Every method I try, leads to a more complicated situation or back somewhere along the path. Here is the expression:
$$\int\sqrt\frac{a\pi+x}{a(a\pi-x)}dx$$
What is the solution for this? is there any?
[updates]
I. U-substitution: $$\int\sqrt\frac{a\pi+x}{a(a\pi-x)}dx = \frac{1}{\sqrt a}\int f(u)u'dx$$ where $u=\frac{a\pi+x}{a\pi-x}$ AND $f(u)=u^{1/2}$
II. Integration by parts
$$\int\sqrt\frac{a\pi+x}{a(a\pi-x)}dx = 1/\sqrt a\int f(x)g'(x) dx$$ where $f(x)=\sqrt{a\pi+x}$ AND $g'(x)=(\sqrt {a\pi-x})^{-1}$
$$g(x)=-2(\sqrt{a\pi-x})$$ $$f'(x)=\frac{1}{2\sqrt{x+a\pi}}$$ $$so:$$ $$\frac{1}{\sqrt a}\int \sqrt\frac{a\pi+x}{a\pi-x} dx = \frac{1}{\sqrt a} \biggl(-2\sqrt{(a\pi)^2-x^2}+\int \sqrt\frac{a\pi-x}{a\pi+x} dx \biggr)$$
In this case, it looks like we are half the way to get back from where we've started...
$${\displaystyle\int}\sqrt{\dfrac{x+{\pi}a}{a\left({\pi}a-x\right)}}\,\mathrm{d}x$$ $$=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{\sqrt{a}}}}{\displaystyle\int}\dfrac{\sqrt{x+{\pi}a}}{\sqrt{{\pi}a-x}}\,\mathrm{d}x$$
Substitute $$u=\dfrac{\sqrt{{\pi}a-x}}{\sqrt{x+{\pi}a}}$$ $$\mathrm{d}x=\dfrac{1}{-\frac{1}{2\sqrt{{\pi}a-x}\sqrt{x+{\pi}a}}-\frac{\sqrt{{\pi}a-x}}{2\left(x+{\pi}a\right)^\frac{3}{2}}}\,\mathrm{d}u$$
$$=-\class{steps-node}{\cssId{steps-node-2}{4{\pi}\sqrt{a}}}{\cdot}{\displaystyle\int}\dfrac{1}{\left(u^2+1\right)^2}\,\mathrm{d}u$$
$${\displaystyle\int}\dfrac{1}{\left(u^2+1\right)^2}\,\mathrm{d}u=\dfrac{u}{2\left(u^2+1\right)}+\class{steps-node}{\cssId{steps-node-6}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u$$ Can you finish it from here?