Is there an analytic solution for the following non-linear equation $x'(t) = a(t) + b\, x(t) + c\,x^3(t)$?

Question

Here $b, c$ are constants. Of course there exists locally a solution by the Picard-Lindelöf theorem but I'm looking for an explicit expression.

In fact, this question comes from here but I changed the notation (the present $a(t)$ corresponds to $f(t)$). There may be another way to solve the matrix equation (of the other post) but solving the present equation seemed the most natural to me.

Attempts and thoughts:

1. Our equation looks like Riccati's but with a third power instead of a second. So I thought about look for a solution of the form $x(t) + y(t)$ with $x$ satisfying $x'(t) = b\, x(t) + c\,x^3(t)$ so that $y$ has to satisfy another equation. However that latter is even more complex than the original one.\ More generally, this idea of introducing a new unknown but additively (I also thought about doing something like variation of constants, or looking at a solution of the form $x(t) y(t)$...) seems to me like an attempt to exploit the different behavior of l.h.s. vs. r.h.s.. The first is linear, the second one should be related to translation of polynomials, i.e. Taylor expansion.
2. At first, assuming that $x(t)$ never vanishes, I had written $a(t)= x^3(t) \times \tilde{a}(t)$ which formally makes the equation look like Bernouilli's with $\tilde{c}:= c + \tilde{a}(t)$, but solving it would just give an integral form of the original equation, once we have reinjected the dependence in the unknown $x(t)$ within $\tilde{a}(t):= \frac{a(t)}{x^3(t)}$...
3. Look for $T:\mathbb{R}^3\to \mathbb{R}$ be differentiable and see such that $$ \frac{d}{dt}T\big(t,x(t),y(t)\big)=a(t) + b\,T\big(t,x(t),y(t)\big) + c\, T\big(t,x(t),y(t)\big)^3 $$ Assuming that $\dot{x}= bx + cx^3$, one has $$ \partial_t T + \dot{x}\, \partial_x T + \dot{y}\, \partial_y T = \partial_t T + \big(bx + cx^3 \big) \partial_x T + \dot{y}\, \partial_y T = \text{R.h.s.}$$ $$ \partial_t T + \dot{y}\, \partial_y T = a(t) + b \big(T - x\, \partial_x T\big) + c \big(T^3 - x^3\, \partial_x T\big) $$ Making the choice to impose $T^3 \overset{!}{=} x^3\, \partial_x T$ let me to try $T(x,y)= \frac{x}{\sqrt{1+x^2 y}}$...