Evaluate $\int_0^\infty \frac{x^2\operatorname{Ti}_2(x^2)}{x^4+1} \space dx$ At first, I was evaluating
$$I=\int_0^{\frac{\pi}{2}}x\sqrt{\tan x}\space dx=\int_0^\infty\frac{2x^2\arctan{x^2}}{x^4+1}dx $$
I substituted $u=\sqrt{\tan x}$ and I followed up by parametrizing
$$F(t)=\int_0^\infty\frac{x^2\arctan{tx^2}}{x^4+1}dx$$$$I=2F(1)$$
I differentiated both sides, evaluated the derived integral and I integrated both sides as normal:
$$F'(t)=\frac{\pi}{2\sqrt2}\frac{1}{(t+1)(\sqrt{t}+1)\sqrt{t}}$$
$$F(t)=\int_0^\infty\frac{x^2\arctan{tx^2}}{x^4+1}dx=\frac{\pi}{2\sqrt2}\ln{(\sqrt t+1)}-\frac{\pi}{4\sqrt2}\ln{(t+1)}+\frac{\pi}{2\sqrt2}\arctan(t)$$
We can check the constant is equivalent to zero by setting $t$ to zero, then the rest is easy. No questions here. Just giving context.
Anyways, I became curious. Using my knowledge of a few select special functions, I divided both sides by $t$ and then integrated both sides from $0$ to a dummy variable $y$ with respect to t.
$$\int_0^\infty\frac{x^2}{x^4+1}\int_0^y\ \frac{\arctan{(tx^2)}}{t}\space dt\space dx$$$$=\frac{\pi}{2\sqrt2}\int_0^y\frac{\ln{(\sqrt t+1)}}{t}dt-\frac{\pi}{4\sqrt2}\int_0^y\frac{\ln{(t+1)}}{t}dt+\frac{\pi}{2\sqrt2}\int_0^y\frac{\arctan(t)}{t}dt$$
I finally ended up with the following:
$$\int_0^\infty\frac{x^2\operatorname{Ti}_2(yx^2)}{x^4+1}dx=-\frac{\pi}{\sqrt2}\operatorname{Li}_2(-\sqrt y)+\frac{\pi}{4\sqrt2}\operatorname{Li}_2(-y)+\frac{\pi}{\sqrt2}\operatorname{Ti}_2(\sqrt y)$$
I plugged in $y=1$ since it's probably the simplest value to evaluate for dilogarithms and the inverse tangent integral alike. I get
$$\int_0^\infty\frac{x^2\operatorname{Ti}_2(x^2)}{x^4+1}dx=\frac{\pi^3}{16\sqrt2}+\frac{\pi G}{\sqrt2}$$
Where $G$ is Catalan's constant
It's a magnificent result. I don't often see $\pi$ and $G$ multiplied together. My question is how else can we get this result?
Addendum:
WolframAlpha seems to be making an error. For a finitely large upper bound, WolframAlpha gives a result far from 0, yet for an infinite upper bound, WA gives 0. Strange

Too long for comments
The more general integral$$I_a=\int_0^\infty\frac{2x^2}{x^4+a^4}\tan^{-1}(x^2)\,dx$$ is quite easy to compute after partial fraction decomposition
$$I_a=\int_0^\infty\Bigg[\frac{\tan^{-1}(x^2)}{x^2+i\,a^2}+\frac{\tan^{-1}(x^2)}{x^2-i\,a^2}\Bigg]\,dx$$
$CAS$ provide
$$\int_0^\infty \frac{\tan^{-1}(x^2)}{x^2+i\,a^2}=\frac{\sqrt[4]{-1} \pi  \left(-\tanh ^{-1}\left(a^2\right)-i \tan
   ^{-1}(a)+\tanh ^{-1}(a)\right)}{a}$$
$$\int_0^\infty \frac{\tan^{-1}(x^2)}{x^2-i\,a^2}=\frac{\sqrt[4]{-1} \pi  \left(i \tanh ^{-1}\left(a^2\right)+\tan
   ^{-1}(a)-i \tanh ^{-1}(a)\right)}{a}$$
This gives
$$\color{blue}{I_a=\frac{\pi\sqrt{2}  }{a} \left(\tanh
   ^{-1}\left(\frac{a}{a^2+a+1}\right)+\tan ^{-1}(a)\right)}$$
