Evaluate $\int_0^1 \frac{\log ^2(x+1) \log \left(x^2+1\right)}{x^2+1} dx$

Question

How can we evaluate $$\int_0^1 \frac{\log ^2(x+1) \log \left(x^2+1\right)}{x^2+1} dx$$ Any kind of help is appreciated.

Answer 1

I found the answer:

$$\small \int_0^1 \frac{\log ^2(x+1) \log \left(x^2+1\right)}{x^2+1} \, dx=\frac{\pi ^2 C}{48}-\frac{15}{4} C \log ^2(2)+20 \Im(\text{Li}_4(1+i))+2 \log (2) \Im(\text{Li}_3(1+i))+\frac{35 \pi \zeta (3)}{64}-\frac{5}{24} \pi \log ^3(2)-\frac{21}{64} \pi ^3 \log (2)-\frac{3}{256} \left(\psi ^{(3)}\left(\frac{1}{4}\right)-\psi ^{(3)}\left(\frac{3}{4}\right)\right)$$

A generalization:

$$\scriptsize \int_0^1 \frac{\log ^3(x+1) \log \left(x^2+1\right)}{x^2+1} \, dx=-\frac{39}{8} C \log ^3(2)+\frac{1}{32} \pi ^2 C \log (2)-27 \beta (4) \log (2)-48 \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)-6 \log ^2(2) \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+6 \log (2) \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{105}{128} \pi \zeta (3) \log (2)+\frac{119 \pi ^5}{1024}+\frac{5}{16} \pi \log ^4(2)+\frac{105}{256} \pi ^3 \log ^2(2)$$

Answer 2

Not an answer.

This integral resists any CAS I tried (with little hope, I must confess). Using a few elementary constants as a basis for identification, it is close to

$$\frac{-3+9 \sqrt{2}-5 \sqrt{3}-9 e-7 \pi +5 \pi ^2-10 \log (2)}{-1+9 \sqrt{3}-9 e-3 \pi -4 \pi ^2-8 \log (2)+4 \log (3)}$$

The relative error is $4.75 \times 10^{-19}$%.