Solution

$$I = \displaystyle \int \frac{(2 \sin \theta + \sin 2 \theta) d\theta}{(\cos \theta - 1) \sqrt{\cos \theta + \cos^2 \theta + \cos^3 \theta}}$$
Put $$\cos \theta = x^2$$
$$\Rightarrow$$ $$- \sin \theta$$ $$d \theta = 2x dx$$
$$= 2 \displaystyle \int \frac{(1 + x^2)}{(1 - x^2)} . \frac{2x dx}{\sqrt{x^2 + x^4 + x^6}}$$
$$= 4 \displaystyle \int \frac{(1 + 1/x^2)dx}{(1/x - x) \sqrt{(1/x - x)^2 + 3}}$$
Put $$\frac{1}{x} - x = t$$
$$\Rightarrow \left ( - \frac{1}{x^2} - 1 \right ) dx = dt$$
$$= 4 \displaystyle \int \frac{dt}{t \sqrt{t^2 + 3}}$$
Again, put $$t^2 + 3 = u^2$$
$$\Rightarrow$$ $$2t$$ $$dt = 2u$$ $$du$$
$$\therefore$$ $$I = 4 \displaystyle \int \frac{- u du}{u(u^2 - 3)} = -4 \displaystyle \int \frac{du}{u^2 - 3}$$
$$= - \frac{2}{\sqrt 3} \log \left | \frac{u - \sqrt 3}{u + \sqrt 3} \right | + C$$
$$= - \frac{2}{\sqrt 3} \log \left | \frac{\sqrt{t^2 + 3} - \sqrt 3}{\sqrt{t^2 + 3} + \sqrt 3} \right | + C$$
$$= - \frac{2}{\sqrt 3} \log \left | \frac{\sqrt{x^2 + 1/x^2 + 1} - \sqrt 3}{\sqrt{x^2 + 1/x^2 + 1} + \sqrt 3} \right | + C$$
$$= - \frac{2}{3} \log \left | \frac{\sqrt{\cos \theta + \sec \theta + 1} - \sqrt 3}{\sqrt{\cos \theta + \sec \theta + 1} + \sqrt 3} \right | + C$$