1. **State the problem:** We need to evaluate the integral $$\int \frac{dx}{\sqrt{\cos x (3 + 2 \cos x)}}.$$\n\n2. **Rewrite the integral:** The integral is $$\int \frac{dx}{\sqrt{\cos x (3 + 2 \cos x)}} = \int \frac{dx}{\sqrt{3 \cos x + 2 \cos^2 x}}.$$\n\n3. **Substitution:** Let $u = \cos x$. Then, $du = -\sin x \, dx$, or $dx = \frac{-du}{\sin x}$. However, this substitution is complicated because of the $\sin x$ in the denominator.\n\n4. **Alternative approach:** Use the substitution $t = \sin x$, so $dt = \cos x \, dx$, and $dx = \frac{dt}{\cos x}$. But this also complicates the integral.\n\n5. **Try rewriting the integral in terms of $\cos x$ and $\sin x$:** Since $\sin^2 x + \cos^2 x = 1$, we can express $\sin x = \sqrt{1 - \cos^2 x}$.\n\n6. **Rewrite the integral:** $$\int \frac{dx}{\sqrt{\cos x (3 + 2 \cos x)}} = \int \frac{dx}{\sqrt{3 \cos x + 2 \cos^2 x}}.$$\n\n7. **Use substitution:** Let $u = \cos x$, then $du = -\sin x \, dx$, so $dx = \frac{-du}{\sin x} = \frac{-du}{\sqrt{1 - u^2}}$.\n\n8. **Rewrite the integral in terms of $u$:** $$\int \frac{dx}{\sqrt{3 u + 2 u^2}} = \int \frac{-du}{\sqrt{1 - u^2} \sqrt{3 u + 2 u^2}} = -\int \frac{du}{\sqrt{(1 - u^2)(3 u + 2 u^2)}}.$$\n\n9. **Simplify the expression under the square root:** $$\sqrt{(1 - u^2)(3 u + 2 u^2)} = \sqrt{3 u + 2 u^2 - 3 u^3 - 2 u^4}.$$\n\n10. **The integral becomes:** $$-\int \frac{du}{\sqrt{3 u + 2 u^2 - 3 u^3 - 2 u^4}}.$$\n\n11. **This integral is complicated and does not simplify easily with elementary functions. It is an elliptic integral.**\n\n**Final answer:** The integral $$\int \frac{dx}{\sqrt{\cos x (3 + 2 \cos x)}}$$ can be expressed as an elliptic integral and does not have a simple closed form in elementary functions.