Here is another solution with a generalization:
Let $r=\sinh\alpha$ and $s=\sinh\beta$. Then
$$\begin{aligned}
&\int_{0}^{\infty} \arctan\left(\frac{2rx}{1+x^2}\right)\arctan\left(\frac{2sx}{1+x^2}\right) \, \mathrm{d}x\\
&= \pi \left( \alpha \sinh\beta+\beta\sinh\alpha+(\cosh\alpha+\cosh\beta)\log\left(\frac{e^{\alpha}+e^{\beta}}{1+e^{\alpha+\beta}}\right) \right)
\end{aligned} \tag{*}$$
Proof. Let $J = J(\alpha,\beta)$ denote the right-hand side of $\text{(*)}$. Then
$$
J(0, \beta) = 0, \qquad
J_{\alpha}(\alpha, 0) = 0, \qquad
J_{\alpha\beta} = \pi \left( \frac{1+\cosh\alpha\cosh\beta}{\cosh\alpha + \cosh\beta} \right).
$$
Now let $I = I(\alpha, \beta)$ denote the left-hand side of $\text{(*)}$. Then by the substitution $x=\tan(\theta/2)$, we get
$$ I = \frac{1}{2}\int_{0}^{\pi} \frac{\arctan(\sinh\alpha\sin\theta)\arctan(\sinh\beta\sin\theta)}{1+\cos\theta} \, \mathrm{d}\theta. $$
From this, we easily check that $I$ also satisfies
$$ I(0, \beta) = 0, \qquad I_{\alpha}(\alpha, 0) = 0. $$
Moreover,
\begin{align*}
\require{cancel}
I_{\alpha\beta}
&= \frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta(1-\cos\theta)}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\theta \\
&= \frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\theta \\
&\quad - \cancelto{0}{\frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\sin\theta}\\
&= \frac{1}{2}\int_{0}^{\infty} \frac{\cosh\alpha\cosh\beta (1 + t^2)}{(t^2 + \cosh^2\alpha)(t^2 + \cosh^2\beta)} \, \mathrm{d}t \tag{$t=\cot\theta$}
\end{align*}
It is not hard to check that the last integral is equal to $J_{\alpha\beta}$. Therefore we get $I = J$.
