How to integrate $\int_0^a\sqrt{\frac{1}{x}-\frac{1}{a}}\cosh\left(\sqrt{x}\right)\textrm{d}x$? How to solve the integral
$$\int_0^a\sqrt{\frac{1}{x}-\frac{1}{a}}\cosh\left(\sqrt{x}\right)\textrm{d}x=\pi I_1(\sqrt{a})$$
where $a>0$ and $I_1$ is the modified Bessel function of $1^\textrm{st}$  kind and order $1$? Mathematica gives the solution but Rubi not, so it seems there is no simple integration procedure.
Edit: Related question concerning Mathematica
 A: From the Digital Library of Mathematical Functions we have the integral formula
$$
I_{\nu}(z) =\frac{\left(\frac{1}{2}z \right)^{\nu}}{\pi^{\frac{1}{2}}\Gamma\left(\nu + \frac{1}{2}\right)}\int_{-1}^{1}\left(1-t^2\right)^{\nu - \frac{1}{2}}e^{\pm zt} \ dt, \qquad \Re(\nu) > -\frac{1}{2}
$$
Which for the case of $\nu=1$ reduces to$$
I_{1}(z) =\frac{z}{\pi}\int_{-1}^{1}\sqrt{1-t^2}e^{\pm zt} \ dt
$$
Notice that because we can choose either the "plus" or "minus" symbol on the exponential and still get the same $I_{1}(z)$, we can say that
\begin{align*}
I_{1}(z) &= \frac{I_{1}(z) + I_{1}(z)}{2} \\
& =\frac{z}{\pi}\int_{-1}^{1}\sqrt{1-t^2}\frac{e^{\color{blue}{+}zt}}{2} \ dt + \frac{z}{\pi}\int_{-1}^{1}\sqrt{1-t^2}\frac{e^{\color{blue}{-}zt}}{2} \ dt\\
& =\frac{z}{\pi}\int_{-1}^{1}\sqrt{1-t^2}\frac{e^{\color{blue}{+}zt}+e^{\color{blue}{-}zt}}{2} \ dt \\
& =\frac{z}{\pi}\int_{-1}^{1}\sqrt{1-t^2}\cosh(zt) \ dt \\
& =\color{purple}{2}\frac{z}{\pi}\int_{\color{purple}{0}}^{1}\sqrt{1-t^2}\cosh(zt) \ dt 
\end{align*}
where on the last step you notice that $\sqrt{1-t^2}\cosh(zt)$ is an even function in $t$. Lastly, we just need to make the change of variable $zt = \sqrt{x}\implies zdt =\frac{1}{2\sqrt{x}} dx$. We thus get
\begin{align*}
\pi I_{1}(z)& =2 z\int_{0}^{z^2}\sqrt{1-\frac{x}{z^2}}\cosh(\sqrt{x})\frac{1}{2z \sqrt{x}} \ dx =\int_{0}^{z^2}\sqrt{\frac{1}{x}-\frac{1}{z^2}}\cosh(\sqrt{x}) \ dx
\end{align*}
which gives your desired integral with $z^2 = a$.
