Let $A$ be a real $n \times d$ matrix, $\vec{b} \in \mathbb{R}^n$, and $\vec{x} \in \mathbb{R}^d$. I'd like to find a simple formula for$$ F(\vec{x}) = \left(\left(A-\vec{b}\vec{x}^T\right)^T \left(A-\vec{b}\vec{x}^T\right)\right)^{-1}. $$ My go-to resource for matrix formulas, The Matrix Cookbook, doesn't have this.
I'm thinking of $A$ and $\vec{b}$ as constants and of $\vec{x}$ as a variable. I'm also interested in the case $n \gg d$. I would like to precompute things so that I can make many calls to $F(\vec{x})$ with run times that are independent of $n$. There is an obvious way to do this:$$ F(\vec{x}) \doteq \left(C - \vec{x} \vec{c}^T - \vec{c} \vec{x}^T + \vec{x}\vec{x}^T\right)^{-1}, $$ where $$ C \doteq A^T A \in \mathbb{R}^{d \times d}\quad\text{and}\quad \vec{c} \doteq \vec{b}^T A \in \mathbb{R}^d. $$ This solves the problem, but it seems like one could do more. E.g., the Sherman–Morrison formula tells us $$ \left(C + \vec{x}\vec{x}^T\right)^{-1} = C^{-1} - \frac{C^{-1}\vec{x}\vec{x}^TC^{-1}}{1+\vec{x}^TC^{-1}\vec{x}}. $$ This allows us to see explicitly how the inverse depends on $\vec{x}$. In principle, we could use Sherman–Morrison repeatedly to get a formula for $F(\vec{x})$, but this turns into a mess of algebra. Usually I'd ask Mathematica to simplify that type of thing, but I've never found a good way to get Mathematica to do noncommutative algebra.
One alternative to repeatedly applying the Sherman-Morrison formula (in general, actually) is to use the Woodbury matrix identity. In particular, we can write $F = A^TA + UV$, where $$ U = -\pmatrix{\vec x & \vec c}, \quad V = \pmatrix{\vec c - \vec x & \vec x}^T. $$ The "mess of algebra" that you refer to is reduced to multiplying by the inverse of a certain $2 \times 2$ matrix.
For another formulation, you could write $F = A^TA + UKV$, where $$ U = V^T = \pmatrix{\vec x & \vec c}, \quad K = \pmatrix{1 & -1\\-1 & 0}. $$ To wit, we end up finding $$ \begin{align} F^{-1} &= [A^TA]^{-1} - [A^TA]^{-1}\pmatrix{\vec x & \vec c} \\ & \qquad (K^{-1} + \pmatrix{\vec x & \vec c}^T[A^TA]^{-1}\pmatrix{\vec x & \vec c})^{-1}\pmatrix{\vec x & \vec c}^T[A^TA]^{-1}. \end{align} $$ The inverse of $K$ is of course easy to compute: $$ K^{-1} = \pmatrix{0&-1\\-1&-1}. $$