Calculating explicit transfer function for two diode voltage dividers

I'm seeking to write explicit, non-approximated voltage transfer functions for some diode voltage dividers. Ideally, I want to obtain \$V_{out}\$ purely as a function of \$V_{in}\$ without using iterative methods or piecewise approximations.

I'm ignoring internal resistance, reverse breakdown, capacitance, and other time-dependent properties like the reverse-recovery effect.

^{simulate this circuit– Schematic created using CircuitLab}

For a single diode and resistor like above, this is not too complicated:

$$I_S \exp \left(\frac{V_{in} - V_{out}}{\eta V_T}\right) - I_S - \frac{V_{out}}{R_1} = 0 $$

Using the Lambert W function, this can be rearranged as an explicit function:$$V_{out}(V_{in}) = \eta V_T W\left(\frac{I_S R_1}{\eta V_T} \exp\left(\frac{V_{in} + I_S R_1}{\eta V_T}\right)\right) - I_S R_1$$

which matches SPICE and the result by T.C. Banwell and A. Jayakumar.

^{simulate this circuit}

However, I'm running into difficulties with two-diode circuits like the above. I can get:

$$ I_S \exp \left(\frac{V_{x} - V_{a}}{\eta V_T}\right) + I_S \exp \left(\frac{V_{x} - V_{out}}{\eta V_T}\right)- 2I_S + \frac{V_x - V_b}{R_1} = 0$$$$I_S \exp \left(\frac{V_{x} - V_{out}}{\eta V_T}\right) - I_S - \frac{V_{out}}{R_2} = 0$$

but I'm not sure how to manipulate it into an explicit function.

By curve fitting SPICE results, I believe the result takes the form:

$$V_{out}(V_{a}, V_{b}) = a_0 W\left(\cdot\right) - a_1 W\left(\cdot\right) + a_2$$

though I'm unclear how to get there (or if this form is even correct).

Any guidance is greatly appreciated!