Consider the following cubic curve defined by:
$$y^3 - 3y + x^3 = c$$
for $c$ a real parameter. The implicit function theorem guarantees three functions, $\phi_1$, $\phi_2$, and $\phi_3$ with domains $(- \infty, (c+2)^{1/3})$, $( (c - 2)^{1/3}, (c+2)^{1/3})$ and $((c-2)^{1/3}), \infty)$ that satisfy:
$$\phi_j(x)^3 - 3 \phi_j(x) + x^3 = c$$
Cardano's formula to the cubic gives an explicit form for $\phi_1$ and $\phi_3$, but only outside the shared domain of $( (c - 2)^{1/3}, (c+2)^{1/3})$, as in this domain if we view the cubic as a function of $y$ we are in the casus irreducibilis with $3$ real roots. In this domain we can use the trigonometric formula for the case of 3 real roots to obtain explicit forms for $\phi_1$, $\phi_2$, and $\phi_3$ on the domain $( (c - 2)^{1/3}, (c+2)^{1/3})$.
I am wondering if there is a way to express these $\phi_j$ as a single formula of some well known functions. I am willing to accept the trigonometric solution for $\phi_2$, but am not satisfied by stitching the solution obtained by Cardano's formula to the solution from the trigonometric formula to obtain $\phi_1$ and $\phi_3$.
Blue curve is the curve with $c = 0$. Purple line $x = (c - 2)^{1/3}$ and green line is $(c + 2)^{1/3}$. Cardano's formula gives an equation for $y$ as a function of $x$ outside of the lines, and the trignometric formula gives equations for the three branches of $y$ as a function of $x$ between the lines.