Arc length to $x$ of a hyperbola? I want to find the relationship between arc length $A$ of a hyperbolic function and it's corresponding horizontal location $a$ relative to the $y$ axis. In this case: the arc length is the input and $a$ is the output.
To find the arc length $A$ of a function $f$ between $x=0$ and $a$, one uses the formula:
$$A(a)=\int_0^a\sqrt{1+(\frac{d}{dx}f(x))^2}dx$$
Though, when calculating it for a hyperbolic function $\sqrt{x^2-1}$, the integral is non-elementary.
$$A(a)=\int_0^a\sqrt{1+\frac{x^2}{x^2-1}}dx$$
Yet I still want to find the inverse relationship between $A$ and $a$ (closed form or not).
i.e:
$$a(A)=?$$
 A: First, you will have to start measuring from somewhere other than $x=0$ because the hyperbola does not reach $x=0$.  It only covers $(-\infty,-1]\cup[1,\infty)$.  It would be a little cleaner to regard $x$ as a function of $y$.  The positive branch goes through $(1,0)$ and has a horizontal instead of vertical slope at the point $(1,0)$.  Now the function is $x=\sqrt{1+y^2}$ Your arc length becomes $\int_0^a \sqrt{1+\frac {y^2}{y^2+1}}dy$ which Alpha can do using an elliptic integral of a hyperbolic sine.
A: For a two sheet parabola, the integral simplifies down to $\int \sqrt{1+tanh^2(u)} du$, which Wolfram Alpha can solve, the answer being equivalent to the elliptic integral, but no imaginary numbers are needed and its all just hyperbolic trig functions and constants.
A: $\def\E{\operatorname E}\def\I{\operatorname I} \def\L{\operatorname L} \def\sgn{\operatorname{sgn}}$
Here is a closed form from

Inverse of elliptic integral of the second kind

using $\E(x,m),m=k^2$, regularized incomplete beta $\I_z(a,b)$ function, and second Lemniscate $\L_2$ constant:
$$A(a)=\int_0^a \sqrt{\frac{x^2}{x^2+1}+1}dx=-i\E(i\sinh^{-1}(a),2)=-i\sgn(x)\L_2\I_{-\sinh^2(2\sinh^{-1}(a))}\left(\frac12,\frac34\right)$$
Unfortunately, inverse beta regularized $\I^{-1}_s(a,b)$ only solves $A(a)=z,\text{Re}(z)=0$:
$$a(A)=\pm i\sin\left(\frac12\sin^{-1}\left(\sqrt{\I^{-1}_{\mp\frac{i A}{\L_2}}\left(\frac12,\frac34\right)}\right)\right)=\pm\frac i2\left(\sqrt{1-\sqrt{\I^{-1}_{\mp\frac{i A}{\L_2}}\left(\frac12,\frac34\right)}}-\sqrt{\sqrt{\I^{-1}_{\mp\frac{i A}{\L_2}}\left(\frac12,\frac34\right)}+1}\right),0\le iA\le\L_2$$
shown here where one of each sign is taken. The plot in the link gives a very small error, but it is a computation limitation. Plot of $a(i w),|w|\le\L_2:$

