yes, please only provide full symbolic computation of the residue.

To compute the residue of the function

\[
f(s) = \frac{-\omega1 \cdot P0 \cdot \sinh\left(\sqrt{cc2 \left(\frac{q1 \cdot s + q0}{p1 \cdot s + 1}\right) - cc1} \cdot x\right) \cdot \exp(st)}{s \cdot \cosh\left(\sqrt{cc2 \left(\frac{q1 \cdot s + q0}{p1 \cdot s + 1}\right) - cc1} \cdot l\right) \cdot \sqrt{cc2 \left(\frac{q1 \cdot s + q0}{p1 \cdot s + 1}\right) - cc1}}
\]

at the point \( s_0 = \frac{-cc2 \cdot q0 + cc1}{-cc1 \cdot p1 + cc2 \cdot q1} \), we will proceed with the symbolic computation as follows.

### Step 1: Express the function near \( s = s_0 \)

Given that the function has a potential pole at \( s_0 \), we focus on the behavior of the denominator at this point:

The denominator of \( f(s) \) is:

\[
h(s) = s \cdot \cosh\left(\sqrt{cc2 \left(\frac{q1 \cdot s + q0}{p1 \cdot s + 1}\right) - cc1} \cdot l\right) \cdot \sqrt{cc2 \left(\frac{q1 \cdot s + q0}{p1 \cdot s + 1}\right) - cc1}
\]

The potential pole arises from the \( \frac{1}{s} \) term in the denominator. This suggests that \( s_0 \) is a simple pole.

### Step 2: Evaluate the residue at \( s = s_0 \)

The residue of \( f(s) \) at \( s_0 \) is given by:

\[
\text{Residue}(f(s), s_0) = \lim_{s \to s_0} (s - s_0)f(s)
\]

Substitute \( f(s) \):

\[
\text{Residue}(f(s), s_0) = \lim_{s \to s_0} \frac{(s - s_0) \cdot \left[-\omega1 \cdot P0 \cdot \sinh\left(\sqrt{cc2 \left(\frac{q1 \cdot s + q0}{p1 \cdot s + 1}\right) - cc1} \cdot x\right) \cdot \exp(st)\right]}{s \cdot \cosh\left(\sqrt{cc2 \left(\frac{q1 \cdot s + q0}{p1 \cdot s + 1}\right) - cc1} \cdot l\right) \cdot \sqrt{cc2 \left(\frac{q1 \cdot s + q0}{p1 \cdot s + 1}\right) - cc1}}
\]

This expression simplifies to:

\[
\text{Residue}(f(s), s_0) = \lim_{s \to s_0} \frac{-\omega1 \cdot P0 \cdot (s - s_0) \cdot \sinh\left(\sqrt{cc2 \left(\frac{q1 \cdot s + q0}{p1 \cdot s + 1}\right) - cc1} \cdot x\right) \cdot \exp(st)}{s \cdot \cosh\left(\sqrt{cc2 \left(\frac{q1 \cdot s + q0}{p1 \cdot s + 1}\right) - cc1} \cdot l\right) \cdot \sqrt{cc2 \left(\frac{q1 \cdot s + q0}{p1 \cdot s + 1}\right) - cc1}}
\]

Since \( s_0 \) is a simple pole, the residue simplifies to the value of the remaining function when the pole is canceled out, giving:

\[
\text{Residue}(f(s), s_0) = \frac{-\omega1 \cdot P0 \cdot \sinh\left(\sqrt{cc2 \left(\frac{q1 \cdot s_0 + q0}{p1 \cdot s_0 + 1}\right) - cc1} \cdot x\right) \cdot \exp(s_0 \cdot t)}{s_0 \cdot \cosh\left(\sqrt{cc2 \left(\frac{q1 \cdot s_0 + q0}{p1 \cdot s_0 + 1}\right) - cc1} \cdot l\right) \cdot \sqrt{cc2 \left(\frac{q1 \cdot s_0 + q0}{p1 \cdot s_0 + 1}\right) - cc1}}
\]

Where \( s_0 = \frac{-cc2 \cdot q0 + cc1}{-cc1 \cdot p1 + cc2 \cdot q1} \).

This is the full symbolic expression for the residue of \( f(s) \) at \( s_0 \).

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**Questions to consider:**
1. Do you want to simplify the symbolic expression further?
2. Would you like to substitute specific values for the constants?
3. Is there any part of the function you'd like to analyze further?
4. Would you like to verify this residue with an example function?
5. Are there other points of interest in this function for which you want the residues calculated?

**Tip:** In symbolic computations involving hyperbolic and exponential functions, always double-check the nature of the poles and branch cuts to ensure accurate residue evaluation.

Learn how to compute the symbolic residue for the complex function f(s) using the residue theorem and symbolic computation methods. This problem involves hyperbolic and exponential functions, suitable for advanced undergraduate students studying complex analysis.

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