Integrate $\int (1+\alpha^{2})^{-3/2} \sin \theta d \theta$where $\alpha = \cos \theta + a \sin \theta$ with a constant $a$

Integrate $$\int (1+\alpha^{2})^{-3/2} \sin \theta d \theta$$where $\alpha = \cos \theta + a \sin \theta$ with a constant $a$.

How could I possibly do that? Trigonometrical manipulations? Or integration parts?

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Hint Making the change of variables $\theta=\arctan(t)$ casts the integral to the form
$$\int \!{\frac {t}{ \left( 2+2\,at+({a}^{2}+1){t}^{2} \right) ^{3/2 }}}{dt}=\frac{1}{\alpha}\int \!{\frac {t}{ \left( (t+\frac{a}{\alpha})^2+\frac{a^2+2}{\alpha} \right) ^{3/2 }}}{dt}\,.$$