Determine contribution quadratic in $B$ of $[\det (A+B) ]^{1/2}$

Let $$A$$ and $$B$$ be $$n\times n$$ matrices over $$\mathbb R$$, with $$n\ge 4$$. $$A$$ is a symmetric matrix with inverse $$A^{-1}$$ and $$B$$ is an antisymmetric matrix.

Consider $$f= \left[\det (A+B) \right]^{1/2}$$ where $$\det$$ is the matrix determinant and we can assume that $$\det (A+B)> 0$$. I want to find an expression for the contributions to $$f$$ that are quadratic in the matrix elements $$B_{ab}$$ of $$B$$.

I believe it must be of the form $$\alpha B_{ab} (A^{-1})^{ac}(A^{-1})^{bd}B_{cd}= -\alpha \, \mathrm{tr}\, (B A^{-1}B A^{-1})$$ where tr is the standard matrix trace and $$\alpha$$ is some coefficient.

Unfortunately I have no idea how to show this in a simple way. Brute force calculation might help, but I don't know how I would have to handle the square root.

Any help on this will be much appreciated.

Turns out the answer is pretty simple. The symmetry properties of $$A$$ and $$B$$ are irrelevant. Just use the expansion $$\det (\mathbb 1 + A) = 1 + \mathrm{Tr}\,{A} + \frac{1}{2} \left(\mathrm{Tr}\,{A}\right)^2 - \frac{1}{2} \mathrm{Tr}\,{A^2} + o(A^3)$$ and then expand the square root. The terms with $$\left(\mathrm{Tr}\,{A}\right)^2$$ cancel out and the term $$\mathrm{Tr}\,{A^2}$$ has a factor $$-1/4$$.