If $\int \frac{\mathrm{dx}}{{\left(2\mathrm{sinx}+\mathrm{secx}\right)}^4}={\mathrm{At}}^5+{\mathrm{Bt}}^6+{\mathrm{Ct}}^7+\mathrm{k},$where $\mathrm{t}={\left(1+\mathrm{tanx}\right)}^{-1},$then $\mathrm{A}+\mathrm{B}+\mathrm{C}=$

$\mathrm{I}=\int \frac{{\sec }^4\mathrm{xdx}}{{\left(2\tan \mathrm{x}+{\sec }^2\mathrm{x}\right)}^4}=\int \frac{{\sec }^4\mathrm{x}}{(1+\tan \mathrm{x}{)}^8}\mathrm{dx}=\int \frac{\left(1+{\tan }^2\mathrm{x}\right){\sec }^2\mathrm{xdx}}{(1+\tan \mathrm{x}{)}^8}$
Put $1+\mathrm{tanx}=y.\text{ Then }{\sec }^2\mathrm{xdx}=dy$
Thus, $$\begin{gathered} \mathrm{I}=\int \frac{1+(y-1{)}^2}{y^8}\mathrm{dt}=\int \left(\frac{1}{y^6}-\frac{2}{y^7}+\frac{2}{{\mathrm{y}}^8}\right)\mathrm{dt}=\frac{y^{-5}}{-5}+\frac{2{\mathrm{y}}^{-6}}{6}+\frac{2y^{-7}}{-7}+\mathrm{k} \\ \mathrm{I}=\frac{{\mathrm{t}}^5}{-5}+\frac{2{\mathrm{t}}^6}{6}+\frac{2{\mathrm{t}}^7}{-7}+\mathrm{k} \end{gathered}$$
Hence, $\mathrm{A}=\frac{-1}{5},\mathrm{B}=\frac{1}{3},\mathrm{C}=\frac{-2}{7}$
$\therefore \mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{-1}{5}+\frac{1}{3}-\frac{2}{7}=\frac{-21+35-30}{105}=\frac{-16}{105}$