<s>[INST] interesting Integral , alternative solution. Show the following relation:
$$\int_{0}^{\infty} \frac{x^{29}}{(5x^2+49)^{17}} \,\mathrm dx = \frac{14!}{2\cdot 49^2 \cdot 5^{15 }\cdot 16!}.$$   
I came across this intgeral on a physics forum and solved by  (1) making a substitution (2) finding a recursive formula (with integration by parts).
I would like to see a clever alternative solution to this one, since the only way I would solve such an integral is by finding a recursive formula. But I'm sure there are other ways/methods to do it (maybe more elegant) that  I want to discover.
 [/INST] $$I_{(n,m)}=\int_0^\infty \frac{x^n}{(ax^2+b)^m} \,\mathrm dx$$
$$=x^{n-1}\int_0^\infty\frac{xdx}{(ax^2+b)^m}-\int_0^\infty\left(\frac{d(x^{n-1})}{dx}\int\frac{xdx}{(ax^2+b)^m}\right)dx $$
$$=x^{n-1}\frac1{-2a(m-1)(ax^2+b)^{m-1}}\big|_0^\infty-\frac{n-1}{-2a(m-1)}\int_0^\infty \frac{x^{n-2}}{(ax^2+b)^{m-1}}dx$$
$\lim_{x\to\infty}x^{n-1}\frac1{(ax^2+b)^{m-1}}=\lim_{x\to\infty}\frac1{aO(x^{2m-2-(n-1)})}=0$ if $2m-2-(n-1)=2m-n-1\ge1\iff n\le 2m-2$
$$\implies I_{(n,m)}=\frac{n-1}{2a(m-1)}I_{(n-2,m-1)}\text{if } n\le 2m-2$$
We have $n=29,m=17$ So, we start with $n=2m-5<2m-2$
$$\implies I_{(2m-5,m)}=\frac{2m-5-1}{2a(m-1)}I_{(2m-5-2,m-1)}=\frac{2m-3}{a(m-1)}I_{(2m-7,m-1)}$$
Replacing $m$ with $m-1,$
$$\implies I_{(2m-5,m)}=-\frac{2m-5-1}{2a(m-2)}I_{(2m-5-2,m-1)}=-\frac{m-3}{a(m-2)}I_{(2m-7,m-1)}$$
$$m=17\implies I_{(29,17)}=-\frac{14}{16a}I_{(27,16)}$$
$$m=16\implies I_{(27,16)}=-\frac{13}{15a}I_{(25,15)}$$
$$\cdots$$
$$m=4\implies I_{(3,4)}=-\frac1{3a}I_{(1,3)}$$
Now, $$I_{(1,3)}=\int_0^\infty \frac x{(ax^2+b)^3}dx=\frac1{2a}\int_b^\infty \frac {du}{u^3}=\frac{1}{2a(-2)u^2}\big|_b^\infty=\frac1{4ab^2}$$
$$\implies I_{(29,17)}=\frac{14}{(16\cdot15\cdots4\cdot3)\cdot a^{14}\cdot 4ab^2}=\frac{14!2!}{16!\cdot4a^{15}b^2}=\frac{14!}{16!\cdot2a^{15}b^2}$$
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