In this paper on page 7 it says that the difference between the whole length of the hyperbola and that of its asymptotes, $\Delta$, is given by $$\Delta=4a\int_0^{\alpha}\sqrt{1-e^2\sin^2\theta}d\theta,$$ where $\sin\alpha=1/e$, $e$ is the eccentricity, and this is for a hyperbola with equation $x^2/a^2-y^2/b^2=1$. No proof was given and I couldn't find one online.
I have tried to find expressions for the length of the asymptote and the length of the hyperbola in the 1st quadrant using polar coordinates with $L=\int_{\theta_0}^{\theta_1}\sqrt{r^2+(dr/d\theta)^2}d\theta$, but the difference was always an ugly expression instead of what was given(I'm not sure if converges if I do it this way). I also tried taking $\theta$ to be the acute angle between the tangent to the hyperbola and its asymptote(in the 1st quadrant) but I do not know how to interpret the integral geometrically or otherwise. I also tried substituting $u=\sqrt{1-e^2\sin^2\theta}$, but I cannot make sense of the resulting integral.
I have no idea how to proceed now, can someone please help?
Let's take just one quarter of the hyperbola and its asymptotes, namely that part lying in the first quadrant. One could try to give a sense to that difference by integrating the difference of line elements, for instance along $y$.
The line element of the asymptote is $dl_a=\sqrt{1+a^2/b^2}dy$ and that of the hyperbola is $dl_h=\sqrt{1+a^4/b^4\cdot y^2/x^2}dy$. The difference in length could then be computed as $$ \begin{align} {\Delta\over4}&=\int_0^{+\infty}\left( \sqrt{1+{a^2\over b^2}}-\sqrt{1+{a^4\over b^4}{y^2\over x^2}} \right)dy\\ &=\sqrt{1+{a^2\over b^2}}\int_0^{+\infty}\left( 1-\sqrt{1-{a^2\over a^2+b^2}{b^2\over b^2+y^2}} \right)dy\\ &=ae\int_0^{+\infty}\left( 1-\sqrt{1-{1\over e^2}{1\over 1+t^2}} \right)dt,\\ \end{align} $$ which is indeed convergent (and can be written in terms of elliptic functions).
I don't know how to transform the above integral into the one you found in that paper, but I checked with Mathematica and they give the same result.