I am attempting to solve a first-order ODE of the form:
$$y'(x) = {\rm sech}^2(x){\rm sech}'(x) \times(a + b\cos(cx))$$
The part with the constant was easy to solve and I get
$$y(x) = a\int {\rm sech}^2(x){\rm sech}'(x) dx = a\frac{{\rm sech}^3(x)}{3} + C,$$
where $C$ is a constant. However, the part with the cosine term is harder to integrate. I tried simple integration-by-parts since I know the integral of ${\rm sech}^2(x){\rm sech}'(x)$, but I didn't see it leading me anywhere. Wolfram Alpha gave me a hypergeometric function, but I wonder if there is a cleaner closed-form expression for the integral.