$\begin{aligned} & I=\int \frac{d x}{(2 \sin x+\sec x)^4}=\int \frac{\sec ^4 x}{\left(2 \tan x+\sec ^2 x\right)^4} d x \\ & =\int \frac{\left(1+\tan ^2 x\right)}{(1+\tan x)^8} \sec ^2 x d x\end{aligned}$
Let $\tan x=t \Rightarrow \sec ^2 x d x=d t$
So,
$$
\begin{aligned}
& I=\int \frac{1+t^2}{(1+t)^8} d t=\int \frac{(1+t)^2-2 t}{(1+t)^8} d t \\
&=\int \frac{d t}{(1+t)^6}-2 \int \frac{1+t-1}{(1+t)^8} d t \\
&=\int \frac{d t}{(1+t)^6}-2 \int \frac{d t}{(1+t)^7}+2 \int \frac{d t}{(1+t)^8} \\
&=-\frac{1}{5}(1+t)^{-5}+\frac{2}{6}(1+t)^{-6}-\frac{2}{7}(1+t)^{-7}+K \\
&=\frac{-1}{5}(1+\tan x)^{-5}+\frac{1}{3}(1+\tan x)^{-6} \\
&-\frac{2}{7}(1+\tan x)^{-7}+k
\end{aligned}
$$
So, $A=\frac{-1}{5}, B=\frac{1}{3}$ and $C=-\frac{2}{7}$
$$
\therefore A+B+C=-\frac{16}{105} \text {. }
$$