Two integrals involving $\sqrt{\cosh x}$ I have two integrals on my table which I can can calculate numerically, but would have been happy to see whether they have closed form or values in terms of "standard" known functions: $$\int_0^{\infty}\frac{x^a}{\sqrt{\cosh x}}dx \quad \text{for} \quad a=1,3.$$
I appreciate your hints. Many thanks.
Note: By standard here I mean any (well-)known function that are in general available in most math software packages.
 A: We have
$$\begin{eqnarray*} \color{red}{I(a)}=\int_{0}^{+\infty}\frac{x^a}{\sqrt{\cosh x}}\,dx &=& \sqrt{2}\int_{0}^{+\infty}\frac{x^a e^{-x/2}}{\sqrt{1+e^{-2x}}}\,dx\\&=&\sqrt{2}\sum_{n\geq 0}\frac{(-1)^n}{4^n}\binom{2n}{n}\int_{0}^{+\infty}x^a e^{-x/2}e^{-2n x}\,dx\\&=&\color{red}{a!\sqrt{2}\sum_{n\geq 0}\frac{(-1)^n}{4^n\left(2n+\frac{1}{2}\right)^{a+1}}\binom{2n}{n}}\tag{1}\end{eqnarray*}$$
where the last series is a hypergeometric series related with the generating function of Catalan numbers. In particular, for $a=1$ we get:
$$ I(1) = 4\sqrt{2}\;\phantom{}_3 F_2\left(\frac{1}{4},\frac{1}{4},\frac{1}{2};\frac{5}{4},\frac{5}{4};-1\right)\tag{2} $$
and for $a=3$ we get:
$$ I(3)= 96\sqrt{2}\;\phantom{}_5 F_4\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{2};\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};-1\right).\tag{3} $$
It is interesting to point out that for $a=0$ we have the simple closed form
$$ I(0) = \color{red}{\frac{4}{\sqrt{\pi}}\,\Gamma\left(\frac{5}{4}\right)^2 }\tag{4}$$
since the associated $\phantom{}_2 F_1$ function is given by an elliptic integral that can be computed through Euler's beta function:
$$ I(0)=\int_{1}^{+\infty}\frac{dt}{\sqrt{t^3+t}}=\int_{0}^{1}\frac{dt}{\sqrt{t^3+t}}=\frac{1}{2}\int_{0}^{+\infty}\frac{dt}{\sqrt{t^3+t}}\stackrel{t=z^2}{=}\color{red}{\int_{0}^{+\infty}\frac{dz}{\sqrt{z^4+1}}}. $$
