If $l=\int {\sec }^{2}x\cos e{c}^{4}xdx=A{\cot }^{3}x+B\tan x+C\cot x+D,$ then 

$I=\int {\sec }^{2}x\cos e{c}^{4}xdx=\int \frac{{({\sin }^{2}x+{\cos }^{2}x)}^2}{{\cos }^{2}x{\sin }^{4}x}dx$

$=\int \frac{{\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x}{{\cos }^{2}x{\sin }^{4}x}dx$

$=\int ({\sec }^{2}x+2\cos e{c}^{2}x+\frac{{\cos }^{2}x}{{\sin }^{4}x})dx$

$=\tan x-2\cot x+\int {\cot }^{2}x\cos e{c}^{2}xdx$

$=\tan x-2\cot x-\frac{{\cot }^{3}x}{3}+D$