Is there a simple way to evaluate
S x^5 / [rt(x^2-ax)] dx ?
That is, the indefinite integral of (x^5) / [square root of (x^2-ax)].
The Attempt at a Solution
My idea was to rewrite it as x^(9/2) / [rt(x-a)] and then do the substitution u = rt(x-a).
Then you get
S 2(u^2+a)^(9/2) du
But this too is difficult to evaluate.