The following integral $$I=\int_0^1\frac{\ln(x)}{1+x}\text{Li}_2\left(\frac{1+x}2\right)\ dx$$
was already evaluated by @Knas here where he found
$$I=-2\operatorname{Li}_4\left(\dfrac{1}{2}\right)-\dfrac{1}{8}\zeta(3)\ln 2 + \dfrac{1}{180}\pi^4 - \dfrac{1}{12}\ln^4 2$$
I am looking for a more elegant solution if possible.
First trial, by series expansion
$$\frac{1}{1+x}\text{Li}_2\left(\frac{1+x}2\right)=\sum_{n=1}^\infty\frac{(1+x)^{n-1}}{n^2 2^n}$$
substitute $$(1+x)^{n-1}=\sum_{k=1}^\infty {n-1\choose k-1}x^{k-1}$$
we have
$$ I=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{{n-1\choose k-1}}{n^2 2^n}\int_0^1 x^{k-1}\ln(x)\ dx=-\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{{n-1\choose k-1}}{n^2k^2 2^n}$$
and I am stuck here. I verified the numeric result for the double sum in case you are curious.
Second trial, by the identity
$$\frac{\text{Li}_2(z)}{z}=-\int_0^1\frac{\ln(y)}{1-zy}\ dy$$
we have
$$\frac{\text{Li}_2\left(\frac{1+x}{2}\right)}{1+x}=\frac12\frac{\text{Li}_2\left(\frac{1+x}{2}\right)}{\frac{1+x}{2}}=-\frac12\int_0^1\frac{\ln(y)}{1-\frac{1+x}{2}y}\ dy$$
$$\Longrightarrow I=-\frac12\int_0^1\ln(y)\left(\int_0^1\frac{\ln(x)}{(1+x)(1+\frac{1+x}{2}y)}\ dx\right)\ dy=\int_0^1\frac{\ln(y)}{y}\text{Li}_2\left(\frac{y}{2-y}\right)\ dy$$
and this integral seems challenging and I dont know how to crack it.
Third trial, by integration by parts we have
$$I=-\frac{\pi^4}{72}+\int_0^1\frac{\text{Li}_2(-x)}{1+x}\ln\left(\frac{1-x}{2}\right)\ dx+\int_0^1\frac{\ln(x)\ln(1+x)}{1+x}\ln\left(\frac{1-x}{2}\right)\ dx$$
IF we use $\frac{\text{Li}_2(-x)}{1+x}=\sum_{n=1}^\infty (-1)^n H_n^{(2)}x^n$ for the first integral and $\frac{\ln(1+x)}{1+x}=-\sum_{n=1}^\infty (-1)^n H_n x^n$ for the second integral, we will come across around 6 alternating harmonic series some of which are advanced and I am not into long tedious solutions.
Any other ideas? can you take advantages of my trials presented above in a way that leads a nice solution? Thank you