The unbeatable $\int e^{1/\cos(x)} dx$ integral

Question

Is there any way to express this in non-elementary functions? $$ \int e^{1/\cos(x)} dx$$ And/or to calculate this definite integrals? $$ \int_{-\pi/2}^{\pi/2} e^{1/\cos(x)} dx$$ $$ \int_{\pi/2}^{3\pi/ 2} e^{1/\cos(x)} dx$$

Added later:

First integral does not converges, but second on seems to be smooth function and WolframAlpha thinks it is approximately $0.656573$.

Really, nothing is known about this function?

$$f(x)=\int_{\pi/2}^x e^{1/\cos(x)} dx; x \in ({\pi/2} ,{3\pi/ 2}]$$

Answer 1

$e^\frac{1}{\cos(x)} \ge e^\frac{1}{\frac{\pi}{2}-x} \ge 1+\frac{1}{\frac{\pi}{2}-x}$ for any $x \in [0,\frac{\pi}{2})$ because $\cos(\frac{\pi}{2}-x) = \sin(x) \le x$ for any $x \ge 0$

Therefore $\int_0^\frac{\pi}{2} e^\frac{1}{\cos(x)} \text{ dx} = \infty$ and hence also the first definite integral.

The second integral is, as you say, continuous and hence integrable. I don't know about its value, nor the indefinite integral.

[Edit: I misread the second indefinite integral in my original answer.]

Answer 2

The indefinite integral is not an elementary function. This can be proven using the Risch algorithm.