Integral of Sqrt[1/x-1/a] Cosh[Sqrt[x]]

Question

MMA delivers: $$\int_0^a\sqrt{\frac{1}{x}-\frac{1}{a}}\cosh\left(\sqrt{x}\right)\textrm{d}x=\pi I_1(\sqrt{a})$$ where $a>0$ and $I_1$ is the modified Bessel function of $1^\textrm{st}$ kind and order $1$.

1. How was this equation by MMA integrated? Rubi cannot integrate it, so it can be assumed that it is not a simple integration.

Edit: Question 1 was also posted in a more suitable forum and also answered there.

2. Why MMA does give a solution only if it is a definite integral and if upper border is $a$? (For indefinite integral or if instead of $a$ we set upper border to $1$ MMA delivers no solution.)

Edit: Question 2 was answered below.

$Assumptions=a>0;
Integrate[Sqrt[(1/x-1/a)] Cosh[Sqrt[x]],{x,0,a}]
(*\[Pi] BesselI[1,Sqrt[a]*)

MMA 12.3

Answer 1

$Version

(* "12.3.1 for Mac OS X x86 (64-bit) (June 19, 2021)" *)

$Assumptions = a > 0;

Integrate[Sqrt[(1/x - 1/a)] Cosh[Sqrt[x]], {x, 0, a}]

(* π BesselI[1, Sqrt[a]] *)

% /. a -> 1

(* π BesselI[1, 1] *)

For a == 1 you need to replace both instances of a: in the argument as well as the integration bounds.

Integrate[Sqrt[(1/x - 1/1)] Cosh[Sqrt[x]], {x, 0, 1}]

(* π BesselI[1, 1] *)