Consider the ODE for $w:\mathbb{C}\to \mathbb{C}$ \begin{align} w''=2w^3+Aw+B &&(1) \end{align} We can multiply it by $w'$ and then integrate to get \begin{align} w'^2=w^4+Aw^2+2Bw+C &&(2) \end{align} ($A,B,C$ are complex constants)
My question is how would one then integrate (2) - i.e. obtain an explicit expression for the solution $w$?
In particular, equation (1) is found in `Ordinary Differential Equations' by Ince. It's listed as equation VIII in section 14.316. Ince states that it is "integrable in terms of elliptic functions". Unfortunately, I cannot see how this would be done.
As an example of what I'm looking for, in a problem related to the one above, one is able to transform the ODE \begin{align} P'^2-P^4+\lambda P^2+\frac{B^2}{P^2}+2A=0 && (3) \end{align} into the ODE defining the Weierstrass elliptic function $$\wp'^2=4\wp^3-g_2\wp-g_3$$ by taking $$\wp(x):=P(x)^2-\frac{\lambda}{3},$$ where $g_2, g_3$ are constants depending on $A,B,\lambda$. And so the general solution to (3) is $$P(x)=\pm\sqrt{\wp(x)+\frac{\lambda}{3}}$$
The equation is $$(w')^2 = (w - w_0) P(w), \quad \deg P = 3.$$ Let $w = 1/\zeta + w_0$, then $$(\zeta')^2 = \zeta^3 P {\left( \frac 1 \zeta + w_0 \right)} = Q(\zeta), \quad \deg Q = 3.$$ If we eliminate the $\zeta^2$ term with $$\zeta = \frac {4 \wp - \frac 1 3 [\zeta^2] Q(\zeta)} {[\zeta^3] Q(\zeta)}$$ and reduce the coefficients modulo $w_0^4 + A w_0^2 + 2 B w_0 + C$ (treating $w_0$ as a variable), we get $$(\wp')^2 = 4 \wp^3 - \frac {A^2 + 12 C} {12} \wp + \frac {A^3 + 54 B^2 - 36 A C} {216}.$$