$\int \frac{dx}{{\cos }^{3}x\sqrt{4\sin x\cos x}}=\int \frac{dx}{2{\cos }^{4}x\sqrt{\tan x}}$

Let, $\tan x=t^2\Rightarrow {\sec }^{2}x=1+t^4$

And ${\sec }^{2}xdx=2tdt$

$\therefore I= \int \frac{{\sec }^{4}xdx}{2\sqrt{\tan x}}= \int \frac{{\sec }^{2}x.{\sec }^{2}xdx}{2\sqrt{\tan x}}$

$=\int \frac{\left(1+t^4\right)2t dt}{2t}=\int \left(1+t^4\right)dt$

$=t+\frac{t^5}{5}+k$, where $k$ is a constant of integration.

$= \sqrt{\tan x}+\frac{1}{5}{\tan }^{\frac{5}{2}}x+k$ 

Where, $t= \sqrt{\tan x}$.

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

$A+B+C=\frac{16}{5}$