<s>[INST] Find $\int \frac{x}{\sin^2x-3}dx$ I started like this
\begin{align*}
\int \frac{x}{\sin^2x-3}dx & =\int \frac{x\sec^2x}{\tan^2x-3\sec^2x}dx\\
& =-\int \frac{x\sec^2x}{2\tan^2x+3}dx\\
& = -\left [ \frac {x\tan^{-1}\left (\frac {\sqrt {2}\tan x}{\sqrt {3}}\right)}{\sqrt {6}}-\int  \frac {\tan^{-1}\left (\frac {\sqrt {2}\tan x}{\sqrt {3}}\right)}{\sqrt {6}}dx\right]
\end{align*}
Now i am unable to solve this further.
 [/INST] As said in comments, after integration by parts, you are left with
$$I=\int \tan ^{-1}\left(\sqrt{\frac{2}{3}} \tan (x)\right)\,dx$$ which is a monster that I should try to avoid.
In order to compute it, I should rather consider series expansions built around $x=k \pi$ and use for 
$$J_k=\int_{k\pi}^{(2k+1)\frac \pi 2} \tan ^{-1}\left(\sqrt{\frac{2}{3}} \tan (x)\right)\,dx$$ the series expansion of the integrand. This is 
$$\tan ^{-1}\left(\sqrt{\frac{2}{3}} \tan (x)\right)=t+\frac{t^3}{6}-\frac{3 t^7}{280}-\frac{t^9}{504}+O\left(t^{11}\right)$$ where $t=\sqrt{\frac{2}{3}} (x-\pi  k)$.
For illustration, this would give
$$J_0=\frac{\pi ^2}{4 \sqrt{6}}\left(1+\frac{\pi ^2}{72}-\frac{\pi ^6}{80640}-\frac{\pi ^8}{3265920} \right)\approx 1.1305$$ while the monster would give as an exact solution
$$\frac{1}{2 \sqrt{6}}\Phi \left(\frac{2}{3},2,\frac{1}{2}\right)+\frac{1}{2} \log
   \left(\frac{3}{2}\right) \tanh ^{-1}\left(\sqrt{\frac{2}{3}}\right)\approx 1.1326$$ For sure, we could have better results pushing the expansion to higher orders.
</s>