1. **Problem Statement:** Given square invertible matrices $A$, $B$, and $C$ of the same size, with $B^T B = I = B B^T$ and $C$ having no eigenvalue equal to $-1$, find the inverse of $(A B^T + A B^T C)$. 2. **Rewrite the expression:** Factor out $A B^T$ from the sum: $$A B^T + A B^T C = A B^T (I + C)$$ 3. **Use the property of inverses:** The inverse of a product of invertible matrices is the product of their inverses in reverse order: $$(A B^T (I + C))^{-1} = (I + C)^{-1} (A B^T)^{-1}$$ 4. **Find $(A B^T)^{-1}$:** Since $A$ and $B$ are invertible and $B^T$ is the transpose of $B$, which is also invertible, we have: $$(A B^T)^{-1} = (B^T)^{-1} A^{-1} = (B^{-1})^T A^{-1}$$ 5. **Check invertibility of $(I + C)$:** Since $C$ has no eigenvalue equal to $-1$, $I + C$ is invertible. 6. **Final answer:** $$\boxed{(A B^T + A B^T C)^{-1} = (I + C)^{-1} (B^{-1})^T A^{-1}}$$