I am looking for an efficient way to evaluate
$$\int_{1/4}^1 \int_{\sqrt{x-x^2}}^{\sqrt x}\left(\frac{x^2-y^2}{x^2}\right)\,\mathrm{d}y\,\mathrm{d}x$$
I have
\begin{align} I&=\int_{1/4}^1\left(\int_{\sqrt{x-x^2}}^{\sqrt x}\,dy-\frac{1}{x^2}\int_{\sqrt{x-x^2}}^{\sqrt x}y^2\,dy\right)\,dx \\&=\int_{1/4}^1\left[\sqrt x-\sqrt{x-x^2}-\frac{(\sqrt{x})^3-(\sqrt{x-x^2})^3}{3x^2}\right]\,dx \\&=\int_{1/4}^1\sqrt x\,dx-\int_{1/4}^1 \sqrt{x-x^2}\,dx-\frac{1}{3}\int_{1/4}^1 \frac{1}{\sqrt x}\,dx+\frac{1}{3}\int_{1/4}^1\frac{(x-x^2)^{3/2}}{x^2}\,dx \\&=\frac{2}{3}\left(1-\frac{1}{4^{3/2}}\right)-\frac{1}{3}-\color{darkred}{\int_{1/4}^1 \sqrt{x-x^2}\,dx}+\frac{1}{3}\color{green}{\int_{1/4}^1\frac{(x-x^2)^{3/2}}{x^2}\,dx} \end{align}
Using this answer,
$$\color{darkred}{\int_{1/4}^1 \sqrt{x-x^2}\,dx}=\frac{1}{4}\int_{-\pi/6}^{\pi/2}\cos^2\,dt=\frac{1}{8}\int_{-\pi/6}^{\pi/2}(1+\cos 2t)\,dt=\frac{1}{96}(3\sqrt 3+8\pi)$$
For $\color{green}{\int_{1/4}^1\frac{(x-x^2)^{3/2}}{x^2}\,dx}$ or even the indefinite integral, nothing comes to mind off the top of my head.
In a different approach, if I try to change the order of integration right at the start, it complicates matters for me to rewrite the region $$\sqrt{x-x^2}<y<\sqrt x\,,\,1/4<x<1$$
keeping a separate range of $y$ free of $x$ and bounding $x$ with $y$.
Any suggestion regarding specific substitution or change of variables would also be helpful.
The substitution $u=\sqrt{x}$ gives
$$ \int_{1/4}^1 \frac{(x-x^2)^{3/2}}{x^2}dx = \int_{1/4}^1 \frac{(1-x)^{3/2}}{\sqrt{x}}dx = 2\int_{1/2}^1 (1-u^2)^{3/2} du $$
Then $u=\sin t$ should do the trick.