Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.