Evaluation of $\int\frac{1}{\sin^2 x+\sin x+1}dx$ 
Evaluation of $\displaystyle \int\frac{1}{\sin^2 x+\sin x+1}dx$

$\bf{My\; Try::}$ Using $$\; \bullet\;  x^2+x+1 = (x-\omega)\cdot (x-\omega^2)\;,$$ where $\omega,\omega^2$ are cube root of unity
So we can write Integal $$\displaystyle I = \int\frac{1}{(\sin x-\omega)\cdot (\sin x-\omega^2)}dx$$
So we get  $$\displaystyle I = \frac{1}{\omega-\omega^2}\int\frac{(\sin x-\omega^2)-(\sin x-\omega)}{(\sin x-\omega)\cdot (\sin x-\omega^2)}dx$$
So $$\displaystyle I = \frac{1}{\omega-\omega^2}\int \left[\frac{1}{\sin x-\omega}-\frac{1}{\sin x-\omega^2}\right]dx$$
Now Substitute $$\displaystyle \sin x= \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$$
Can we solve it above method our we directly put $$\displaystyle \sin x= \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$$
in $$\displaystyle \int\frac{1}{\sin^2 x+\sin x+1}dx$$ and then solve it.
Or is there is any other process by which we can solve it.
Help me , Thanks

By using Weierstrass substitution $x=2\arctan t$ the problem boils down to computing
$$ \int\frac{1+t^2}{1+2t+6t^2+2t^3+t^4}\,dt $$
through partial fraction decomposition. The roots of that palyndromic $4$th-degree polynomial are located at $t=-\frac{1}{2}\pm \frac{i \sqrt{3}}{2}-\sqrt{\frac{1}{2} \left(-3\pm i \sqrt{3}\right)}$. The remaining part is just tedious work.
