I've found it.
We have
$$
\int_0^{\!\Large \frac{\pi}{2}}\!\!\left(\frac{1}{\log(\tan x)}+\frac{1}{1-\tan x}\right)^3\! \mathrm dx= \color{blue}{\frac92\ln2-\frac{6}{\pi}G -\frac34 -\frac\pi8} \tag1
$$
where $G$ is Catalan's constant.
Proof. Set
$$
I_n:=\int_0^{\!\Large \frac{\pi}{2}}\!\!\left(\frac{1}{\log(\tan x)}+\frac{1}{1-\tan x}\right)^n\! \mathrm dx, \, n=0,1,2... . \tag2
$$
Clearly
$$
I_0=\frac\pi2.
$$
Recall that
$$
\tan \!\left(\!\frac\pi2 -x\!\right) = \frac{1}{\tan x}
$$ giving
$$
\begin{align}
\frac{1}{\log \tan \!\left(\!\frac\pi2 -x\!\right) }+\frac{1}{1-\tan \!\left(\!\frac\pi2 -x\!\right)}& =\frac{1}{\log \left( \dfrac{1}{\tan x}\right) }+\frac{1}{1-\dfrac{1}{\tan x}} \\\\
& =-\frac{1}{\log \left( \tan x \right) }-\frac{1}{1-\tan x}+1 \tag3
\end{align}
$$
Hence, by the change of variable $\displaystyle x \rightarrow \frac\pi2 -x$, we readily have
$$
\begin{align}
I_1&=\int_0^{\!\Large \frac{\pi}{2}}\!\!\left(-\frac{1}{\log(\tan x)}-\frac{1}{1-\tan x}+1\right)\! \mathrm dx =-I_1+\frac\pi2 \tag4
\end{align}
$$
$$
I_1=\frac\pi4 .
$$
Similarly,
$$
\begin{align}
\int_0^{\!\Large \frac{\pi}{2}}\!\!\left(\frac{1}{\log(\tan x)}+\frac{1}{1-\tan x}\right)^3\! \mathrm dx & =\int_0^{\!\Large \frac{\pi}{2}}\!\!\left(\!-\frac{1}{\log \left( \tan x\! \right) }-\frac{1}{1-\tan x}+1 \! \right)^3\! \mathrm dx
\end{align} \tag5
$$
and, by the binomial expansion,
$$
\begin{align}
I_3 & =-I_3+3I_2-3I_1+I_0
\end{align}
$$
$$
I_3 =-I_3+3I_2-3\frac{\pi}{4}+\frac\pi2
$$
$$
I_3 =\frac32 I_2- \frac{\pi}{8} \tag6
$$
we may conclude with this value obtained for $I_2$.
A numerical approximation is
$$
\color{blue}{I_3=0.22709780663611705673940738484148718263....}
$$
