This is example $2$ in my "Integration Using Some Euler-Like Identities" blog post.
$$\int\frac{1}{\tan\left(\frac{1}{2}\csc^{-1} (x)\right) - \tan\left(\frac{1}{2}\sec^{-1}(x)\right)} \, dx\tag{1}$$
Wolfram Alpha is unable to simplify and solve it by itself. By applying the following general transformation formula:
$$\int f\left(x,\tan{\frac{\beta}{2}}, \tan{\frac{\gamma}{2}} \right)\,dx=\int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, e^{i\alpha}, \frac{1-e^{i\alpha}}{1+e^{i\alpha}}\right)\,\frac{e^{-i\alpha}-e^{i\alpha}}{2i}a\,d\alpha,\tag{2}$$
where $\alpha=\cos^{-1}\left(\frac{x}{a}\right)$, $\beta=\csc^{-1}\left(\frac{x}{a}\right)$ and $\gamma=\sec^{-1}\left(\frac{x}{a}\right)$, the integral $(1)$ becomes
$$=-\frac{i}{2}\int\frac{e^{-i\alpha} - e^{i\alpha} - e^{2i\alpha} + 1}{e^{2i\alpha}+2e^{i\alpha}-1}\,d\alpha.$$
Then, let $u=e^{i\alpha}\implies du=ie^{i\alpha}\,d\alpha$. Then, putting terms over a common denominator, and factoring, we have
$$-\frac{i}{2}\int\frac{e^{-i\alpha} - e^{i\alpha} - e^{2i\alpha} + 1}{e^{2i\alpha}+2e^{i\alpha}-1}\,d\alpha=\frac12\int\frac{(u - 1) (u + 1)^2}{u^2 (u^2 + 2 u - 1)}\,du$$
At this point, we can proceed by applying partial fraction decomposition.
What other method do you have in mind to solve $(1)$?
Edit. The integral $(1)$ is defined for $(−\infty,−1) \cup [1,\infty) \setminus \{\sqrt{2}\}$. The method described above (now refined in the summary's blog post) provides a systematic approach to handle both subdomains. Indeed, for an integral of the form $$\int f\Biggl(x, \tan\Bigl(\tfrac{\beta}{2}\Bigr), \tan\Bigl(\tfrac{\gamma}{2}\Bigr)\Biggr) dx,$$ the transformation is $$\boxed{\begin{aligned}\int f\Biggl(x, \tan\Bigl(\tfrac{\beta}{2}\Bigr), \tan\Bigl(\tfrac{\gamma}{2}\Bigr)\Biggr) dx = \int f\Biggl(\frac{t^2+1}{2t}a - b,\; t,\; \pm\frac{1-t}{1+t}\Biggr)\,\frac{t^2-1}{2t^2}a\, dt,\end{aligned}}\tag{3}$$ where the choice of sign $\pm$ corresponds to the domain of $x$: Use the upper sign when $\tfrac{x+b}{a} \geq 1 $, use the lower sign when $ \tfrac{x+b}{a} \leq -1$.
To revert to $x$-terms after integrating with respect to $t$, use $$t = e^{\pm i\alpha} = \frac{x+b \mp \sqrt{(x+b)^2-a^2}}{a},$$ selecting the sign consistent with the branch used during substitution.
Solution. For $1\leq x<\sqrt{2}$ or $x>\sqrt{2}$, we apply the transformation formula $(3)$ in terms of $t$ to obtain $$\begin{aligned}\int\frac{1}{\tan\left(\frac{1}{2}\csc^{-1} (x)\right) - \tan\left(\frac{1}{2}\sec^{-1}(x)\right)} \, dx&=\int \frac{1}{t - \left(\frac{1 - t}{1 + t}\right)} \cdot \frac{t^2 - 1}{2t^2}\,dt\\&=\frac12\int\frac{(t - 1) (t + 1)^2}{t^2 (t^2 + 2 t - 1)}\,dt.\end{aligned}$$ At this point, we can proceed by applying partial fraction decomposition.
By formula $(3)$, if $e^{i\alpha} = \tfrac1t$ for $x < -1$, then an analogous calculation (using the lower sign) leads to $$\begin{aligned}\int\frac{1}{\tan\left(\frac{1}{2}\csc^{-1} (x)\right) - \tan\left(\frac{1}{2}\sec^{-1}(x)\right)} \, dx&=\int \frac{1}{\frac{1}{t} + \left(\frac{1 - t}{1 + t}\right)} \cdot \frac{t^2 - 1}{2t^2}\,dt\\&=\frac12\int\frac{(t + 1) (1 - t^2)}{t (t^2 - 2 t - 1)}\,dt.\end{aligned}$$ Again, one may proceed by partial fraction decomposition to evaluate the resulting integral.
I don't see a simpler method than the one described above to do this, considering the entire domain of $(1)$. Do you?
