Is there a closed form for: $$I(p)=\int_0^\infty \frac{1+z^2}{1+z^4} \frac{z^p}{1+z^{2p}} dz$$
The integral has a number of nice properties:
$$I(p)=I(-p)$$
$$I(p)=2\int_0^1 \frac{1+z^2}{1+z^4} \frac{z^p}{1+z^{2p}} dz=2\int_1^\infty \frac{1+z^2}{1+z^4} \frac{z^p}{1+z^{2p}} dz$$
$$I(p)=\int_{0}^{\infty} \frac{\cosh y}{\cosh 2y} \frac{dy}{\cosh p y}$$
I'm pretty sure it has a closed form expression for every rational $p$, because then we will be able to get a rational function under the integral either directly or with a simple substitution.
Some examples of closed form expressions:
$$I(0)=\frac{\pi}{2 \sqrt{2}}$$
$$I(1)=\frac{\pi}{4}$$
$$I(2)=\frac{\pi}{4 \sqrt{2}}$$
$$I(3)=\left(\frac{2}{3 \sqrt{3}}-\frac{1}{4} \right)\pi$$
$$I(4)=\left(\sqrt{\frac{1}{2} \left(1+\frac{1}{\sqrt{2}}\right)}-\frac{1}{\sqrt{2}} \right)\frac{\pi}{2}$$
$$I(5)=\left(\frac{\sqrt{10(5+\sqrt{5})}}{25}-\frac{1}{4} \right)\pi$$
$$I \left( \frac{1}{2} \right)=\left(\sqrt{1+\frac{1}{\sqrt{2}}}-1 \right)\pi$$
$$I \left( \frac{1}{3} \right)=\left(\frac{1}{\sqrt{3}}-\frac{1}{4} \right)\pi$$
$$I \left( \frac{1}{4} \right)=\left(2-\sqrt{2+\sqrt{2-\sqrt{2}}}\right)\pi$$
$$I \left( \frac{1}{5} \right)=\left(\frac{2\sqrt{5}-1}{4}-\frac{1}{2}\sqrt{2-\frac{2}{\sqrt{5}}}\right)\pi$$
Some not very nice closed forms also appear:
$$I \left( \frac{2}{7} \right)=\frac{ \pi}{8} \left(7 \sqrt{2}+ 4 \sin ^3\left(\frac{\pi }{14}\right) \sin \left(\frac{3 \pi }{14}\right) \csc ^6\left(\frac{\pi }{7}\right) \times \\ \times \left(48 \sin ^2\left(\frac{\pi }{14}\right) \sin \left(\frac{5 \pi }{28}\right) \sin ^2\left(\frac{3 \pi }{14}\right)-4 \sin \left(\frac{\pi }{28}\right) \sin \left(\frac{3 \pi }{14}\right)+\sin \left(\frac{\pi }{14}\right) \times \\ \times \left(2 \sin \left(\frac{3 \pi }{28}\right)-\sin \left(\frac{3 \pi }{14}\right) \left(2 \csc \left(\frac{\pi }{28}\right)+\csc \left(\frac{3 \pi }{28}\right)-3 \csc \left(\frac{5 \pi }{28}\right)\right)\right)\right)\right)$$
Because we can only consider the case $z<1$ a series solution is possible, but the series contain the polygamma function and it's not very useful.
A similar integral has a simple closed form:
$$\int_0^{\infty } \frac{1+z^2}{1+z^4} z^p \, dz=\frac{\pi}{4} \left(\csc \left(\frac{\pi}{4} (p+1)\right)+\sec \left(\frac{\pi}{4} (p+1)\right)\right)$$
This strongly hints on the Beta function, however the series expansion of the original integral will not work, beacuse the limits will be either $\int_0^1$ or $\int_1^\infty$, thus even the series of trigonometric functions doesn't seem possible here.
Mathematica gives:
$$\int_0^1 \frac{1+z^2}{1+z^4} z^p \, dz=\frac{1}{8} \left(\psi \left(\frac{p+5}{8}\right)+\psi\left(\frac{p+7}{8}\right)-\psi\left(\frac{p+1}{8}\right)-\psi \left(\frac{p+3}{8}\right)\right)$$
Is a closed form possible for $I(p)$ for general $p$, not represented as an infinite series of polygamma functions, but something nicer?
Maybe contour integration can help here?
As a separate question, could it be possible to find an ODE for $I(p)$ as a function? It would be almost as good as a closed form to me.
The motivation is this question.
An asymptotic series for $p \to 0$ (obtained by integration of the Taylor series by term) starts as:
$$I(p) \approx \frac{\pi}{2 \sqrt{2}} \left( 1-\frac{3 \pi ^2 p^2}{32}+\frac{95 \pi ^4 p^4}{2048}-\frac{18727 \pi ^6 p^6}{327680}+\frac{23151383 \pi ^8 p^8}{176160768}-\dots \right)=$$
$$=\sum_{n=0}^m (-1)^n \frac{A000364(n) A000281(n)}{(2n)!} \frac{\pi^{2n}}{4^{2n}}p^{2n}$$
The numerator is a product of A000364 and A000281. However, this series does not converge - the numerator grows faster than the denominator for all $p$ as far as I can see.
Update:
Thanks to this great answer we can state the following:
$$I(p)=\frac{\sqrt{2}}{p}I \left( \frac{2}{p} \right)$$
From this functional equation we can see that $p=\sqrt{2}$ is a special value.