Calculating this integral without trigonometric functions

Question

I have a problem with this integral and all the online calculators use t=atan(a). Is there a different way without using trigonometry?

$\int\frac{r^2}{(r^2+a^2)^\frac{5}{2}}dr$

Answer 1

Indeed, the optimal substitution is:

$$ t = \arctan(r/a) \quad \quad \Rightarrow \quad \quad r = a\tan(t) \quad \quad \Rightarrow \quad \quad \text{d}r = \frac{a}{\cos^2(t)}\,\text{d}t $$

from which:

$$ \begin{aligned} \int \frac{r^2}{\sqrt{\left(r^2+a^2\right)^5}}\,\text{d}r & = \int \frac{\sin^2(t)}{a^2}\left(\cos(t)\,\text{d}t\right) \\ & = \frac{\sin^3(t)}{3\,a^2} + c \\ & = \frac{\left(\frac{r/a}{\sqrt{1+(r/a)^2}}\right)^3}{3\,a^2} + c \\ & = \frac{r^3}{3\,a^2\sqrt{\left(r^2+a^2\right)^3}} + c\,. \end{aligned} $$

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Otherwise, with reference to Euler substitutions:

$$ t = r + \sqrt{r^2+a^2} \quad \quad \Rightarrow \quad \quad r = \frac{t^2-a^2}{2\,t} \quad \quad \Rightarrow \quad \quad \text{d}r = \frac{t^2+a^2}{2\,t^2}\,\text{d}t $$

we get:

$$ \begin{aligned} \int \frac{r^2}{\sqrt{\left(r^2+a^2\right)^5}}\,\text{d}r & = \int \frac{4\,t\left(t^2-a^2\right)^2}{\left(t^2+a^2\right)^4}\,\text{d}t \\ & = \int \frac{4\,t}{\left(t^2+a^2\right)^2}\,\text{d}t - \int \frac{16\,a^2\,t}{\left(t^2+a^2\right)^3}\,\text{d}t + \int \frac{16\,a^4\,t}{\left(t^2+a^2\right)^4}\,\text{d}t \\ & = -\frac{2}{t^2+a^2} + \frac{4\,a^2}{\left(t^2+a^2\right)^2} - \frac{8\,a^4}{3\left(t^2+a^2\right)^3} + k \\ & = - \frac{2\left(3\,t^4+a^4\right)}{3\left(t^2+a^2\right)^3} + k \\ & = \frac{r^3}{3\,a^2\sqrt{\left(r^2+a^2\right)^3}} - \frac{1}{3\,a^2} + k \\ & = \frac{r^3}{3\,a^2\sqrt{\left(r^2+a^2\right)^3}} + c\,. \\ \end{aligned} $$

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Otherwise, with reference to Rubi Rule 270:

$$ \int (c\,x)^m\left(a + b\,x^n\right)^p\,\text{d}x = \frac{(c\,x)^{m+1}(a+b\,x^n)^{p+1}}{a\,c\,(m+1)} + k $$

provided that $a\,c\,(m+1) \ne 0$ and $\frac{m+1}{n}+p+1 = 0$.