Method for evaluating integrals with squared roots in denominator?

Question

In numerical astrophysics, one uses density power-law profiles as an approximation for dark matter halos and spherically stellar system to derive characteristics of that system. However, solving these equation analytically can be challenging. I am evaluating one of these density profiles and trying to replicate results from a paper. Unfortunately I am stumped on the process of efficiently evaluating one integral: $$ \int_{r}^{\infty}\frac{dr'}{\sqrt{r'}(r'+a)^{3/2}} $$ Quickly running it through wolfram-alpha and mathematica does give it the correct from $$ \int_{r}^{\infty}\frac{dr'}{\sqrt{r'}(r'+a)^{3/2}} = \left[ \frac{2}{a} \frac{\sqrt{r'}}{\sqrt{r'+a}} \right]_{r}^{\infty} = \frac{2}{a}\left[1-\frac{\sqrt{r}}{\sqrt{r+a}}\right] $$

But I do not know how this software is solving this. I tried evaluating it as a contour via Cauchy's theorem with no avail and tried to apply several unique u-substitutions. Surely it is possible to evaluate analytically, but I am stumped on the process.

** If I am breaking any rules with this post, let me know and I will change it.

Answer 1

Consider indefinite integral:

$$\int \frac{dx}{\sqrt{x}(x+a)^{3/2}}$$

Obvious substitution would be:

$$x=y^2$$

$$\int \frac{dx}{\sqrt{x}(x+a)^{3/2}}=2 \int \frac{dy}{(y^2+a)^{3/2}} \color{blue}{=}$$

Let's make another substitution:

$$y^2+a=z^2, \qquad y=\sqrt{z^2-a}, \qquad dy=\frac{z dz}{\sqrt{z^2-a}}$$

$$\color{blue}{=}2 \int \frac{z dz}{z^3 \sqrt{z^2-a}} =2 \int \frac{dz}{z^2 \sqrt{z^2-a}}\color{blue}{=}$$

Nothing good so far, but let's try another substitution:

$$z=\frac{1}{t} ,\qquad dz= -\frac{dt}{t^2}$$

$$\color{blue}{=} -2 \int \frac{dt}{\sqrt{\frac{1}{t^2}-a}}=-2 \int \frac{t dt}{\sqrt{1-at^2}}=-\int \frac{dt^2}{\sqrt{1-at^2}}=-\frac{1}{a} \frac{dat^2}{\sqrt{1-at^2}}\color{blue}{=}$$

Finally substituting:

$$at^2=p$$

$$\color{blue}{=}-\frac{1}{a} \int \frac{dp}{\sqrt{1-p}}=\frac{1}{a} \int \frac{d(1-p)}{\sqrt{1-p}}=\frac{1}{a} \int \frac{dq}{\sqrt{q}}=\frac{2}{a} \sqrt{q}+const$$

Now getting back through all our substitions:

$$q=1-p=1-at^2=1-\frac{a}{z^2}=1-\frac{a}{y^2+a}=1-\frac{a}{x+a}=\frac{x}{x+a}$$

So we get:

$$\int \frac{dx}{\sqrt{x}(x+a)^{3/2}}=\frac{2}{a} \sqrt{\frac{x}{x+a}}+const$$

Answer 2

Consider the following integral. $$\int \frac{dr}{\sqrt{r}(r+a)^{3/2}}$$

Lets start with $r = a\tan^2 (u)$ (this kills two birds with one stone) and $dr = 2\tan (u) \sec^2 (u) du$

$$\int \frac{2\tan (u) \sec^2 (u) du}{\sqrt{a} \tan(u) (a \tan^{2} u+a)^{3/2}} = \int \frac{2\sec^2 (u) du}{a^2 \sec^3(u)} du\\ = \frac{2 \sin(u)}{a^2} +C$$

Simplify it and convert it in terms of $r$.