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.

The given integral is \begin{align}
I=&\int_{0}^{\infty}\frac{x^{29}}{(5x^2+49)^{17}}dx\\
\ =&\frac{1}{2\cdot 5^{15}}\int_{0}^{\infty}\frac{z^{14}}{(z+49)^{17}}dz\quad(\mbox{substituting}\ 5x^2=z)\\
\ =&\frac{1}{2\cdot 5^{15}}\int_{0}^{\pi/2}\frac{(49)^{14}\tan^{28}\theta\cdot2\cdot49\cdot\tan \theta\sec^2\theta}{49^{17}\sec^{34}\theta}dz\quad(\mbox{substituting}\ z=49\tan^2\theta)\\
\ =&\frac{1}{2\cdot 5^{15}\cdot49^2}\cdot2\int_{0}^{\pi/2}\frac{\sin^{29}\theta}{\cos^{29}\theta}\cos^{32}\theta\ d\theta\\
\ =&\frac{1}{2\cdot 5^{15}\cdot 49^2}\cdot\beta\left(15,2\right)\\
\ =&\frac{\Gamma(15)\Gamma(2)}{2\cdot5^{15}\cdot49^2\cdot\Gamma(17)}\\
\ =&\frac{14!}{2\cdot5^{15}\cdot49^2\cdot16!}
\end{align}
Along similar lines of argument, the integral $I(a,b,c,d)$($a,b,c,d$ positive integres) as defined by  Hagen von Eitzen becomes(if I have not committed any mistake) $$\frac{1}{2\cdot c^d}\left(\frac{c}{b}\right)^{(n+1)/2}\beta\left(\frac{n+1}{2},d-\frac{n-1}{2}-1\right)\\ =\frac{1}{2\cdot c^{d-\frac{n+1}{2}}\cdot b^{\frac{n+1}{2}}}\cdot\frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(d-1-\frac{n-1}{2}\right)}{(d-1)!}$$
