left(\frac{3}{2}\right)= -\operatorname{Li}_2\left(-\frac{1}{2}\right)+\frac{\pi ^2}{6}+\log (2) \log \left(\frac{3}{2}\right)-i \pi  \log \left(\frac{3}{2}\right)$$
$$\operatorname{Li}_2(2)= \frac{\pi ^2}{6}+\frac{\pi ^2}{12}-i \pi  \log (2)$$
The final result then becomes (I'm busy simplifying this) 
$$i_{1} = -\operatorname{Li}_4(-2)+\operatorname{Li}_4\left(-\frac{1}{2}\right)-\frac{1}{2} \operatorname{Li}_2\left(\frac{1}{3}\right) \log ^2(2)-2 \operatorname{Li}_2\left(\frac{2}{3}\right) \log ^2(2)+\frac{1}{4} \operatorname{Li}_2\left(\frac{1}{4}\right) \log ^2(2)-\frac{1}{2} \operatorname{Li}_2\left(\frac{1}{3}\right) \log (3) \log (2)+\frac{1}{2} \operatorname{Li}_2\left(\frac{2}{3}\right) \log (27) \log (2)-\frac{1}{2} \operatorname{Li}_2\left(\frac{3}{4}\right) \log (8) \log (2)+\frac{1}{2} \operatorname{Li}_2\left(\frac{3}{4}\right) \log (9) \log (2)-\frac{1}{2} \operatorname{Li}_3(-2) \log (2)-\frac{1}{2} \text{Li}_3\left(\frac{1}{3}\right) \log (2)+\frac{3}{2} \operatorname{Li}_3\left(\frac{2}{3}\right) \log (2)+\frac{1}{4} \operatorname{Li}_3\left(\frac{1}{4}\right) \log (2)-\operatorname{Li}_3\left(\frac{3}{4}\right) \log (2)+\frac{1}{2} \operatorname{Li}_3(-2) \log (4)-\frac{1}{2} \zeta (3) \log (2)-\frac{3}{8} \zeta (3) \log (4)-\frac{1}{24} 35 \log ^4(2)+\frac{5}{2} \log (3) \log ^3(2)+\frac{1}{3} \log ^3(3) \log (2)+\frac{1}{2} \zeta(2) \log ^2(2)-2 \log ^2(3) \log ^2(2)+\frac{1}{6} \log (3) \log (64) \log ^2(2)$$
This is a linear combination of the components $\log(2)$, $\log(3)$, $\zeta(2)$, $\zeta(3)$, $\operatorname{Li}_{2,3,4}$. Notice that there is no $\pi$ or (Euler)-$\gamma$ appearing.