Very spookie integral involving powers of 3/4 and hypergeomteric functions.

Question

What is the anti derivative of $(2x^2+9)^{3/4}$? I tried standard methods and even more complex methods but couldn't get it to work. I know that wolfram and maple give something in terms of hypergeometric functions but how do they arrive at this?

Answer 1

Use $x=\frac{3}{\sqrt{2}}\tan t$. \begin{align} \int \frac{dx}{(2x^2+9)^{3/4}} &= \frac{1}{\sqrt{6}}\int \frac{\sec^2t\ dt}{(\tan^2t+1)^{3/4}} \\ & = \frac{1}{\sqrt{6}}\int \frac{dt}{\sqrt{\cos t}} \\ & = \frac{1}{\sqrt{6}}\int \frac{dt}{\sqrt{1-2\sin^2 (\frac{t}{2})}} \\ & = \frac{2}{\sqrt{6}}\int \frac{du}{\sqrt{1-2\sin^2 u}} \\ & = \frac{2}{\sqrt{6}}F(x,2) \end{align}

where the answer is an elliptic integral of the first kind.