TITLE: Difficult improper integral: $\int_0^\infty \frac{x^{23}}{(5x^2+7^2)^{17}}\,\mathrm{d}x$
QUESTION [5 upvotes]: How can I find a closed-form expression for the following improper integral in a slick way?

$$\mathcal{I}= \int_0^\infty \frac{x^{23}}{(5x^2+7^2)^{17}}\,\mathrm{d}x$$

REPLY [2 votes]: $$\mathcal{I}= \int_0^{\infty} \frac{x^{23}}{(5x^2+49)^{17}}\,\mathrm{d}x$$
Let $y=\left(\frac{49b}{1-b}\right)^{1/2}$ which is the sum total of some 4 or 5 substitutions in one go, it is not advisable to use this directly but go with $y=x^2$ then $z=5+49/y$ then $a=1-5/z$ and then $b=1-a$. Note that result could be evaluated at a but the $(1-a)^{11}$ is tedious to expand so I changed the form.Anyways by beta function there is no difference.
Now:
$$\mathcal{I}= \frac{1}{2*49^5*5^{12}}\int_{0}^{1} b^{11}(1-b)^4\,\mathrm{d}b$$
Now using beta function or expanding and integrating and adding:
$$\int_{0}^{1} b^{11}(1-b)^4\,\mathrm{d}b=\frac{11!*4!}{16!}$$
So answer is:
$$\boxed{\displaystyle\large\qquad\mathcal{I}=\frac{11!*4!}{16!*2*49^5*5^{12}}=\frac1{3012333710039062500000}\qquad}$$