Can you force Integrate[] to find a complete symbolic solution for all variables?

(As I've asked on the Math StackExchange and on a related previous question), I am interested in getting a complete symbolic solution to the integral of an expression with a lot of unassigned variables. If you combine some of the variables, the integral can be reduced to the form:

$$\int_{-\infty}^\infty \frac{\text{A} \Delta +\text{B}}{\left(\Delta ^2+W^2\right) \left(\text{C}+\text{D}\Delta +\text{E}\Delta ^2 \right)}d\Delta$$

Mathematica claims that the solution to this integral is:

$$\frac{\pi (\text{B}-i \text{A} W)}{W (\text{C}-W (\text{E} W+i \text{D}))} \text{if: } \Im\left(\frac{E\pm\sqrt{E^2-4 C E}}{E}\right)<0$$

Shown as code:

Integrate[
(A1 Δ + B1)/((W^2 + Δ^2) (C1 + D1 Δ + E1 Δ^2)),
{Δ, -∞, ∞}, Assumptions -> {W > 0}]

Which returns:

ConditionalExpression[(π (B1 - I A1 W))/(
W (C1 - W (I D1 + E1 W))),
Im[(D1 - Sqrt[D1^2 - 4 C1 E1])/E1] < 0 &&
Im[(D1 + Sqrt[D1^2 - 4 C1 E1])/E1] < 0 && Re[W] > 0]

Mathematica generates a conditional-expression, but doesn't specify if this is a "full" answer. For instance what if we consider the integral under the domain of parameters with the opposite inequalities: $$\Im\left(\frac{E\pm\sqrt{E^2-4 C E}}{E}\right)>0$$? Is there a solution in this domain of parameters?

I can try to force Mathematica to spit out an answer under different conditions. For example:

Integrate[
(A1 Δ + B1)/((W^2 + Δ^2) (C1 + D1 Δ + E1 Δ^2)),
{Δ, -∞, ∞},
Assumptions -> {W > 0, Im[(D1 - Sqrt[D1^2 - 4 C1 E1])/E1] > 0,
Im[(D1 + Sqrt[D1^2 - 4 C1 E1])/E1] < 0}]

I can get another symbolic answer for this new parameter space:

$$\frac{i \pi \left(\text{A1} D W+(-i) \text{B1} \sqrt{D^2-4 C E}-2 \text{B1} E W\right)}{W \sqrt{D^2-4 C E} \left(C+W \left(E W+i \sqrt{D^2-4 C E}\right)\right)}$$

Is there an option to do this automatically and generate a solution for the entire set of possible combinations in the domain space? I'm honestly pretty surprised that it does not automatically return a combined piece-wise function with these different integrated results.

This is what GenerateConditions$$\to$$All is supposedly supposed to do, but I find it doesn't usually work. From the documentation GenerateConditions$$\to$$All should