Evaluating $\int \cos(x) \sqrt{\sin(2 x)} dx$ 
Evaluate the following indefinite integral:
$$\int \cos(x) \sqrt{\sin(2 x)} dx$$

Only hint I have is from W|A that expresses the integral in terms of a hypergeometric function and it looks rather ugly.
Can we solve it in a simpler way and get a nicer form?  Thanks.

Alternatively, rewrite
$$I=\sqrt{2}\int(1-(\sin x)^2)^\frac{1}{4}(\sin{x})^\frac{1}{2}d(\sin{x})\\=\sqrt{2}\int(1-t^2)^{\frac{1}{4}}t^\frac{1}{2}dt\\=\sqrt{2}\int\left(\frac{1}{t^2}-1\right)^\frac{1}{4}tdt \\=\frac{1}{\sqrt{2}}\int\left(\frac{1}{z}-1\right)^\frac{1}{4}dz$$
Now let
$$\frac{1}{z}-1=u^2$$
$$-\frac{dz}{z^2}=2udu$$
$$dz=-\frac{2udu}{(1+u^2)^2}=d\left(\frac{1}{1+u^2}\right)$$
$$\sqrt{2}I=\int u^\frac{1}{2}d\left(\frac{1}{1+u^2}\right)=\frac{u^\frac{1}{2}}{1+u^2}-\int\frac{d\left(u^\frac{1}{2}\right)}{1+u^2}$$
Where the last integral is equivalently $\int\frac{dv}{1+v^4}$ to which there exist various approaches.
