# Closed form of the integral $\int_{0}^\infty \frac{\tan^{-1}{x} \tan^{-1}{\frac{x}{2}} \tan^{-1}{\frac{x}{3}}}{x^2+1} \, \text{d}x$ [closed]

This integral was found on a dead forum so I cannot obtain any further information on it.

$$\int_{0}^\infty \frac{\tan^{-1}{x} \tan^{-1}{\frac{x}{2}} \tan^{-1}{\frac{x}{3}}}{x^2+1} \, \text{d}x$$

I could not find a closed form using ISC++

The equivalent form

$$\int_0^{\frac{\pi}{2}} \theta \tan^{-1}{\left(\frac{\tan{\theta}}{2}\right)} \tan^{-1}{\left(\frac{\tan{\theta}}{3}\right)} \, \text{d}x$$

does not seem any more promising.

Perhaps a carefully considered contour could solve this problem but I do not have enough familiarity with complex analysis to come up with an argument.

## closed as off-topic by Carl Mummert, A. Goodier, The Phenotype, José Carlos Santos, SaadMar 20 '18 at 0:04

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• Paging @Cleo $\mbox{}$ – Umberto P. Feb 23 '18 at 23:40
• That looks so horrendous that you should have let it rest in peace in that forum... – DonAntonio Feb 23 '18 at 23:46
• Have you been able to find any identities that relate $\arctan(x)$ to $\arctan\left(\frac{x}{k}\right)$? – Michael McGovern Feb 24 '18 at 0:04
• @DonAntonio I guess that's why they call it Morbid Curiousity... :P – Jack Lam Feb 24 '18 at 0:11
• If you know how to compute $\int_{0}^{+\infty}\frac{\arctan(ax)}{1+x^2}\,dx$ by differentiation under the integral sign, you just have to apply a partial fraction decomposition to compute $$\int_{0}^{+\infty}\frac{x^3}{(1+x^2)(1+a^2 x^2)(1+b^2 x^2)(1+c^2 x^2)}\,dx$$ then integrate such thing over the proper parallelepiped in $da\,db\,dc$ to get a nasty combination of dilogarithms. It is not difficult, it is just boring and more suited for a CAS. – Jack D'Aurizio Feb 24 '18 at 2:05