# Evaluating $\int \sqrt{\frac{x^a}{(x+b)^c}} dx$ for $a,b,c\in\mathbb N$ and $a \lt c$

For $a,b,c\in\mathbb N$ are they any procedures how to calculate the following indefinite integral? $$\int \sqrt{\frac{x^a}{(x+b)^c}} dx$$

it is also possible to assume that $a \lt c$ but it would be great if it works more general.

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The general solution requires the Gauss Hypergeometric Function; I don't believe that will change for $a<c$. Wolfram gives this solution: wolframalpha.com/input/?i=sqrt%28x%5Ea%2F%28x%2Bb%29%5Ec%29 – Eric Stucky May 28 '12 at 11:00

It seems in any particular case, Maple knows how to do this. $$\int \frac{x^{\frac{3}{2}}}{(x + 5)^{\frac{5}{2}}} d x = \frac{-2\sqrt{5} x^{\frac{5}{2}} \Biggl(4 x^{\frac{7}{2}} + 15 x^{\frac{5}{2}} - 15 \sqrt{5} \operatorname{arcsinh} \biggl(\frac{\sqrt{5} \sqrt{x}}{5}\biggr) x^{2} \biggl(\frac{x}{5} + 1\biggr)^{\frac{3}{2}}\Biggr)}{ 75 x^{\frac{9}{2}} \biggl(\frac{x}{5} + 1\biggr)^{\frac{3}{2}}}$$ So I assume you can get some reduction formulas through integration by parts, then do a few beginning cases, like $a=1,2$, $c=1,2$
added substitution $y=\mathrm{arcsinh}\sqrt{x/b}$ leads to
$$\int \frac{x^{\bigl(\frac{a}{2}\bigr)}}{(x + b)^{\bigl(\frac{c}{2}\bigr)}} d x = 2 b^{\Bigl(1 + \frac{a}{2} - \frac{c}{2}\Bigr)} \int {\operatorname{sinh} (y)^{(1 + a)}}{\operatorname{cosh} (y)^{(1-c)}} d y$$ and the method for this trig-type integral is known.