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..

• 1. Please avoid post titles that consists entirely of a formula. 2.Please use the trig function macros in MathJax. – Arturo Magidin May 26 at 19:32
• trig function macros? what are those – DDD4C4U May 26 at 19:33
• The title is literally 150 words I can not put text – DDD4C4U May 26 at 19:34
• \tan to get $\tan$ instead of $tan$; \sec, \sin, \cos. – Arturo Magidin May 26 at 19:34
• Putting $z=\tan(x)$ returned $$\int \frac{1-z^2}{\sqrt{z^2+z+1} z^{3/2}+\sqrt[3]{z^2+z+1} z^{5/3}} \, dz;$$Mathematica couldn't handle that (or the original). I'm beginning to think this is hopeless. – Integrand May 27 at 2:36

$$\int\frac{(1-\tan^4\theta) d\theta}{\tan^{\frac{9}{6}}\theta(\sec^2 \theta+\tan\theta)^{\frac{1}{2}}+\tan^{\frac{10}{6}}\theta (\sec^2 \theta+ \tan\theta)^{\frac{1}{3}}}$$

$$\rightarrow \int \frac{ \sec^2 \theta (1- \tan^2 \theta) d\theta}{ \tan^{\frac{3}{2}} \theta \left[\sec^2 \theta + \tan \theta \right]^{\frac{1}{2}} + \tan^{ \frac{5}{3} } \theta \left[\sec^2 \theta + \tan \theta \right]^{\frac{1}{3}}}$$

$$\rightarrow \int \frac{ \sec^2 \theta (1- \tan^2 \theta) d\theta}{ \tan^2 \theta \left[ ( \frac{\sec^2 \theta}{\tan \theta} +1 )^{\frac{1}{2}} + ( \frac{\sec^2 \theta}{\tan \theta} +1 )^{\frac{1}{3}} \right]}$$

$$k= \frac{\sec^2 \theta}{\tan \theta} +1$$

$$dk = - \frac{ \sec^2 \theta (1- \tan^2 \theta)}{ \tan^2 \theta} d\theta$$

$$\int \rightarrow -\frac{dk}{ k^{\frac{1}{2}} + k^{\frac{1}{3}}}$$

$$k^{\frac{1}{6} } = y$$

$$dk = 6y^5 dy$$

$$\rightarrow \int \frac{ - 6 y^5 dy}{y^3 + y^2}$$

$$=-2y^3 +3y^2-6y +6log(y+1) +C$$

And sub back $$y= \left( \frac{\sec^2 \theta +1}{ \tan \theta} \right)^6$$