TITLE: Evaluating $\int_0^1\frac{\arctan x\ln\left(\frac{2x^2}{1+x^2}\right)}{1-x}dx$
QUESTION [9 upvotes]: Here is a nice problem proposed by Cornel Valean

$$
I=\int_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{192}\,\pi^{3} +
2\Im\left\{%
\text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}
$$

My Trial: By subbing $x=\frac{1-t}{1+t}$ we have
$$I=\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln\left(\frac{(1-x)^2}{1+x^2}\right)}{x(1+x)}dx$$
$$=2\underbrace{\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln(1-x)}{x(1+x)}dx}_{x\to (1-x)/(1+x)}-\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln(1+x^2)}{x(1+x)}dx$$
$$=2\int_0^1\frac{\arctan x\ln(\frac{2x}{1+x})}{1-x}dx-\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln(1+x^2)}{x(1+x)}dx$$
and got stuck here. Any idea? thanks.

REPLY [2 votes]: Some generalizations. Enjoy!

$\small \int_0^1 \frac{\log ^3\left(\frac{2 x^2}{x^2+1}\right) \tan ^{-1}(x)}{1-x} \, dx=-\frac{192}{19} \sqrt{2} \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{2}\right)-\frac{15 }{19456}\pi  \, _7F_6\left(1,1,1,1,1,\frac{3}{2},\frac{3}{2};2,2,2,2,2,2;1\right)+\frac{105 C \zeta (3)}{16}-\frac{3 \pi  C^2}{2}-\frac{3}{8} \pi ^2 C \log (2)-\frac{3}{4} \pi ^2 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{480}{19} \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{39}{8} \pi  \text{Li}_4\left(\frac{1}{2}\right)+\frac{1905 \pi  \zeta (3) \log (2)}{1216}-\frac{881 \pi ^5}{29184}+\frac{203 \pi  \log ^4(2)}{1216}-\frac{49 \pi ^3 \log ^2(2)}{2432}$

$\small \int_0^1 \frac{\log \left(\frac{2 x^2}{x^2+1}\right) \tan ^{-1}(x)^3}{1-x} \, dx=-\frac{48}{19} \sqrt{2} \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{2}\right)-\frac{15 }{77824}\pi  \, _7F_6\left(1,1,1,1,1,\frac{3}{2},\frac{3}{2};2,2,2,2,2,2;1\right)+\frac{105 C \zeta (3)}{64}-\frac{3 \pi  C^2}{8}-\frac{3}{32} \pi ^2 C \log (2)+\frac{3}{16} \pi ^2 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{108}{19} \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{9}{32} \pi  \text{Li}_4\left(\frac{1}{2}\right)+\frac{1905 \pi  \zeta (3) \log (2)}{4864}+\frac{537 \pi ^5}{48640}+\frac{51 \pi  \log ^4(2)}{4864}+\frac{103 \pi ^3 \log ^2(2)}{9728}$

$\scriptsize \int_0^1 \frac{\log ^2\left(\frac{2 x^2}{x^2+1}\right) \tan ^{-1}(x)^2}{1-x} \, dx=-\frac{1}{8} \, _7F_6\left(1,1,1,1,1,1,\frac{5}{4};\frac{3}{2},2,2,2,2,2;1\right)+\frac{15 }{19456}\pi  \, _7F_6\left(1,1,1,1,1,\frac{3}{2},\frac{3}{2};2,2,2,2,2,2;1\right)+\frac{192}{19} \sqrt{2} \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{2}\right)+4 C \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{21 C \zeta (3)}{8}+\frac{\pi ^3 C}{96}+\pi  C^2+\frac{1}{6} C \log ^3(2)+\frac{3}{8} \pi  C \log ^2(2)+2 C^2 \log (2)-\frac{5}{24} \pi ^2 C \log (2)-2 \pi  \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{584}{19} \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\pi  \log (2) \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{3 \text{Li}_5\left(\frac{1}{2}\right)}{4}-\frac{211 \pi ^2 \zeta (3)}{768}-\frac{3317 \zeta (5)}{512}+\frac{7}{8} \zeta (3) \log ^2(2)+\frac{89}{304} \pi  \zeta (3) \log (2)+\frac{1}{64} \pi  \zeta \left(4,\frac{1}{4}\right)-\frac{1}{64} \pi  \zeta \left(4,\frac{3}{4}\right)+\frac{1}{32} \zeta \left(4,\frac{1}{4}\right) \log (2)-\frac{1}{32} \zeta \left(4,\frac{3}{4}\right) \log (2)-\frac{15697 \pi ^5}{145920}+\frac{\log ^5(2)}{120}+\frac{3}{608} \pi  \log ^4(2)-\frac{35}{576} \pi ^2 \log ^3(2)-\frac{175 \pi ^3 \log ^2(2)}{1216}-\frac{1307 \pi ^4 \log (2)}{23040}$

$\scriptsize \int_0^1 \frac{\log ^4\left(\frac{2 x^2}{x^2+1}\right)}{1-x} \, dx=-3 \, _7F_6\left(1,1,1,1,1,1,\frac{5}{4};\frac{3}{2},2,2,2,2,2;1\right)+\frac{45 \pi  }{2432}\, _7F_6\left(1,1,1,1,1,\frac{3}{2},\frac{3}{2};2,2,2,2,2,2;1\right)+\frac{4608}{19} \sqrt{2} \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{2}\right)+96 C \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-63 C \zeta (3)-\frac{\pi ^3 C}{4}+24 \pi  C^2+4 C \log ^3(2)+9 \pi  C \log ^2(2)+48 C^2 \log (2)-5 \pi ^2 C \log (2)+48 \pi  \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{14016}{19} \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+24 \pi  \log (2) \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-78 \text{Li}_5\left(\frac{1}{2}\right)-\frac{35 \pi ^2 \zeta (3)}{4}+\frac{1605 \zeta (5)}{64}+21 \zeta (3) \log ^2(2)+\frac{267}{38} \pi  \zeta (3) \log (2)+\frac{21}{64} \pi  \zeta \left(4,\frac{1}{4}\right)-\frac{21}{64} \pi  \zeta \left(4,\frac{3}{4}\right)+\frac{3}{4} \zeta \left(4,\frac{1}{4}\right) \log (2)-\frac{3}{4} \zeta \left(4,\frac{3}{4}\right) \log (2)-\frac{15697 \pi ^5}{6080}+\frac{3 \log ^5(2)}{5}+\frac{9}{76} \pi  \log ^4(2)-\frac{13}{8} \pi ^2 \log ^3(2)-\frac{525}{152} \pi ^3 \log ^2(2)-\frac{277}{320} \pi ^4 \log (2)$