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

As was done in the answer HERE by the user Setness Ramesory, we can use the duplication formula for the dilogarithm to put the integrand into a form that lends itself to an evaluation using contour integration.
The key point, as explained in the linked answer, is that the branch cut for $\operatorname{Li}_{2}(e^{i \pi/4}z)$ is entirely in the lower half-plane, while the branch cut for $\operatorname{Li}_{2}(-e^{i \pi/4}z) $ is entirely in the upper half-plane.
And since $\operatorname{Li}_{2}(z) \sim -\frac{\ln^{2}(-z)}{2}$ as $|z| \to \infty$ (see here), we have
$$ \begin{align} &\int_{0}^{\infty} \frac{x^{2}\operatorname{Ti}_{2}(x^{2})}{1+x^{4}} \, \mathrm dx \\ &= \frac{1}{2} \int_{-\infty}^{\infty}  \frac{x^{2}\operatorname{Ti}_{2}(x^{2})}{1+x^{4}} \, \mathrm dx \\ &= \frac{1}{2} \, \Im \int_{-\infty}^{\infty} \frac{x^{2}\operatorname{Li}_{2}(ix^{2})}{1+x^{4}} \, \mathrm dx \\ &= \Im \int_{-\infty}^{\infty} \frac{x^{2} \operatorname{Li}_{2}(e^{\pi i /4}  x)}{1+x^{4}} \, \mathrm dx + \Im \int_{-\infty}^{\infty} \frac{x^{2} \operatorname{Li}_{2}(-e^{\pi i /4} x)}{1+x^{4}} \, \mathrm dx \\ &=  \Im \int_{-\infty}^{\infty} f(x) \, \mathrm dx + \Im \int_{-\infty}^{\infty} g(x) \, \mathrm dx  \\ &= \small\Im  \left( 2 \pi i \left(\operatorname{Res} \left[f(z), e^{\pi i/4}\right] + \operatorname{Res} \left[f(z), e^{3 \pi i/4}\right]\right) \right) - \Im \left(  \, 2 \pi i \left(\operatorname{Res} \left[g(z), e^{-\pi i/4}\right] + \operatorname{Res} \left[g(z), e^{-3 \pi i/4}\right]\right) \right)  \\ &= \small\Im \left(  2 \pi i \left(\frac{i e^{-3 \pi i/4} \operatorname{Li}_{2}(i)}{4} - \frac{i e^{- \pi i/4}\operatorname{Li}_{2}(-1)}{4} \right)\right)- \Im \left(  2 \pi i \left(- \frac{ie^{3 \pi i /4}\operatorname{Li}_{2}(-1)}{4} + \frac{ie^{\pi i /4} \operatorname{Li}_{2}(i)}{4} \right) \right) \\ &= \Im \left( \frac{\pi}{\sqrt{2}} \, \operatorname{Li}_{2}(i)  \right)  + \Im \left(\frac{\pi i }{\sqrt{2}} \, \operatorname{Li}_{2}(i)  \right)- \frac{\pi}{\sqrt{2}} \, \operatorname{Li}_{2}(-1) \\ &=  \frac{\pi }{\sqrt{2}} \, \operatorname{Ti}_{2}(1) + \frac{ \pi}{4 \sqrt{2}} \,  \operatorname{Li}_{2}(-1) - \frac{\pi}{\sqrt{2}} \, \operatorname{Li}_{2}(-1) \\ &= \frac{\pi G}{\sqrt{2}} - \frac{\pi^{3}}{48 \sqrt{2}}+ \frac{\pi^{3}}{12 \sqrt{2}} \\ &= \frac{\pi G}{\sqrt{2}} + \frac{\pi^{3}}{16 \sqrt{2}} . \end{align}$$
