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 $(1)?$