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
 A: 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
$$\color{blue}{I_a=\frac{\pi\sqrt{2}  }{a} \left(\tanh
   ^{-1}\left(\frac{a}{a^2+a+1}\right)+\tan ^{-1}(a)\right)}$$
A: 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
$$ \begin{align} &\int_{0}^{\infty} \frac{x^{2}\operatorname{Ti}_{2}(x^{2})}{1+x^{4}} \, \mathrm dx \\ &= \frac{1}{2} \int_{-\infty}^{\infty}  \frac{x^{2}\operatorname{Ti}_{2}(x^{2})}{1+x^{4}} \, \mathrm dx \\ &= \frac{1}{2} \, \Im \int_{-\infty}^{\infty} \frac{x^{2}\operatorname{Li}_{2}(ix^{2})}{1+x^{4}} \, \mathrm dx \\ &= \Im \int_{-\infty}^{\infty} \frac{x^{2} \operatorname{Li}_{2}(e^{\pi i /4}  x)}{1+x^{4}} \, \mathrm dx + \Im \int_{-\infty}^{\infty} \frac{x^{2} \operatorname{Li}_{2}(-e^{\pi i /4} x)}{1+x^{4}} \, \mathrm dx \\ &=  \Im \int_{-\infty}^{\infty} f(x) \, \mathrm dx + \Im \int_{-\infty}^{\infty} g(x) \, \mathrm dx  \\ &= \small\Im  \left( 2 \pi i \left(\operatorname{Res} \left[f(z), e^{\pi i/4}\right] + \operatorname{Res} \left[f(z), e^{3 \pi i/4}\right]\right) \right) - \Im \left(  \, 2 \pi i \left(\operatorname{Res} \left[g(z), e^{-\pi i/4}\right] + \operatorname{Res} \left[g(z), e^{-3 \pi i/4}\right]\right) \right)  \\ &= \small\Im \left(  2 \pi i \left(\frac{i e^{-3 \pi i/4} \operatorname{Li}_{2}(i)}{4} - \frac{i e^{- \pi i/4}\operatorname{Li}_{2}(-1)}{4} \right)\right)- \Im \left(  2 \pi i \left(- \frac{ie^{3 \pi i /4}\operatorname{Li}_{2}(-1)}{4} + \frac{ie^{\pi i /4} \operatorname{Li}_{2}(i)}{4} \right) \right) \\ &= \Im \left( \frac{\pi}{\sqrt{2}} \, \operatorname{Li}_{2}(i)  \right)  + \Im \left(\frac{\pi i }{\sqrt{2}} \, \operatorname{Li}_{2}(i)  \right)- \frac{\pi}{\sqrt{2}} \, \operatorname{Li}_{2}(-1) \\ &=  \frac{\pi }{\sqrt{2}} \, \operatorname{Ti}_{2}(1) + \frac{ \pi}{4 \sqrt{2}} \,  \operatorname{Li}_{2}(-1) - \frac{\pi}{\sqrt{2}} \, \operatorname{Li}_{2}(-1) \\ &= \frac{\pi G}{\sqrt{2}} - \frac{\pi^{3}}{48 \sqrt{2}}+ \frac{\pi^{3}}{12 \sqrt{2}} \\ &= \frac{\pi G}{\sqrt{2}} + \frac{\pi^{3}}{16 \sqrt{2}} . \end{align}$$
A: Just like it was mentioned in the comments, the integrand in question is strictly positive. I don't know how Wolfram Alpha is interpreting your input. (Maybe it is interpreting it as the Generalized Inverse Tangent Function?)
Your $F'(t)$ is correct. I simplified it down to
$$F'(t) = \frac{\pi}{2\sqrt{2}}\frac{\sqrt{t}-1}{\sqrt{t}\left(t^{2}-1\right)}.$$
Using the Fundamental Theorem of Calculus, we can integrate both sides with respect to $t$ on the interval $\left[0,1\right]$ to get
$$\frac{\pi}{\sqrt{2}}\int_{0}^{1}\frac{\sqrt{t}-1}{\sqrt{t}\left(t^{2}-1\right)}dt.$$
Solving this integral should be straightforward. The final answer is

 \begin{align} \dfrac{\pi}{2\sqrt{2}}\left(\ln\left(2\right)+\dfrac{\pi}{2}\right). \end{align}

Hopefully, this answer helps. Please let me know if there are any questions.
