Distance between points lying on a hyperbola?

Question

The question is rather simple but I can't find the answer I'm looking for anywhere. On an ordinary 1-dimensional hyperbola, given two points on the hyperbola, what is the length of the path between the points that lies on the hyperbola? The first way I tried to solve this was with the hyperbola curve given by $y=\frac{1}{x}$. Essentially, one can use the formula learned in first year single variable calculus: $$l=\int^{b}_{a}\sqrt{1+(\frac{dy}{dx})^2}dx$$ In this case, $\frac{dy}{dx}=-\frac{1}{x^2}$, so $$l=\int^{b}_{a}\sqrt{1+\frac{1}{x^4}}dx$$ This should find the length of the path on the hyperbola between $(a,\frac{1}{a})$ and $(b,\frac{1}{b})$. The problem here is that $\sqrt{1+\frac{1}{x^4}}$ has a highly complicated antiderivative (according to wolfram alpha) and I'm not sure how its found or if it would yield the explicit formula I'm looking for if it was applied. The other approach I took was to consider the hyperbola parametrized by $x=cosh(t)$ and $y=sinh(t)$. By constructing the distance vector: $$r=cosh(t)\hat{i}+sinh(t)\hat{j}$$ And then with $$\frac{dr}{dt}=sinh(t)\hat{i}+cosh(t)\hat{j}$$ One can construct a metric of sorts for the 1-manifold: $$\frac{dr}{dt} \cdot \frac{dr}{dt}=sinh^2(t)+cosh^2(t)=cosh(2t)$$ So then the integral naturally becomes $$l=\int^a_b \sqrt{(cosh(2t)} dt$$ Which should be the path length between $(cosh(a),sinh(a))$ and $(cosh(b),sinh(b))$. Again, the function in question has a highly compicated antidervative (at least according to wolfram alpha) and what I'm looking for is a somewhat concise formula-I assumed such a thing exists but I could be wrong. Thank you.