How to show that $\int_{0}^{\pi/2}{\arctan(\tan^2 x)\over \sin^2 x\sqrt{\tan x}}\cdot(3\pm\tan x)\mathrm dx=2\pi\sqrt{2\pm \sqrt{2}}?$ A bit of messy integral but seem to yield a simple closed form
Given that:

$$\int_{0}^{\pi/2}{\arctan(\tan^2 x)\over \sin^2 x\sqrt{\tan x}}\cdot(3\pm\tan x)\mathrm dx=2\pi\sqrt{2\pm \sqrt{2}}\tag1$$

Simplifying this part doesn't yield a simple from.
$${3+\tan x\over \sin^2 x\sqrt{\tan x}}$$
Else we can split the integral $(1)\implies$
$$3\int_{0}^{\pi/2}{\arctan(\tan^2 x)\over \sin^2 x\sqrt{\tan x}}\mathrm dx+\int_{0}^{\pi/2}{\sqrt{\tan x}\arctan(\tan^2 x)\over \sin^2 x}\mathrm dx=I+J\tag2$$
$$2\sin^2x=1-{1-\tan^2 x\over 1+\tan^2 x}$$
$$\sin^2x={\tan^2x\over 1+\tan^2x}$$
$$I=3\int_{0}^{\pi/2}{1+\tan^2x\over \tan^2x}\cdot{\arctan(\tan^2 x)\over \sqrt{\tan x}}\mathrm dx$$
$$J=\int_{0}^{\pi/2}{1+\tan^2x\over \tan^2x}\cdot{\sqrt{\tan x}\arctan(\tan^2 x)}\mathrm dx$$
$$u=\tan^2x\implies du=2\tan x\sec^2xdx=2u^{1/2}+2u^{3/2}$$
$I+J\implies$
$${3\over 2}\int_{0}^{\infty}{1+u\over u^{7/4}+u^{11/4}}\arctan(u)\mathrm du+{1\over 2}\int_{0}^{\infty}{1+u\over u^{5/4}+u^{9/4}}\arctan(u)\tag3$$
simplify to
$${3\over 2}\int_{0}^{\infty}{1\over u^{7/4}}\arctan(u)\mathrm du+{1\over 2}\int_{0}^{\infty}{1\over u^{5/4}}\arctan(u)\tag4$$
$${1\over 2}\int_{0}^{\infty}(3+u^{1/2}){\arctan(u)\over u^{7/4}}\mathrm du\tag5$$
$$u=v^4\implies du=4v^3dv$$
$$2\int_{0}^{\infty}(3+v^2){\arctan(v^4)\over v^4}\cdot{\mathrm dv}\tag6$$
Q: How can we prove $(1)?$
 A: By setting $x=\arctan t$ we are left with
$$ \int_{0}^{+\infty}\frac{\arctan(t^2)}{t^{5/2}}(3\pm t)\,dt \stackrel{t\mapsto\sqrt{u}}{=} \frac{1}{2}\int_{0}^{+\infty}\frac{3\pm\sqrt{u}}{u^{7/4}}\arctan(u)\,du \tag{1}$$
and we may apply Feynman's trick to $\int_{0}^{+\infty}\frac{3\pm\sqrt{u}}{u^{7/4}}\arctan(\alpha u)\,du$. We have:
$$ \int_{0}^{+\infty}\frac{(3\pm\sqrt{u})u}{u^{7/4}(1+\alpha^2 u^2)}\,du = \frac{\pi}{2\alpha^{3/4}}\left(3\sqrt{\alpha}\csc\frac{\pi}{8}\pm\sec\frac{\pi}{8}\right) \tag{2}$$
for any $\alpha<0$ by the residue theorem, hence the claim readily follows by integrating $(2)$ over $(0,1)$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\int_{0}^{\pi/2}{\arctan\pars{\tan^{2}\pars{x}} \over
\sin^{2}\pars{x}\root{\tan\pars{x}}}\,\bracks{3 \pm \tan\pars{x}}\,\dd x =
2\pi\root{2 \pm \root{2}}:\ {\large ?}}$.

\begin{align}
&\int_{0}^{\pi/2}{\arctan\pars{\tan^{2}\pars{x}} \over
\sin^{2}\pars{x}\root{\tan\pars{x}}}\,\bracks{3 \pm \tan\pars{x}}\,\dd x
\\[1cm] & =
\int_{x\ =\ 0}^{x\ =\ \pi/2}
{\tan^{2}\pars{x} + 1 \over \tan^{2}\pars{x}}
{\arctan\pars{\tan^{2}\pars{x}} \over
\bracks{\tan^{2}\pars{x}}^{1/4}}\,\bracks{3 \pm \bracks{\tan^{2}\pars{x}}^{1/2}}
\,\times
\\[3mm] & \phantom{=\int_{x\ =\ 0}^{x\ =\ \pi/2}}
{\dd\bracks{\tan^{2}\pars{x}} \over 2\bracks{\tan^{2}\pars{x}}^{1/2}\bracks{\tan^{2}\pars{x} + 1}}
\\[1cm] \stackrel{\tan^{2}\pars{x}\ \mapsto\ x}{=}\,\,\,&
{1 \over 2}\int_{0}^{\infty}{\arctan\pars{x} \over x^{7/4}}
\,\pars{3 \pm x^{1/2}}\,\dd x
\\[5mm] = &\
{1 \over 2}\int_{x\ =\ 0}^{x\ \to\ \infty}\!\!\!\!\!\arctan\pars{x}
\,\dd\bracks{\pars{-4x^{-3/4}} \pm \pars{-4x^{-1/4}}}
\\[5mm] \stackrel{\mbox{IBP}}{=}\,\,\,&
2\int_{0}^{\infty}{x^{-3/4} \pm x^{-1/4} \over x^{2} + 1}\,\dd x
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
\int_{0}^{\infty}{x^{-7/8} \pm x^{-5/8} \over x + 1}\,\dd x
\\[5mm] \stackrel{t\ =\ 1/\pars{x + 1} \iff x = 1/t - 1}{=}\,\,\,&
\int_{1}^{0}t\bracks{\pars{{1 \over t} - 1}^{-7/8} \pm
\pars{{1 \over t} - 1}^{-5/8}}\pars{-\,{\dd t \over t^{2}}}
\\[5mm] = &\
\int_{0}^{1}t^{-1/8}\,\pars{1 - t}^{-7/8}\,\dd t \pm
\int_{0}^{1}t^{-3/8}\,\pars{1 - t}^{-5/8}\,\dd t
\\[5mm] = &\
{\Gamma\pars{7/8}\Gamma\pars{1/8} \over \Gamma\pars{1}} \pm
{\Gamma\pars{5/8}\Gamma\pars{3/8} \over \Gamma\pars{1}} =
{\pi \over \sin\pars{\pi/8}} \pm {\pi \over \sin\pars{3\pi/8}}
\\[5mm] = &\
\pi\bracks{{1 \over \sin\pars{\pi/8}} \pm {1 \over \cos\pars{\pi/8}}}
\\[5mm] = &\
2\pi\,\braces{%
{\root{\bracks{1 + \cos\pars{\pi/4}}/2} \over \sin\pars{\pi/4}} \pm
{\root{\bracks{1 - \cos\pars{\pi/4}}/2} \over \sin\pars{\pi/4}}}
\\[5mm] & =
2\pi\,{\root{2 + \root{2}} \pm \root{2 - \root{2}} \over \root{2}}
= \bbx{\root{2 \pm \root{2}}}
\end{align}
