Step 1: Plan - verify that $ \frac{d}{dx}[\arctan(\sec x+\cos x)] $ matches the numerator.
If true, the integral is simply $ \int \frac{1}{v}dv = \ln|v|+C $ where $ v=\arctan(\sec x+\cos x) $.
Step 2: Differentiate $ u=\sec x+\cos x $.
\[ \frac{du}{dx} = \sec x\tan x-\sin x = \frac{\sin x}{\cos^2 x}-\sin x = \frac{\sin^3 x}{\cos^2 x} \]
Step 3: Compute $ 1+u^2 $.
$ u^2=\sec^2 x+2+\cos^2 x $, so $ 1+u^2=\sec^2 x+\cos^2 x+3 $. Multiplying by $ \cos^2 x $: $ (1+u^2)\cos^2 x = \cos^4 x+3\cos^2 x+1 $.
Step 4: Find $ \frac{d}{dx}[\arctan u] $.
\[ \frac{d}{dx}[\arctan u]=\frac{u'}{1+u^2}=\frac{\sin^3 x/\cos^2 x}{(\cos^4 x+3\cos^2 x+1)/\cos^2 x}=\frac{\sin^3 x}{\cos^4 x+3\cos^2 x+1} \] This exactly matches the numerator of the integrand.
Step 5: Evaluate the integral.
With $ v=\arctan(\sec x+\cos x) $: \[ \int\frac{dv/dx}{v}dx=\int\frac{dv}{v}=\ln|v|+C \] So $ f(x)=\ln|\arctan(\sec x+\cos x)| $.
Step 6: Compute $ e^{f(x)} $.
\[ e^{f(x)}=e^{\ln|\arctan(\sec x+\cos x)|}=\arctan(\sec x+\cos x) \] \[ \boxed{\arctan(\sec x+\cos x)} \]