9.1 Problem number 82

$$\int \frac{(b\sec (c+dx){)}^n(A+B\sec (c+dx)+C{\sec }^2(c+dx))}{{\sec }^{\frac{3}{2}}(c+dx)}dx$$

Optimal antiderivative $$-\frac{2C{\left(b\sec (dx+c)\right)}^n\sin (dx+c)}{d(1-2n)\sqrt{\sec (dx+c)}}-\frac{4(A+C(3-2n)-2An)\mathrm{hypergeom}([\frac{1}{2},\frac{5}{4}-\frac{n}{2}],[\frac{9}{4}-\frac{n}{2}],{\cos }^2(dx+c)){\left(b\sec (dx+c)\right)}^n\sin (dx+c)}{d(4n^2-12n+5)\sec {(dx+c)}^{\frac{5}{2}}\sqrt{2-2\cos (2dx+2c)}}-\frac{4B\mathrm{hypergeom}([\frac{1}{2},\frac{3}{4}-\frac{n}{2}],[\frac{7}{4}-\frac{n}{2}],{\cos }^2(dx+c)){\left(b\sec (dx+c)\right)}^n\sin (dx+c)}{d(3-2n)\sec {(dx+c)}^{\frac{3}{2}}\sqrt{2-2\cos (2dx+2c)}}$$

command

Integrate[((b*Sec[c + d*x])^n*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(3/2),x]

Mathematica 13.1 output

$$\text{\$Aborted}$$

Mathematica 12.3 output

$$-\frac{i{2}^{n+\frac{1}{2}}{e}^{-\frac{1}{2}i(4c+d(2n+1)x)}{\left(\frac{e^{i(c+dx)}}{1+e^{2i(c+dx)}}\right)}^{n+\frac{1}{2}}{(1+e^{2i(c+dx)})}^{n+\frac{1}{2}}{\sec }^{-n-2}(c+dx)(b\sec (c+dx){)}^n(A+B\sec (c+dx)+C{\sec }^2(c+dx))(e^{2ic}(\frac{e^{\frac{1}{2}i(2c+d(2n+3)x)}(A(2n+3)e^{i(c+dx)}{}_2{F}_1(n+\frac{1}{2},\frac{1}{4}(2n+5);\frac{1}{4}(2n+9);-e^{2i(c+dx)})+2B(2n+5){}_2{F}_1(n+\frac{1}{2},\frac{1}{4}(2n+3);\frac{1}{4}(2n+7);-e^{2i(c+dx)}))}{d(2n+3)(2n+5)}+\frac{2(A+2C)e^{\frac{1}{2}id(2n+1)x}{}_2{F}_1(n+\frac{1}{2},\frac{1}{4}(2n+1);\frac{1}{4}(2n+5);-e^{2i(c+dx)})}{2dn+d})+\frac{A{e}^{\frac{1}{2}id(2n-3)x}{}_2{F}_1(n+\frac{1}{2},\frac{1}{4}(2n-3);\frac{1}{4}(2n+1);-e^{2i(c+dx)})}{d(2n-3)}+\frac{2B{e}^{\frac{1}{2}i(2c+d(2n-1)x)}{}_2{F}_1(n+\frac{1}{2},\frac{1}{4}(2n-1);\frac{1}{4}(2n+3);-e^{2i(c+dx)})}{d(2n-1)})}{A\cos (2c+2dx)+A+2B\cos (c+dx)+2C}$$