TITLE: General rule for integrals of irrational functions involving $r = \sqrt{a^2 + x^2}$
QUESTION [6 upvotes]: Browsing the List of integrals of irrational functions I found out that if
$$r=\sqrt{(a^2+x^2)}$$
then
$$\int{rdx}=\frac12(xr+a^2\ln{(x+r)})$$
$$\int{r^3dx}=\frac14xr^3+\frac38a^2xr+\frac38a^4\ln(x+r)$$
$$\int{r^5dx}=\frac16xr^5+\frac{5}{24}a^2xr^3+\frac{5}{16}a^4xr+\frac{5}{16}a^6\ln(x+r)$$
Is there any general rule for defining $$\int{r^ndx}, \quad n=2k+1, \quad k\in\mathbb{N}$$
If so, then how to derive the formula? Thank you.

REPLY [3 votes]: Assume $a\ge0$. From $r^2=a^2+x^2$ we obtain $rdr=xdx$. Therefore bu substituting we can write$$\int r^ndx=\int {r^{n+1}dr\over \sqrt{r^2-a^2}}$$Now by defining $r=a\cosh u$ we have$$I{=\int a^{n+1}\cosh^{n+1}udu\\=\left({a\over 2}\right)^{n+1}\int (e^u+e^{-u})^{n+1}du\\=\left({a\over 2}\right)^{n+1}\int \sum_{k=0}^{n+1}\binom{n+1}{k} e^{(2k-n-1)u}du\\=\left({a\over 2}\right)^{n+1}\left[\sum_{k=0\\k\ne {n+1\over 2}}^{n+1}\binom{n+1}{k} {e^{(2k-n-1)u}\over 2k-n-1}+\binom{n+1}{{n+1\over 2}}u\right]}$$also $$r=a\cosh u\iff e^u={r+x\over a}$$which by substituting yields to

$$I=\left({a\over 2}\right)^{n+1}\sum_{k=0\\k\ne {n+1\over 2}}^{n+1}\binom{n+1}{k} {\left({r+x\over a}\right)^{2k-n-1}\over 2k-n-1}+\left({a\over 2}\right)^{n+1}\binom{n+1}{n+1\over 2}\ln {r+x\over a}+C$$