In my post, I found that $$ \int_{0}^{\frac{\pi}{2}} x\tan 2 x \ln (\tan x) d x = -\frac{\pi^{3}}{32},$$ then I want to generalise it to $$ I_n=\int_{0}^{\frac{\pi}{2}} x\tan 2 x \ln^n (\tan x) d x $$ Again, letting $x\mapsto \frac{\pi}{2}-x$ transform the integral $$ \begin{aligned} I_n=&\frac{(-1)^{n+1} \pi}{2} \int_0^{\frac{\pi}{2}} \tan (2 x) \ln ^n(\tan x) d x +(-1)^n \int_0^{\frac{\pi}{2}} x \tan (2 x) \ln ^n(\tan x) d x \end{aligned} $$ Fortunately, when $n$ is odd, $$ I_n=-\frac{\pi}{4} \underbrace{\int_0^{\frac{\pi}{2}} \tan (2 x) \ln ^n(\tan x) d x}_{J_n} $$ Helpfully, the substitution $y=\tan^2 x$ simplifies the integral
$$ \begin{aligned} J_n &=\int_0^{\frac\pi2} \frac{2 \tan x}{1-\tan ^2 x} \ln ^n(\tan x) dx =\frac{1}{2^n} \int_0^{\infty} \frac{\ln ^n y}{1-y^2} d y \\ &=\frac{1}{2^{n-1}} \int_0^1 \frac{\ln ^n y}{1-y^2} d y=\frac{n !}{2^{n-1}}\left(1-\frac{1}{2^{n+1}}\right)\zeta(n+1) \end{aligned} $$
(For the last integral please refer to the footnote)
Now we can conclude that $$\boxed{\int_{0}^{\frac{\pi}{2}} x\tan 2 x \ln^n (\tan x) d x =-\frac{n ! \pi }{2^{n+1}}\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1)} $$ where $n$ is a odd natural numbers.
For examples, $$ \begin{aligned} I_1 &=-\frac{\pi \cdot 1 !}{2^2}\left(1-\frac{1}{2^2}\right) \zeta(2)=-\frac{\pi^3}{32} \\ I_3 &=-\frac{\pi \cdot 3 !}{2^4}\left(1-\frac{1}{2^4}\right) \zeta(4)=-\frac{\pi^5}{256} \\ I_5 &=-\frac{\pi \cdot 5 !}{2^6}\left(1-\frac{1}{2^6}\right) \zeta(6)=-\frac{\pi^7}{512} \\ I_{23} &=-\frac{\pi \cdot 23 !}{2^{24}}\left(1-\frac{1}{2^{24}}\right) \zeta(24) =-\frac{968383680827 \pi^{25}}{536870912} \end{aligned} $$
Succeeding in finding a formula for $I_{2n+1}$ with odd powers, I am eager to find one for the $I_{2n}$ with even powers as the technique just used fails when $n$ is even. Could you help?
Footnote:
$$\int_0^1 \frac{\ln ^n y}{1-y^2} d y =\sum_{k=0}^{\infty} \int_0^1 y^{2 k} \ln ^n y d y =\left.\frac{\partial^n}{\partial a^n} \int_0^1 y^a d y\right|_{a=2 k} \\= \sum_{k=0}^{\infty}\frac{(-1)^n n !}{(2 k+1)^{n+1}} =n!\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1) $$
