Tables and histories of methods of finding $\int\sec x\,dx$? The famously most difficult among completely elementary antiderivatives is that of the secant function.
Has someone tabulated all the ways it can be done, or written a somewhat comprehensive history of them, or an account of logical connections among them?
Does the feeling that this particular antiderivative can be found only by methods that are unexpected except by hindsight correspond in some way to some precisely stateable mathematical fact?
Parenthesis: Look at the tangent half-angle formula in the form
$$
f(x)= \tan\left(\frac x 2 + \frac\pi4\right) = \tan x + \sec x.
$$
Differentiating both sides yields
$$
f'(x) = 2\sec^2\left(\frac x 2 + \frac\pi4\right) = \sec^2 x + \tan^2 x = (\sec x)(\sec x + \tan x) = (\sec x)f(x).
$$
So $f$ satisfies a differential equation
$$
f'(x) = (\sec x)f(x).
$$
$$
\frac{df}{f} = \sec x.
$$
Antidifferentiating both sides gives $\log|f(x)|= \text{the thing sought}$.  Is this one "out there" somewhere (in published source or on the web)?
Later edit: Perhaps some ways of finding this antiderivative are neither "unexpected" (in the sense of being things that can be seen to work only by hindsight) nor applicable only to this one integral.  But a fact persists: Lots of ways of doing this that are out there in the literature do match that description.  Probably far more so than with all other elementary antiderivatives.  So there's a question of whether there's some mathematical fact that expains why that should be true only of this one integral.
 A: Since you ask for references, the original way this integral was found, via cartography, should be kept in mind. See V. F. Rickey and P. M. Tuchinsky, "An Application of Geography to Mathematics: History of the Integral of the Secant", Mathematics Magazine 53 (1980), 162-166. It is available for free on JSTOR.
A: I don't agree that evaluating this integral only involves methods that are unexpected.  In particular, it is possible to integrate any rational expression involving trigonometric functions using the substitution
$$
u = \tan(x/2).
$$
This substitution has the nice property that
$$
\sin x \;=\; \frac{2u}{1+u^2},\qquad \cos x \;=\; \frac{1-u^2}{1+u^2}, \qquad\text{and}\qquad dx \;=\; \frac{2\,du}{1+u^2}.
$$
This is a perfectly standard technique, though it usually isn't taught in calculus classes anymore.
Applying this to the integral of secant gives
$$
\int \sec x\,dx \;=\; \int \frac{dx}{\cos x} \;=\; \int \frac{2\,du}{1-u^2} \;=\; \ln\left|\frac{1+u}{1-u}\right|+C \;=\; \ln\left|\frac{1+\tan(x/2)}{1-\tan(x/2)}\right|+C
$$
Most computer algebra systems use this technique as part of their integration algorithm, so this is the answer that you tend to get if you ask a computer for the integral of secant.
By the way, if this trick strikes you as clever, be aware that it works just as well to integrate rational expressions of sine and cosine using Euler's identity and the substitution $u=e^{ix}$. This tends to involve a lot of complex numbers, but it might seem more straightforward than the above substitution.
In any case, the only sense in which the integral of secant is difficult is that it can't be evaluated easily using the bag of tricks that we tend to teach in calculus classes nowadays.  However, I don't think there's anything mathematically "natural" about the set of tricks that we teach, so I don't think the difficulty of integrating secant has any real mathematical significance.
Edit: By the way, the substitution $u = \tan(\theta/2)$ corresponds to a certain parameterization of the circle by rational functions.  In particular, this is essentially the stereographic projection of the unit circle from the point $(-1,0)$ to $y$-axis, with the $\theta/2$ coming from the fact that an inscribed angle is half of the corresponding central angle.  This same parameterization can be used to enumerate all Pythagorean triples.
Edit 2: To illustrate the point further, here's a way of integrating secant that only involves an "obvious" substitution.  Let $u = \cos x$.  Then
$$
du \;=\; - \sin x\,dx \;=\; -\sqrt{1-u^2}\,dx
$$
so
$$
\int \sec x\,dx \;=\; \int \frac{dx}{\cos x} \;=\; \int -\frac{du}{u\sqrt{1-u^2}} \;=\; \mathrm{sech}^{-1}u+ C \;=\; \mathrm{sech}^{-1}(\cos u) + C.
$$
Now, you might object to using the derivative formula for $\mathrm{sech}^{-1}$, on the grounds that this isn't usually covered in a first-year calculus course.  But again, it seems arbitrary to me that we cover inverse trig functions but not inverse hyperbolic trig functions in first-year calculus.
