If $I=\int \frac{\sin x+{\sin }^{3}x}{\cos 2x}dx=A\cos x+B\log \left|f\left(x\right)\right|+C$, then

$I=\int \frac{\sin x+{\sin }^{3}x}{\cos 2x}dx=\int \frac{(1+1-{\cos }^{2}x)\sin x}{2{\cos }^{2}x-1}dx$

$=\int \frac{2-{\cos }^{2}x}{2{\cos }^{2}x-1}.\left(\sin x\right)dx$

$Let \cos x=t\Rightarrow \sin xdx=-dt$

$=\int \frac{(2-t^2)(-dt)}{2t^2-1}=\frac{1}{2}\int \frac{2(t^2-2)}{2t^2-1}dt$

$=\frac{1}{2}\int \frac{2t^2-4}{2t^{2-1}}dt=\frac{1}{2}\int \frac{2(t^2-1-3)}{2t^2-1}dt$

$=\frac{1}{2}\left[\int 1dt-\frac{3}{2}\int \frac{1}{t^2-1/2}dt\right]+c$

$=\frac{1}{2}[t-\frac{3}{2}.\frac{1}{2.\frac{1}{\sqrt{2}}}\log \left|\frac{t-1/\sqrt{2}}{t+1/\sqrt{2}}\right|+c$

$=\frac{1}{2}\left[\cos x\right]-\frac{3}{4\sqrt{3}}.\log \left|\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}\right|+c$

Comparing with given expression.

$\Rightarrow A=\frac{1}{2},B=\frac{-3}{4\sqrt{3}},f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}$