Evaluating ${\int}_{0}^{\frac{\pi}{2}}{(\frac{1}{\mathrm{log}(\mathrm{tan}x)}+\frac{1}{1-\mathrm{tan}(x)})}^{3}dx$

Using the method shown here, I have found the following closed form.

${\int}_{0}^{\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi}{2}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{(\frac{1}{\mathrm{log}(\mathrm{tan}x)}+\frac{1}{1-\mathrm{tan}x})}^{2}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x=3\mathrm{ln}2-\frac{4}{\pi}G-\frac{1}{2},$

where $G$ is Catalan's constant.

I can see that replicating the techniques for the following integral could be rather lenghty.

${\int}_{0}^{\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi}{2}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{(\frac{1}{\mathrm{log}(\mathrm{tan}x)}+\frac{1}{1-\mathrm{tan}x})}^{3}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x$

My question: Could someone have, ideally, a different idea to evaluate the latter integral?

Using the method shown here, I have found the following closed form.

${\int}_{0}^{\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi}{2}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{(\frac{1}{\mathrm{log}(\mathrm{tan}x)}+\frac{1}{1-\mathrm{tan}x})}^{2}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x=3\mathrm{ln}2-\frac{4}{\pi}G-\frac{1}{2},$

where $G$ is Catalan's constant.

I can see that replicating the techniques for the following integral could be rather lenghty.

${\int}_{0}^{\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi}{2}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{(\frac{1}{\mathrm{log}(\mathrm{tan}x)}+\frac{1}{1-\mathrm{tan}x})}^{3}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x$

My question: Could someone have, ideally, a different idea to evaluate the latter integral?