Evaluate $\int \frac{1}{\sqrt{1-\sin^4{x}}}dx$ $$f(x) = \int \frac{1}{\sqrt{1-\sin^4{x}}}dx$$


I tried this by breaking the denominator as $\sqrt{(\cos^2x)(1+\sin^2x)}$ and then trying to make the integral in forms of $\sec x$ and $\tan x$. But I couldn't succeed. 

Can somebody please help me out?
$$I=\int \frac{\sec{x}}{\sqrt{1+\sin^2{x}}} \; dx$$
Multiply the top and bottom by $\sec{x}$:
$$I=\int \frac{\sec^2{x}}{\sqrt{\tan^2{x}+\sec^2{x}}} \; dx$$
$$I=\int \frac{\sec^2{x}}{\sqrt{2\tan^2{x}+1}} \; dx$$
Let $u=\tan{x}$:
$$I=\int \frac{du}{\sqrt{2u^2+1}}$$
Let $t=u\sqrt{2}$:
$$I=\frac{\sqrt{2}}{2} \int \frac{dt}{\sqrt{t^2+1}}$$
Let $t=\tan{w}$:
$$I=\frac{\sqrt{2}}{2} \int \sec{w} \; dw$$
$$I=\frac{\sqrt{2}}{2} \ln {\big | \sec{w}+\tan{w} \big |}+C$$
$$I=\frac{\sqrt{2}}{2} \ln {\big | \sqrt{1+2\tan^2{x}}+\sqrt{2}\tan{x} \big |}+C$$
Taking $t = \tan \frac{x}{2}$ and having $\sin x = \frac{2t}{1+t^2}$ we come to irrational function.