The reference for this question is Coates and Schmidt, Iwasawa theory for the symmetric square.

Let $G = \textrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}))$ and let $D_r \supseteq I_r$ be a decomposition group and inertia group at $r$. Let $E/\mathbb{Q}$ be an elliptic curve, and for a prime $\ell \neq r$, consider the representations $\rho_r : G\to \textrm{Sym}^2(H^1_\ell(E))^{I_r}$ and $\rho_r' : G\to \textrm{Sym}^2(H^1_\ell(E)^{I_r})$.

The primitive symmetric square of $E$ is the $L$-series defined by the Euler factors

$\mathcal D_r (X) = \textrm{det}(1-\rho_\ell(\textrm{Frob}_r^{-1})X)$

and the imprimitive symmetric square of $E$ is the $L$-series defined by the Euler factors $D_r (X) = \textrm{det}(1-\rho_\ell'(\textrm{Frob}_r^{-1})X).$

Since $\rho'_r$ is a submodule of $\rho_r$, we have that $D_r(X) | \mathcal{D}_r(X)$

for all $r$. Furthermore, if $r$ is a prime where $E$ has good reduction, since $H^1_\ell(E)$ is unramified at $r$ we have $D_r(X) = \mathcal{D}_r(X)$ .

If $E$ has bad multiplicitive reduction at $r$, then by calculations in Coates and Schmidt, $\mathcal{D}_r(X) = 1-X$, and if $E$ has bad additive reduction at $r$, then $\mathcal{D}_r(X)$ is either equal to $(1-\alpha_r^2X)(1-\beta_r^2X)(1-rX)$, $1+rX$, $1-rX$ or $1$ depending on the image of the inertia group $I_r$, and I can calculate this using Mark Watkins's 'sympow' computer program.

My question is:

How do I find the imprimitive Euler factors $D_r(X)$ at primes where $E$ has bad reduction?