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

$\begin{array}{l}\text{ Let }I={\int }_0^{\pi /4} 4x\left(\frac{\pi }{4}-x\right)\log (1+\tan x)dx\to \text{ (1). }\\ \mathrm{Apply} {\int }_a^b f\left(x\right)dx={\int }_a^b f(a+b-x)dx \\ \Rightarrow I={\int }_0^{\pi /4} 4\left(\frac{\pi }{4}-x\right)x\log \left(1+\tan \left(\frac{\pi }{4}-x\right)\right)dx\to \left(2\right)\\ \left(1\right)+\left(2\right)\Rightarrow 2I=(\log 2)\cdot 4{\int }_0^{\pi /4} x\left(\frac{\pi }{4}-x\right)dx\\ \Rightarrow I=2\log 2{\left(\frac{\pi }{4}\cdot \frac{x^2}{2}-\frac{x^3}{3}\right)}_0^{\pi /4}\\ =2\log 2\left(\frac{\pi }{8}\frac{{\pi }^2}{16}-\frac{1}{3}\frac{{\pi }^3}{64}\right)\end{array}$

      $=\frac{{\pi }^3}{192}.\log 2$