Generalizing $\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{\operatorname dx}{x^{2}+1} = \frac{5\pi^{2}}{96}$

Solution 1:

The following is only a partial answer, but it might be useful.

Assuming that all the parameters are positive, the integral $$ I(p,q,r) = \int_{0}^{1} \frac{\operatorname{arccot} q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \, dx $$ can be expressed in terms of $I \left(\frac{1}{q}, \frac{1}{p}, \frac{1}{r} \right)$.

$$ \small \begin{align} & \int_{0}^{1} \frac{\operatorname{arccot} q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \, dx \\ &= \int_{0}^{1} \int_{0}^{1} \frac{1}{t^{2}+p^{2}q^{2}x^{2}+q^{2}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \,dt \, dx \\&= \int_{0}^{1} \frac{(r^{2}+1)pqr}{q^{2}r^{2}+(r^{2}+1)t^{2}} \int_{0}^{1} \frac{1}{(r^{2}+1)p^{2}x^{2}+1} \, dx \, dt - \int_{0}^{1} \int_{0}^{1} \frac{pq^{3}r}{q^{2}r^{2}+(r^{2}+1)t^{2}} \frac{1}{t^{2}+p^{2}q^{2}x^{2}+q^{2}} \,dx \, dt \\ &= (r^{2}+1)pqr \, \frac{\arctan \left(\frac{\sqrt{r^{2}+1}}{qr} \right)}{qr \sqrt{r^{2}+1}} \frac{\arctan (p\sqrt{r^{2}+1})}{p \sqrt{r^{2}+1)}} - \int_{0}^{1} \frac{pq^{3}r}{q^{2}r^{2}+(r^{2}+1)t^{2}} \frac{\operatorname{arccot} \left(\frac{1}{p}\sqrt{\frac{t^{2}}{q^{2}}+1} \right)}{pq^{2}\sqrt{\frac{t^{2}}{q^{2}}+1}} \, dt \\ &= (r^{2}+1)pqr \, \frac{\arctan \left(\frac{\sqrt{r^{2}+1}}{qr} \right)}{qr \sqrt{r^{2}+1}} \frac{\arctan (p\sqrt{r^{2}+1})}{p \sqrt{r^{2}+1)}} - \int_{0}^{1} \frac{\frac{1}{pqr}}{\frac{1+r^{2}}{r^{2}} \frac{t^{2}}{q^{2}}+1}\frac{\operatorname{arccot} \left(\frac{1}{p}\sqrt{\frac{t^{2}}{q^{2}}+1} \right)}{p\sqrt{\frac{t^{2}}{q^{2}}+1}} \, dt \\ &= (r^{2}+1)pqr \, \frac{\arctan \left(\frac{\sqrt{r^{2}+1}}{qr} \right)}{qr \sqrt{r^{2}+1}} \frac{\arctan (p\sqrt{r^{2}+1})}{p \sqrt{r^{2}+1)}} - I \left(\frac{1}{q}, \frac{1}{p}, \frac{1}{r} \right).\end{align}$$

And by making the substitution $u= \frac{1}{x}$ followed by the substitution $w^{2}= p^{2}+u^{2}$, one can show that

$$ \begin{align} &\int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \,dx \\ &= \frac{\pi}{2} \int_{0}^{1} \frac{1}{ q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} dx - \int_{0}^{1} \frac{\text{arccot} \, q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \,dx \\ &= \frac{\pi}{2} \, \text{arctan} \left(\frac{pr}{\sqrt{p^{2}+1}} \right) -I(p,q,r). \end{align}$$