The period for an inverted pendulum on which some external forces act is expressed in terms of its initial angle, $\Phi_o\in[0º,70º]$ (higher angles are unstable), as
$$\mathcal{T}(\Phi_o)=2\sqrt 2\sqrt{\dfrac{\ell}{g}}\int_0^{\Phi_o}\dfrac{d\Phi}{\sqrt{(\cos\Phi_o-\cos\Phi)+\zeta(\Phi_o-\Phi)}},$$
where $g$ is gravity, and $\ell$ and $\zeta$ are pendulum-related constants.
In brief, we're interested in
$$\int_0^{\Phi_o}\dfrac{d\Phi}{\sqrt{A-\cos\Phi-B\Phi}}.$$
Is there any series expansion for this integral or any elliptical type function describing it maybe? Using Desmos to visualize the period in terms of the initial angle, looks like there shouldn't be any problem with convergence of this integral within the initial angle's interval and zeta's interval, $\zeta\in[0,2]$.
If it were too hard to find an expression in terms of $\Phi_o$, at least it'd want to try it for $\Phi_o=\frac{\pi}{4}$, which is the default inital value I use the most.
Best Answer
$\def\C{\operatorname C} \def\S{\operatorname S} \def\J{\operatorname J} $
$$\int_0^c\frac{dx}{\sqrt{a-\cos(x)-bx}}=\frac1{\sqrt b}\int_{-c-\frac\pi2}^{-\frac\pi2}\frac{dx}{\sqrt{x-\left(-\frac1b\right)\sin(x)+\frac\pi2+\frac ab}}$$
Substitute: $$x-\left(-\frac1b\right)\sin(x)\to x-\frac\pi2-\frac ab\implies dx\to f’_{-\frac1b}\left(x-\frac\pi2-\frac ab\right)dx$$
where $f_v(x)$ is the Kepler equation Fourier series solution and $\J_n(x)$ is Bessel J
$$f_v(x)=x+2\sum_{n=1}^\infty\frac{\J_n(vn)}n\sin(nx)\\f’_v(x)=\sum_{n\in\Bbb Z}^\infty\J_n(vn)\cos(nx)$$
Therefore:
$$\int_0^c\frac{dx}{\sqrt{a-\cos(x)-bx}}=\int_{\frac{a-\cos(c)}b-c}^\frac{a-1}b\frac {dx}{\sqrt{bx}}+2\sum_{n=1}^\infty\J_n\left(-\frac nb\right)\int_{\frac{a-\cos(c)}b-c}^\frac{a-1}b\cos\left(n\left(x-\frac\pi2-\frac ab\right)\right)\frac{dx}{\sqrt{bx}}$$
The integral requires normalized Fresnel functions:
$$\boxed{\int_0^c\frac{dx}{\sqrt{a-\cos(x)-bx}}=\left.\left[2\sqrt\frac xb+2\sum_{n=1}^\infty\sqrt\frac{2\pi}{bn}\J_n\left(-\frac nb\right)\left(\cos\left(n\left(\frac ab+\frac\pi 2\right)\right)\C\left(\sqrt{\frac{2n}\pi x}\right)+ \sin\left(n\left(\frac ab+\frac\pi 2\right)\right)\S\left(\sqrt{\frac{2n}\pi x}\right)\right)\right]\right|_{\frac{a-\cos(c)}b-c}^\frac{a-1}b}$$
or via unnormalized Fresnel integrals $\int_0^x e^{it^2}dt=\C(x)+i\S(x)$:
$$\boxed{\int_0^c\frac{dx}{\sqrt{a-\cos(x)-bx}}=\left.\left[2\sqrt\frac xb+\sum_{n=1}^\infty\frac4{\sqrt{bn}}\J_n\left(-\frac nb\right)\left(\cos\left(n\left(\frac ab+\frac\pi 2\right)\right)\C\left(\sqrt{nx}\right)+ \sin\left(n\left(\frac ab+\frac\pi 2\right)\right)\S\left(\sqrt{nx}\right)\right)\right]\right|_{\frac{a-\cos(c)}b-c}^\frac{a-1}b}$$
shown here: