"Milk" the integral $\int_0^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx$ I found the following integral in chapter $13$ of Irresistible Integrals, and I would like to see which conclusions you can reach from it. My goal in asking this question is to see which methods I can employ in the future to generalize/"milk" cool integrals like this. I admit this post is very similar to the original "Integral Miking" post, but since this post is concerning a specific integral, it is not a duplicate.
\begin{align}
\int_0^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm  dx&= \int_0^1\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2}\mathrm dx\\
&=\frac12\int_0^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2}\mathrm dx\\
&=\int_0^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\mathrm dx\\
&=\sqrt{\frac{\pi(a+1)}{2}}\frac{\Gamma(r-\frac12)}{(2a+2)^r\Gamma(r)}
,\end{align}
Which works for $r>\frac12$ and all(?) $s$, because as the authors showed, the integral is independent of $s$.
This question wouldn't be complete without my attempts:
Setting $a=1$, we have 
$$\int_0^\infty\left(\frac{x}{x^2+1}\right)^{2r}\mathrm dx=\frac{\sqrt{\pi}\,\Gamma(r-\frac12)}{2^{2r}\Gamma(r)}.$$
Taking $\frac{d}{dr}$ on both sides,
$$\int_0^\infty\left(\frac{x}{x^2+1}\right)^{2r}\log\left(\frac{x}{x^2+1}\right)\mathrm dx=\frac{\sqrt{\pi}}{2}\frac{d}{dr}\frac{\Gamma(r-\frac12)}{2^{2r}\Gamma(r)}.$$
And it can be shown, in a somewhat similar way, that 
$$\int_0^\infty\left(\frac{x}{x^2+1}\right)^{2r}\log^n\left[\frac{x}{x^2+1}\right]\frac{\mathrm dx}{(x^2+1)^2}=\frac{\sqrt\pi}{2^{n+4}}\left(\frac{d}{dr}\right)^n\frac{\Gamma(r+\frac32)}{4^r\Gamma(r+2)}.$$
Unfortunately, I feel as if my creative well has run dry, and I would like to see what you can get from this integral. Have fun!
Edit: Context
The authors of Irresistible Integrals called this integral a "Master Formula" because it apparently could produce a plethora of identities. I would like to see which identities you can derive from said integral.
 A: First I will like to give some steps and maybe  some more insight of this integral.$$I=\int_0^1\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx+\int_1^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx$$
With $x\rightarrow \frac{1}{x}$ in the second one we get:
$$\int_1^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx=\int_0^1 \left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2\left(\frac{1}{x^s}+1\right)}\mathrm dx$$
Now if we add with first part of the integral that was splited using: $\displaystyle{\frac{1}{x^s+1}+\frac{1}{\frac{1}{x^s}+1}=1}\,$ this being the reason why the $s$ doesn't affect our integral.
$$I=\int_0^1\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2}dx$$
Again via $x\rightarrow \frac{1}{x}$ we get:
$$I=\int_1^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2}dx$$
$$\Rightarrow I=\frac12\int_0^\infty \left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2}dx=\frac14 \int_{-\infty}^\infty \left(\frac{1}{x^2+\frac{1}{x^2}+2a}\right)^r\left(1+\frac{1}{x^2}\right)dx$$
And now by writting $\displaystyle{x^2+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)^2+2}$ and do a $x-\frac{1}{x}=t$ we get: $$I=\frac12 \int_{-\infty}^\infty \frac{1}{(t^2+2(a+1))^r}dx$$ By letting $t=x\sqrt{2(a+1)}$  we get rather easy using beta function the result.

But the substitution $x-\frac{1}{x}$ reminds us of Glasser's Master theorem. Of course in order to milk it we can take the original integral and apply this theorem how many times we want.
$$I=\frac12 \int_{-\infty}^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx= \frac12 \int_{-\infty}^\infty \left(\frac{x^6-2x^4+x^2}{x^8+2ax^6-4x^6-4ax^4+7x^4+2ax^2-4x^2+1}\right)^r \frac{x^4-x^2+1}{(x^2-1)^2}dx$$
Where I have used $x\rightarrow x-\frac{1}{x}$ and $s=0$. Of course we can be mean and  in the simplest form use $\int_{-\infty}^\infty f(x)dx=\int_{-\infty}^\infty f(x-\cot x)dx$ to get :$$I=\frac12 \int_{-\infty}^\infty \frac{1}{(x^2+2(a+1))^r}dx=\frac12 \int_{-\infty}^\infty \frac{1}{((x-\cot x)^2+2(a+1))^r}dx$$ And by setting $a+1=\frac12$ and $r=2$ to get: $$\int_0^\infty \frac{1}{((x-\cot x)^2 +1)^2}dx=\sqrt 2 \pi$$
One can do the same thing when there is $x^4$ in the denominator, but that's quite evilish.

Or another thing would be to let $x^2=t$ in order to get some Mellin transforms, for example: $$I=\frac12\int_0^\infty x^{r-1} \frac{1}{(x^2+2ax+1)^r}\left(\sqrt x+\frac{1}{\sqrt x}\right)dx$$
Also those two theorems might also do some milk with this integral: Laplace transform property and  Plancherel theorem.
A: This is not an answer just a pause for thought on your question:
Math. Operation A, Integrate w.r.t $a$, $n$ times consecutively
$$\frac{ \Gamma \left(r-\frac{1}{2}\right) } {2^r\, \Gamma (r)} \sqrt{\frac{\pi }{2}}\int_n...\int (a+1)^{1-r} \, da^n=\frac{ \Gamma \left(r-\frac{1}{2}\right) } {2^r\, \Gamma (r)} \sqrt{\frac{\pi }{2}}\left(  \frac{ (a+1)^{1+n-r} }{(1-r+1)(2-r+1)...(n-r+1)}\right)$$
Math. Operation B, Differentiate w.r.t $a$, $n$ times
$$\frac{ \Gamma \left(r-\frac{1}{2}\right) } {2^r\, \Gamma (r)} \sqrt{\frac{\pi }{2}}\,\left(\frac{d}{da}\right)^n (a+1)^{1-r} =\frac{ \Gamma \left(r-\frac{1}{2}\right) } {2^r\, \Gamma (r)} \sqrt{\frac{\pi }{2}}  \left( -(1-r-1) (1-r-2)...(1-r-n) \,(a+1)^{1-n-r}  \right)$$
Math. Operation C, Differentiate w.r.t $r$, $n$ times as you have indicated in your example
Integrating with respect to $r$ doesn't seem to be possible (at least for me).
So I have couple of thoughts for you while your creative well refills: Starting with $n=1$, are all basic permutations of repeated elementary operations 1,2 and 3 possible and does it matter in what order the desired mathematical operations are applied? Are there any other possibilities?
