Step 1: Verify by differentiating
Let $F=\log(\tan^{-1}(\sec x+\cos x))$. Then $F'=\dfrac{1}{\tan^{-1}(\cdot)}\cdot\dfrac{(\sec x\tan x-\sin x)}{1+(\sec x+\cos x)^2}$.
Step 2: Simplify
$\sec x\tan x-\sin x=\dfrac{\sin x(1-\cos^2x)}{\cos^2x}=\dfrac{\sin^3x}{\cos^2x}$, and $1+(\sec x+\cos x)^2=\dfrac{\cos^4x+3\cos^2x+1}{\cos^2x}$. The ratio is $\dfrac{\sin^3x}{\cos^4x+3\cos^2x+1}$, which matches the integrand. Option (A).
Final Answer:
Option (A) is the antiderivative.
\[ \boxed{\log\left(\tan^{-1}(\sec x+\cos x)\right)+c} \]