Calculating a crazy trigonometric integral What is the fastest and best way to do this crAZy integral? 
$$ \int\frac{1-\tan^4\theta d\theta}{\tan^{\frac{9}{6}}\theta(\sec^2 \theta+\tan\theta)^{\frac{1}{2}}+\tan^{\frac{10}{6}}\theta (\sec^2 \theta+ \tan\theta)^{\frac{1}{3}}}$$
I tried substituting:
$ \tan x = t$
but that comes out to something more ugly..
 A: $$ \int\frac{(1-\tan^4\theta) d\theta}{\tan^{\frac{9}{6}}\theta(\sec^2 \theta+\tan\theta)^{\frac{1}{2}}+\tan^{\frac{10}{6}}\theta (\sec^2 \theta+ \tan\theta)^{\frac{1}{3}}}$$
$$ \rightarrow \int \frac{ \sec^2 \theta (1- \tan^2 \theta) d\theta}{  \tan^{\frac{3}{2}} \theta \left[\sec^2 \theta + \tan \theta \right]^{\frac{1}{2}} + \tan^{ \frac{5}{3} } \theta  \left[\sec^2 \theta + \tan \theta \right]^{\frac{1}{3}}}$$
$$ \rightarrow \int \frac{ \sec^2 \theta (1- \tan^2 \theta) d\theta}{ \tan^2 \theta \left[ ( \frac{\sec^2 \theta}{\tan \theta} +1 )^{\frac{1}{2}}  +  ( \frac{\sec^2 \theta}{\tan \theta} +1 )^{\frac{1}{3}}  \right]}$$
$$  k=  \frac{\sec^2 \theta}{\tan \theta} +1 $$
$$ dk = - \frac{ \sec^2 \theta (1- \tan^2 \theta)}{ \tan^2 \theta} d\theta$$
$$ \int \rightarrow -\frac{dk}{ k^{\frac{1}{2}} + k^{\frac{1}{3}}}$$
$$k^{\frac{1}{6} } = y$$
$$ dk = 6y^5 dy$$
$$ \rightarrow \int \frac{ - 6 y^5 dy}{y^3 + y^2}$$
$$=-2y^3 +3y^2-6y +6log(y+1) +C$$
And sub back $$ y= \left( \frac{\sec^2 \theta +1}{ \tan \theta} \right)^6$$
