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

Let I
\(\begin{aligned}
& =\int \frac{d x}{\cos ^3 x \sqrt{2 \sin 2 x}} \\
& =\int \frac{d x}{\cos ^3 x \sqrt{2(2 \sin x \cos x)}} \\
& =\frac{1}{2} \int \frac{\sec ^3 x}{\sqrt{\sin x \cos x}} \mathrm{~d} x \\
& =\frac{1}{2} \int \frac{\sec ^4 x}{\sqrt{\tan x}} \mathrm{~d} x
\end{aligned}\)
Put \(\tan x=\mathrm{t} \Rightarrow \sec ^2 x \mathrm{~d} x=\mathrm{dt}\)
\(\begin{aligned}
I & =\frac{1}{2} \int \frac{1+t^2}{\sqrt{t}} d t \\
& =\frac{1}{2} \int t^{-\frac{1}{2}} d t+\frac{1}{2} \int t^{\frac{3}{2}} d t
\end{aligned}\)
\(\begin{aligned} & =\frac{1}{2}\left(\frac{\mathrm{t}^{\frac{1}{2}}}{\frac{1}{2}}\right)+\frac{1}{2}\left(\frac{\mathrm{t}^{\frac{5}{2}}}{\frac{5}{2}}\right)+\mathrm{k} \\ & =(\tan x)^{\frac{1}{2}}+\frac{1}{5}(\tan x)^{\frac{5}{2}}+\mathrm{k} \\ \therefore \quad A & =\frac{1}{2}, B=\frac{5}{2}, C=\frac{1}{5} \\ \Rightarrow & A+B+C=\frac{16}{5}\end{aligned}\)