Computation of $\int_{0}^{\infty}\frac{x}{\left(1+y^2x^2\right)\left(1+a^2x^2 \right)}dx$ The following integral is simple to compute
$$I=\int_{0}^{\infty}\frac{\arctan(x)}{1+x^2}dx \,\,\tag{1}$$
letting $$u=\arctan x \Rightarrow du=\frac{dx}{1+x^2}$$
$$\int\frac{\arctan(x)}{1+x^2}dx=\int u \, du = \frac{u^2}{2}$$
Therefore
$$\int_{0}^{\infty}\frac{\arctan(x)}{1+x^2}dx=\frac{\arctan^2 (x)}{2}\Big|_{0}^{\infty}=\frac{\pi^2}{8}$$

But if we introduce a parameter $a$ in (1) such that the integral becomes
$$I(a)=\int_{0}^{\infty}\frac{\arctan(ax)}{1+x^2}dx \,\,\tag{2}$$
the same technique above does not apply anymore.
My attempt for this integral is differentiating (2) w.r. to a
$$I^{\prime}(a)=\int_{0}^{\infty}\frac{x}{\left(1+x^2\right)\left(1+(ax)^2 \right)}dx \,\,$$
From this point I got stuck, altough I tried one more step using the fact that
$$\int_{0}^{\infty}e^{-xt}\cos(t)dt=\frac{x}{1+x^2}$$
$$I^{\prime}(a)=\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-xt}\cos(t)}{\left(1+(ax)^2 \right)}\,dt\,dx $$
$$I^{\prime}(a)=\int_{0}^{\infty}\cos(t)\int_{0}^{\infty}\frac{e^{-xt}}{\left(1+(ax)^2 \right)}\,dx\,dt $$
But it seems too hard.

Alternatively, integrating by parts we obtain
$$I(a)=\int_{0}^{\infty}\frac{\arctan(ax)}{1+x^2}dx=-a\int_{0}^{\infty}\frac{\arctan(x)}{1+(ax)^2}dx$$
now using the following integral representation
$$\arctan (x)=\int_{0}^{1}\frac{x}{1+x^2y^2}dy$$
$$I(a)=-a\int_{0}^{\infty}\int_{0}^{1}\frac{x}{\left(1+y^2x^2\right)\left(1+a^2x^2 \right)}dy \,dx$$
$$I(a)=-a\int_{0}^{1}\int_{0}^{\infty}\frac{x}{\left(1+y^2x^2\right)\left(1+a^2x^2 \right)}dx \, dy$$

I dont know the solution of this integral, however I saw these results below and I wanted to proof them
$$\int_{0}^{\infty} \frac{\arctan \frac{x}{\phi}}{1+x^{2}} d x=\frac{\pi^{2}}{12}+\frac{3 \ln ^{2} \phi}{4} $$
$$\int_{0}^{\infty} \frac{\arctan \phi x}{1+x^{2}} d x =\frac{\pi^{2}}{6}-\frac{3 \ln ^{2} \phi}{4}$$
 A: Note that
$$I’(a)=\int_{0}^{\infty}\frac{xdx }{\left(1+x^2\right)\left(1+a^2x^2 \right)}
\overset{t=x^2}= \frac12\int_{0}^{\infty}\frac{dt}{\left(1+t\right)\left(1+a^2 t\right)}
= \frac{\ln a}{a^2-1}
$$
Then
\begin{align}
I (a)=&\int_0^a I’(s) ds= \int_0^a \frac{\ln s}{s^2-1} ds
\overset{s=ay}=\int_0^1 \frac{a\ln a}{a^2y^2-1}dy
+\int_0^1 \frac{a\ln y}{a^2y^2-1}dy\\
=& \frac12 \ln a\ln\frac{1-a}{1+a} +\frac12 (\text{Li}_2(a)- \text{Li}_2(-a))
\end{align}
A: I've left some ideas below. However I'm having difficulty confirming them numerically.
I'll consider the integral in your title, namely
$$F(x,y)=\int_0^\infty \frac{t}{(1+x^2t^2)(1+y^2t^2)}\mathrm{d}t$$
What we do first is use a substitution $s=t^2\implies \mathrm ds/2=t~\mathrm dt$. So,
$$F(x,y)=\frac{1}{2}\int_0^\infty\frac{\mathrm ds}{(1+x^2s)(1+y^2s)}$$
Though it's tempting to use partial fractions to split the integral up here, it fails, since the individual pieces will fail to converge. So we need to get a little more creative. What we do is expand the denominator, then complete the square.
$$(1+x^2s)(1+y^2s)=x^2y^2\left(\frac{1}{x^2y^2}+\frac{x^2+y^2}{x^2y^2}s+s^2\right)=x^2y^2\left[\left(s+\frac{x^2+y^2}{2x^2y^2}\right)^2+\left(\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}\right)\right]$$
Which leads us to make the substitution
$$z=s+\frac{x^2+y^2}{2x^2y^2}$$
Hence our integral is now
$$F(x,y)=\frac{1}{2x^2y^2}\int\limits_{\frac{x^2+y^2}{2x^2y^2}}^\infty \frac{\mathrm dz}{z^2+\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}$$
Finally now this is a well known integral:
$$F(x,y)=\frac{1}{2x^2y^2\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\arctan\left(\frac{z}{\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\right)~\Bigg|^{z=\infty}_{z=\frac{x^2+y^2}{2x^2y^2}}$$
So
$$F(x,y)=\frac{1}{2x^2y^2\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\left[\frac{\pi}{2}-\arctan\left(\frac{(x^2+y^2)/2x^2y^2}{\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\right)\right]$$
