I tried to use integration by parts but it led to results which I could not integrate further. Also substituting $\sec x$ in terms of $\tan x$ or vice-versa led me to nothing proper.
Update: Taking $\cos x$ as $u$ made this an elementary problem. Thanks to @MyGlasses
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$\begingroup$Note that$$\sqrt{\sec x} \tan x = \frac{1}{\sqrt{\cos x}} \cdot \frac{\sin x}{\cos x}= \frac{\sin x}{(\cos x)^{3/2}}$$ So let $u= \cos x$ then $du= -\sin x \, dx$ and thus $$\int \sqrt{\sec x} \tan x \,dx =\int \frac{\sin x}{(\cos x)^{3/2}} \,dx = \int \frac{-1}{u^{3/2}} \, du = \frac{2}{\sqrt u}+C = 2 \sqrt{\sec x}+C$$
$\endgroup$ $\begingroup$$$\int \sqrt{\sec x} \tan x \; dx = \int \frac{\sec x \tan x}{\sqrt{\sec x}} \; dx = \int \frac{du}{\sqrt{u}} =2\sqrt{u}+C =2\sqrt{\sec x}+C. $$
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