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.
On
Just for your curiosity and the fun.
Let $$I_a=\int_0^{\infty}\frac{x^a}{\sqrt{\cosh x}}dx $$
Here are the "nice" formulas given by a CAS $$I_0=4 \sqrt{\frac{2}{\pi }} \Gamma \left(\frac{5}{4}\right)^2$$ $$I_1=\sqrt{2} \left(\frac{8}{25} \, _3F_2\left(\frac{5}{4},\frac{5}{4},\frac{3}{2};\frac{9}{4},\frac{9}{4};-1\right) -\frac{4}{5} \sqrt[4]{58 \sqrt{2}-82} \, _2F_1\left(\frac{3}{4},\frac{5}{4};\frac{9}{4};\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+4\right)$$ $$I_2=16 \sqrt{2} \, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{2};\frac{5}{4},\frac{5}{ 4},\frac{5}{4};-1\right)$$ $$I_3=96 \sqrt{2} \, _5F_4\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{2};\frac{5}{ 4},\frac{5}{4},\frac{5}{4},\frac{5}{4};-1\right)$$ $$I_4=768 \sqrt{2} \, _6F_5\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{ 2};\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};-1\right)$$ $$I_5=7680 \sqrt{2} \, _7F_6\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{ 4},\frac{1}{2};\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac {5}{4};-1\right)$$
$$I_6=92160 \sqrt{2} \, _8F_7\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{ 4},\frac{1}{4},\frac{1}{2};\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac {5}{4},\frac{5}{4},\frac{5}{4};-1\right)$$
$$I_7=1290240 \sqrt{2} \, _9F_8\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{ 4},\frac{1}{4},\frac{1}{4},\frac{1}{2};\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac {5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};-1\right)$$ $$I_8=20643840 \sqrt{2} \, _{10}F_9\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{ 1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{2};\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4 };-1\right)$$
For $a\geq 2$, there is a nice pattern. Concerning the leading coefficient, it corresponds to sequences $A032184$ and $A066318$ at OEIS.
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:
$$ I(0)=\int_{1}^{+\infty}\frac{dt}{\sqrt{t^3+t}}=\int_{0}^{1}\frac{dt}{\sqrt{t^3+t}}=\frac{1}{2}\int_{0}^{+\infty}\frac{dt}{\sqrt{t^3+t}}\stackrel{t=z^2}{=}\color{red}{\int_{0}^{+\infty}\frac{dz}{\sqrt{z^4+1}}}. $$