finding $\int {\tan^{3/2} 3x\sec 3x\,dx}$ How do I integrate this expression?
$$\int \tan^{3/2} 3x\sec 3x dx$$
 A: Notice, $$\int \tan^{3/2} 3x\sec 3x\ dx=\int \sqrt{\tan3x}.\tan3x.\sec3xdx$$

So if we let $\sec 3x=t$, $3\sec3x\tan3xdx=dt$, 
 We get,
$$\frac13\int\sqrt{\sqrt{t^2-1}}dt$$
Wolfram: 
 But as you can see, and deduce, there is not a closed form of it, so, I don't think there is an answer of your question in elementary functions. 
Edit: Lets try something. Put $t=\sin \theta;dt=\cos \theta d\theta$ 
Our integral becomes $$\int\sqrt{-\cos\theta} \cos \theta d\theta$$
$$=\int i\cos^{\frac32}\theta d\theta$$
Wolfram Says:  
We can treat $i$ as a constant, but still there won't be an integral in terms of an elementary function. This tells that even if we switch from real plane to complex plane, integral will not exist in an elementary form.
A: Well, I'll let you know that no CAS system I have access to can solve this problem. The obvious first step is to change the equation from $\tan^{\frac{3}{2}}(3x)\sec(3x)$ to $3x\tan^{\frac{3}{2}}(x)\sec(3x)$, but that yields no more results. We do know that $$\int \tan^{3/2}(x)dx$$
$$= \frac{1}{4} (2 \sqrt{2} \,\tan^{-1}(1-\sqrt{2}\,\tan^{1/2}(x))-2 \sqrt{2} \,\tan^{-1}(\sqrt{2}\,\tan^{1/2}(x)+1)+ 8\tan^{1/2}(x)+\sqrt{2} \,\log(\tan(x)-\sqrt{2} \,\tan^{1/2}(x)+1)-\sqrt{2}\, \log(\tan(x)+\sqrt{2} \,\tan^{1/2}(x)+1))+constant$$
But I haven't been able to get anywhere with this monstrosity.  
(I can't explain to you how much of my life was spent making that statement look pretty in MathJax.)
A: With some changes of variables, it is easy to obtain an Incomplete Beta function as closed form :

