Is there a quicker way to solve this integral: $\int \frac{3-\cos(x)}{(1+2\cos(x))\sin^2(x)}dx$? The integral is:
$$ \int \frac{3-\cos(x)}{(1+2\cos(x))\sin^2(x)}dx$$
This is the way I approached it:
$$ \tan\left(\frac{x}{2}\right)=u\\dx=\frac{2}{\sec^2\left(\frac{x}{2}\right)}du$$
By using trigonometric identities we get:
$$ \sin(x)=\frac{2u}{1+u^2};\ \cos(x)=\frac{1-u^2}{1+u^2};\ \sec^2\left(\frac{x}{2}\right)=1+u^2
$$
Therefore the integral now becomes:
$$ 2\int \frac{3-\frac{1-u^2}{1+u^2}}{\left(1+2\left(\frac{1-u^2}{1+u^2}\right)\right)\left(\frac{2u}{1+u^2}\right)^2(1+u^2)}du=$$ $$\int\frac{(1+2u^2)(1+u^2)}{u^2(3-u^2)}du$$ By dividing the two polynomials we get: $$\int\left(-2-\frac{9u^2+1}{u^2(3-u^2)}\right)du$$ Using partial fractions we get to the simplified form: $$\int\left(-2-\frac{1}{3u^2}+\frac{28}{3(3-u^2)}\right)du$$ $$-2u-\frac{1}{3u}+\frac{28\sqrt3}{9}\int\frac{1}{1-\left(\frac{u}{\sqrt3}\right)^2}du \\ -2u-\frac{1}{3u}+\frac{28\sqrt3\tanh^{-1}{\left(\frac{u}{\sqrt3}\right)}}{9}+C$$
By substituting back in for $u$, we get the solution: $$ \bbox[5px,border:2px solid black]{\frac{28\sqrt3\tanh^{-1}{\left(\frac{\tan\left(\frac{x}{2}\right)}{\sqrt3}\right)}}{9}-2\tan\left(\frac{x}{2}\right)-\frac{1}{3}\cot\left(\frac{x}{2}\right)+C}$$
My question is, as you can understand from the title, is there any easier and faster way to solve this integral? If so, how? Thank you.

You can simplify the calculation by using a trigonometric identity and partial fractions before doing the Weierstrass substitution: since $\sin^2{x}=1-\cos^2{x}$, the initial fraction simplifies to
$$ \frac{3-\cos{x}}{(1+2\cos{x})(1+\cos{x})(1-\cos{x})} = \frac{14/3}{1+2\cos{x}} + \frac{-2}{1+\cos{x}} + \frac{-1/3}{1-\cos{x}}. $$
The middle fraction simplifies to a multiple of $\frac{1}{2}\sec^2{\left(\frac{1}{2}x\right)}$, which easily integrates to $-\tan{\frac{1}{2}x}$, and similarly the last fraction is a multiple of $\frac{1}{2}\csc^2{\left(\frac{1}{2}x\right)}$, which integrates to $-\cot{\frac{1}{2}x}$; the other fraction really does need the Weierstrass substitution, but at least the algebra is now much simpler!
