I am stuck!

$\displaystyle \int x^{3}\cos{\tan^{-1}{x}}$

I tried

$\displaystyle \mbox{Let } u=\tan^{-1}{x}, du=\frac{1}{x^{2} +1}$

Then with some substituting,

$\displaystyle \int x^{3}\cos{\tan^{-1}{x}} = \int\sin^{3}{u}$

but I don't know how to integrate the second form any more than the first. Can anyone explain how to do this?