I found this integral: $$\int_{0}^{1}\frac{x\log^{3}{(x+1)}}{x^2+1}dx$$
And it seems look like this problem but i don't know how to process with this one.
First, i tried to use series of $\frac{x}{x^2+1}=\sum_{k=0}^{\infty}(-1)^kx^{2k+1}$, and use the representation $\log^3{(x+1)}=\frac{d^3}{da^3}(x+1)^a$ at $a=0$. Then it ended up with hypergeometric function. Can I ask some ideas from everyone? Thank you.
\begin{align} I&= \int_{0}^{1}\frac{x\ln^{3}{(x+1)}}{x^2+1}dx\\ & =6\ \Re\left(\operatorname{Li}_4(1+i)-\operatorname{Li}_4(\frac{1+i}2)\right)-\frac{105}{32}\ln2\ \zeta(3) \end{align}
which is to be derived with the substitute $t=\frac1{1+x}$
$$\begin{align} I =& \int_{\frac12}^1 \frac{(t-1)\ln^3t}{(t^2+(1-t)^2)\ t}dt =-\int_{\frac12}^1 \Re \frac{(1+i)\ln^3t}{1-(1+i)t}+ \frac{\ln^3t }t \ dt\\ =& -\Re \int_{0}^1 \frac{(1+i)\ln^3t}{1-(1+i)t}dt + \Re \int_{0}^{\frac12} \frac{(1+i)\ln^3t}{1-(1+i)t}\ \overset{t\to \frac t2}{dt} + \frac14\ln^42\\ =& \ 6\ \Re\operatorname{Li}_4(1+i)+\Re \int_0^1 \frac{\frac{1+i}2(\ln t-\ln2)^3}{1-\frac{1+i}2 t}dt+ \frac14\ln^42\\ =& \ 6\ \Re\operatorname{Li}_4(1+i)-6\ \Re\operatorname{Li}_4(\frac{1+i}2)\\ &\>\>\> - 6\ln2\ \Re\operatorname{Li}_3(\frac{1+i}2)-3\ln^22\ \Re\operatorname{Li}_2(\frac{1+i}2)-\frac14\ln^42\\ \end{align}$$ To reduce it to the stated result, utilize the identities \begin{align} &\Re\operatorname{Li}_2(\frac{1+i}2)=\frac{5\pi^2}{96}-\frac1{8}\ln^22\\ &\Re\operatorname{Li}_3(\frac{1+i}2)=\frac{35}{64}\zeta(3)-\frac{5\pi^2}{192}\ln2+\frac1{48}\ln^32 \end{align}
