Find the length of the curve $y=\sinh(x)$, $0\leq x\leq1$ [closed]

How to calculate

$$L= \int\limits_0^1\sqrt{1+\cosh^2(x)}dx$$?

I tried substituting $$t=e^x$$, but it did not help.

• I think the indefinite integral cannot be expressed in terms of elementary functions. See WA for example. Jul 30 '21 at 23:45
• Mathematica: $$-i \sqrt{2} E\left(i\left|\frac{1}{2}\right.\right) ,$$ where $E()$ is the elliptic integral. Jul 31 '21 at 0:15

The problem would be the same with $$\sin(x)$$, the integration of it leading to elliptic integrals.
By Taylor $$\sqrt{1+\cosh^2(x)}=\sqrt 2 \,\Bigg[1+\sum_{n=1}^\infty a_n\,x^{2n}\Bigg]$$ where the first coefficients are $$\left\{\frac{1}{4},\frac{5}{96},-\frac{11}{5760},-\frac{11}{129024},\frac{18121}{11 6121600},-\frac{216599}{6131220480},\cdots\right\}$$ Integrated between $$0$$ and $$1$$, this truncated series would give $$\int_0^1\sqrt{1+\cosh^2(x)}\,dx\sim \frac{435783525797}{199264665600 \sqrt{2}} =1.546413085\cdots$$ while the exact solution given by @David G. Stork $$\int_0^1\sqrt{1+\cosh^2(x)}\,dx=-i \sqrt{2} E\left(i \left|\frac{1}{2}\right.\right)=1.546413264\cdots$$