If $\int \frac{1}{\cos ^3 x \sqrt{2 \sin 2 x}} d x=(\tan x)^A+C(\tan x)^B+k$, where k is a constant of integration, the A + B + C equals

$\frac{16}{5}$

Let $I=\int \frac{1}{\cos ^3 x \sqrt{2 \sin 2 x}} d x$. Then,

$I=\frac{1}{2} \int \frac{1}{\cos ^{7 / 2} x \sin ^{1 / 2} x} d x=\frac{1}{2} \int \frac{\frac{1}{\cos ^4 x}}{\frac{\cos ^{7 / 2} x \sin ^{1 / 2} x}{\cos ^4 x}} d x$

$\Rightarrow I=\frac{1}{2} \int \frac{\sec ^4 x}{\tan ^{1 / 2} x} d x=\frac{1}{2} \int \frac{1+t^2}{\sqrt{t}}$, where $t=\tan x$

$\Rightarrow I=\frac{1}{2}\left\{\frac{t^{1 / 2}}{1 / 2}+\frac{t^{5 / 2}}{5 / 2}\right\}+k \Rightarrow I=\sqrt{\tan x}+\frac{1}{5}(\tan x)^{5 / 2}+k$

∴   $A=\frac{1}{2}, B=\frac{5}{2}$ and $C=\frac{1}{5}$

Hence, $A+B+C=\frac{1}{2}+\frac{5}{2}+\frac{1}{5}=\frac{16}{5}$