How do you integrate
$$
1/(\sin2x(\tan^5x+\cot^5x))
$$
with respect to $x$?
I tried writing tan and cot in terms of sin and cos but when I take $\mathrm{LCM}$ I get powers of $10$ for $\sin$ and $\cos$ in the denominator. I can't think of how I would simplify this without the use of binomial expansion but that might make things more complicated.
Is there another method I could use to approach this problem?
Hint
I would say:
$\displaystyle \int \frac{dx}{\sin2x\cdot(tan^5x+cot^5x)} = \frac12 \int \frac{\cot x \,dx}{\cos^2x\cdot(\tan^5x+\cot^5x)}$
and substitution:
$\displaystyle \tan x = t, \frac{dx}{\cos^2x}=dt$