${\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(\pi x-4x^2\right)\ln \left(1+\tan x\right)dx=}$

$I=4{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}x\left(\frac{\pi }{4}-x\right)\ln \left(1+\tan x\right)dx,x\to \frac{\pi }{4}-x}$

$I=4{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(\frac{\pi }{4}-x\right)x\ln \left(1+\tan \left(\frac{\pi }{4}-x\right)\right)dx}$

$I=4{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}x\left(\frac{\pi }{4}-x\right)x\ln \left(\frac{2}{1+\tan x}\right)dx}$

$2I=4\ln 2{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}x\left(\frac{\pi }{4}-x\right)dx}$

$I=2\ln 2\left[{\left(\frac{\pi }{4}\right)}^3\cdot \frac{1}{2}-{\left(\frac{\pi }{4}\right)}^3\cdot \frac{1}{3}\right]$ $=\frac{{\pi }^3}{192}\ln 2$