Improper Integral from Gradshteyn and Ryzhik [duplicate]

This is the integral one can find in the Introduction of 'Special Integrals of Gradshteyn and Ryzhik the Proofs - Volume I' by Victor H. Moll:

$$\int_0^{+\infty} \frac{dx}{(1+x^2)^{3/2} \left[ \phi(x) + \sqrt{\phi(x)} \right]^{1/2}}, \quad \phi(x) = 1 + \frac{4x^2}{3(1+x^2)^2} \;.$$ The author doesn't know the final answer. It is claimed that it is $\pi / 2 \sqrt{6}$, though numerical integration contradicts this.

Any ideas how to solve it or where to find clues?

marked as duplicate by David H, Joel Reyes Noche, Joey Zou, Henrik, Daniel W. FarlowAug 10 '16 at 22:49

• I agree with you that there is a problem somewhere since numerical integration leads to $0.66637711426883385640$ and inverse symbolic calculators do not find anything. – Claude Leibovici Aug 6 '16 at 14:45
• The $4/3$ ratio hints that the triple angle formula might come at play. Hence i would make a hyperbolic substitution and adjust the function $\phi(x)$ accordingly. I'll try that when i get home. – nospoon Aug 6 '16 at 16:01
Just out of curiosity, I worked numerically$$I(k)=\int_0^{\infty } \frac{dx}{\left(x^2+1\right)^{3/2} \sqrt{\phi (x)+\sqrt{\phi (x)}}}\qquad, \qquad \phi(x) = 1 + \frac{kx^2}{(1+x^2)^2}$$ and tried to find $k$ such that $$I(k)=\frac{\pi }{2 \sqrt{6}}$$ The closest value I found is $k_*=0.5923509316314110643$ which inverse symbolic calculators did not find any equivalent. $$I(k_*)\approx 0.6412749150809320477747$$ $$\frac{\pi }{2 \sqrt{6}}\approx0.6412749150809320477720$$