How to approach an Hyperbolic Integral that doesn't appear to be solvable in closed form.

Question

I'm interested in tackling the following integral:

$$\int_{-\ln (2+\sqrt 5)}^{\ln (2+\sqrt 5)} \sqrt{4+\sinh^2(x)} dx$$

While I've attempted various techniques, it appears challenging to find a closed-form solution for this integral. I'm beginning to suspect that it might not have one.

Do you have any insights into expressing it as an infinite series or in terms of special functions?

Any guidance or suggestions on alternative approaches would be greatly appreciated. Thank you for your assistance!

Answer 1

Let's first translate out of hyperbolic language by substituting $y=e^x$ and integrating by parts. $\newcommand{\arsinh}{\operatorname{arsinh}}$

$$\begin{align*} I &= \int_{-\log (2+\sqrt 5)}^{\log (2+\sqrt 5)} \sqrt{4+\sinh^2(x)} \, dx \\ &= \frac12 \int_{-\log(2+\sqrt5)}^{\log(2+\sqrt5)} \sqrt{e^{2x} + e^{-2x} + 14} \, dx \\ &= \frac12 \int_{-2+\sqrt5}^{2+\sqrt5} \frac{\sqrt{y^4 + 14y^2 + 1}}{y^2} \, dy \\ &= \int_{-2+\sqrt5}^{2+\sqrt5} \frac{y^2+7}{\sqrt{y^4+14y^2+1}} \, dy \end{align*}$$

Now consult this answer which demonstrates how to approach the integrand $\dfrac{ax^2+b}{\sqrt{cx^4+dx^2+e}}$.

WolframAlpha produces the following antiderivatives, which I've adjusted to agree with the non-Wolfram conventions for $E,F$ used in the linked answer.

$$\begin{align*} \int \frac{dy}{\sqrt{y^4+\alpha y^2+1}} &= -i A\, F\left(-\arsinh^2(Ay) ; \frac1{A^4}\right) +C \\ \int \frac{y^2\,dy}{\sqrt{y^4+\alpha y^2+1}} &= -i A\, \left[E\left(-\arsinh^2(Ay) ; \frac1{A^4}\right) - F\left(-\arsinh^2(Ay) ; \frac1{A^4}\right)\right] +C \end{align*}$$

where $A=\sqrt{\frac2{\alpha+\sqrt{\alpha^2-4}}} \implies \alpha=\dfrac{1+A^4}{A^2}$, so that $\alpha=14\implies A=2-\sqrt3$.

Answer 2

$$I=\int\sqrt{a+\sinh^2(x)}\, dx$$ $$x=it \quad \implies \quad I=i\int \sqrt{a-\sin^2(t)}\, dt$$ So, we face a typical elliptic integral $$I=i\, \sqrt{a} \,E\left(t\left|\frac{1}{a}\right.\right)$$ Back to $x$ $$I=-i\, \sqrt{a}\, E\left(i x\left|\frac{1}{a}\right.\right)$$

$$J=\int_{-p}^{+p}\sqrt{a+\sinh^2(x)}\, dx=-2 i\, \sqrt{a}\, E\left(i p\left|\frac{1}{a}\right.\right)$$

Answer 3

Observe that the value $a=\log(2+\sqrt 5)$ satisfies $\sinh a=\frac 12((\sqrt 5+2)-(\sqrt 5-2))=2$. So the substitution $t=\sinh x$ with (formally) $dt=\cosh\; dx$, $\cos h=\sqrt{1+\sinh^2 x}=\sqrt {1+t^2}$, leads to $$ \begin{aligned} I &=\int_{-a}^a\sqrt{4+\sinh^2x}\; dx =2\int_0^a\sqrt{4+\sinh^2x}\; dx \\ &=2\int_0^2\sqrt{4+t^2}\; \frac{dt}{\sqrt {1+t^2}} =4\int_0^2\sqrt{\frac{1+\frac 14t^2}{1+t^2}}\; dt \\ &\qquad\text{ now consider the above as an integral on a path in $\Bbb C$, set $t=-iu$, $u=it$, $dt=-i\; du$} \\ &=-4i\int_0^{2i}\sqrt{\frac{1-\frac 14t^2}{1-t^2}}\; dt \ , \end{aligned} $$ and compare with 19.2.5, Legendre's integrals, which is explicitly $\displaystyle E(\varphi,k)=\int_0^{\sin\varphi}\frac{\sqrt{1-k^2t^2}}{\sqrt{1-t^2}}\; dt$, note that $\sin^{-1}(2i)=i\sinh^{-1} 2=ia$, to obtain the formula $$ I =-4iE\left(\varphi=ia,k= \frac 12\right) =-4iE\left(\varphi=ia,m= \frac 14\right) \ . $$ Here, the second argument of the elliptic integral $m$ is in some sources $k$, in other $m=k^2$. Depending on the convention, one has to take the one or the other value. Wolfram alpha works with the $m$-parameter, and gives the following answer that confirms the last formula.