$$\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} $$

=  $$\int {{{dx} \over {{{\cos }^3}x\sqrt {4\sin x\cos x} }}} $$

=  $$\int {{{dx} \over {2{{\cos }^4}x\sqrt {\tan x} }}} $$

Let tan x   =   t²

$$ \Rightarrow $$$$\,\,\,$$ sec²xdx = 2t dt

as   sec²x = 1 + tan²x = 1 + t⁴

=  $$\int {{{{{\sec }^4}x\,dx} \over {2\sqrt {\tan x} }}} $$

=  $$\int {{{{{\sec }^2}x\left( {{{\sec }^2}x\,dx} \right)} \over {2\sqrt {\tan x} }}} $$

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

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

=  t + $${{{t^5}} \over 5}$$ + k

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

By comparing with the given equation, we get

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

$$\therefore\,\,\,$$ A + B + C = $${{16} \over 5}$$