Integrand size = 43, antiderivative size = 294 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^3 (10439 A+9230 B+8368 C) \tan (c+d x)}{6435 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (2717 A+2522 B+2224 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (10439 A+9230 B+8368 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac {2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac {2 a (10439 A+9230 B+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d} \]
2/15015*a*(10439*A+9230*B+8368*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/14 3*a*(13*B+5*C)*sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/13*C*sec (d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/6435*a^3*(10439*A+9230*B+8 368*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/9009*a^3*(2717*A+2522*B+2224* C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/45045*a^2*(10439*A+9 230*B+8368*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/1287*a^2*(143*A+182*B+ 136*C)*sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 3.16 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.74 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 (322751 A+325910 B+343612 C+70 (4576 A+5083 B+5552 C) \cos (c+d x)+14 (32747 A+31850 B+30334 C) \cos (2 (c+d x))+141570 A \cos (3 (c+d x))+138450 B \cos (3 (c+d x))+125520 C \cos (3 (c+d x))+156585 A \cos (4 (c+d x))+138450 B \cos (4 (c+d x))+125520 C \cos (4 (c+d x))+20878 A \cos (5 (c+d x))+18460 B \cos (5 (c+d x))+16736 C \cos (5 (c+d x))+20878 A \cos (6 (c+d x))+18460 B \cos (6 (c+d x))+16736 C \cos (6 (c+d x))) \sec ^6(c+d x) \tan (c+d x)}{180180 d \sqrt {a (1+\sec (c+d x))}} \]
Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
(a^3*(322751*A + 325910*B + 343612*C + 70*(4576*A + 5083*B + 5552*C)*Cos[c + d*x] + 14*(32747*A + 31850*B + 30334*C)*Cos[2*(c + d*x)] + 141570*A*Cos [3*(c + d*x)] + 138450*B*Cos[3*(c + d*x)] + 125520*C*Cos[3*(c + d*x)] + 15 6585*A*Cos[4*(c + d*x)] + 138450*B*Cos[4*(c + d*x)] + 125520*C*Cos[4*(c + d*x)] + 20878*A*Cos[5*(c + d*x)] + 18460*B*Cos[5*(c + d*x)] + 16736*C*Cos[ 5*(c + d*x)] + 20878*A*Cos[6*(c + d*x)] + 18460*B*Cos[6*(c + d*x)] + 16736 *C*Cos[6*(c + d*x)])*Sec[c + d*x]^6*Tan[c + d*x])/(180180*d*Sqrt[a*(1 + Se c[c + d*x])])
Time = 1.97 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.05, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4576, 27, 3042, 4506, 27, 3042, 4506, 27, 3042, 4504, 3042, 4287, 27, 3042, 4489, 3042, 4279}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec ^3(c+d x) (\sec (c+d x) a+a)^{5/2} (a (13 A+6 C)+a (13 B+5 C) \sec (c+d x))dx}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec ^3(c+d x) (\sec (c+d x) a+a)^{5/2} (a (13 A+6 C)+a (13 B+5 C) \sec (c+d x))dx}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (13 A+6 C)+a (13 B+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {2}{11} \int \frac {1}{2} \sec ^3(c+d x) (\sec (c+d x) a+a)^{3/2} \left ((143 A+78 B+96 C) a^2+(143 A+182 B+136 C) \sec (c+d x) a^2\right )dx+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{11} \int \sec ^3(c+d x) (\sec (c+d x) a+a)^{3/2} \left ((143 A+78 B+96 C) a^2+(143 A+182 B+136 C) \sec (c+d x) a^2\right )dx+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{11} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((143 A+78 B+96 C) a^2+(143 A+182 B+136 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {2}{9} \int \frac {1}{2} \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (715 A+598 B+560 C) a^3+(2717 A+2522 B+2224 C) \sec (c+d x) a^3\right )dx+\frac {2 a^3 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (715 A+598 B+560 C) a^3+(2717 A+2522 B+2224 C) \sec (c+d x) a^3\right )dx+\frac {2 a^3 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (715 A+598 B+560 C) a^3+(2717 A+2522 B+2224 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {2 a^3 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 4504 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+9230 B+8368 C) \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^4 (2717 A+2522 B+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+9230 B+8368 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^4 (2717 A+2522 B+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 4287 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+9230 B+8368 C) \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^4 (2717 A+2522 B+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+9230 B+8368 C) \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^4 (2717 A+2522 B+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+9230 B+8368 C) \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^4 (2717 A+2522 B+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+9230 B+8368 C) \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^4 (2717 A+2522 B+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (10439 A+9230 B+8368 C) \left (\frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^4 (2717 A+2522 B+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {2 a^2 (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}+\frac {1}{11} \left (\frac {2 a^3 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}+\frac {1}{9} \left (\frac {2 a^4 (2717 A+2522 B+2224 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{7} a^3 (10439 A+9230 B+8368 C) \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )\right )\right )}{13 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}\) |
(2*C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(13*d) + ((2* a^2*(13*B + 5*C)*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/( 11*d) + ((2*a^3*(143*A + 182*B + 136*C)*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(9*d) + ((2*a^4*(2717*A + 2522*B + 2224*C)*Sec[c + d*x ]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (3*a^3*(10439*A + 9230* B + 8368*C)*((2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d) + ((14*a^ 2*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*Sqrt[a + a*Sec[c + d *x]]*Tan[c + d*x])/(3*d))/(5*a)))/7)/9)/11)/(13*a)
3.5.100.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 ))), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B , e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b *(m + 1), 0] && !LtQ[m, -2^(-1)]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)) Int[Sqrt[a + b*Csc[e + f* x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ [A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && !LtQ[n, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m , n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Time = 0.57 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.79
\[\frac {2 a^{2} \left (83512 A \cos \left (d x +c \right )^{6}+73840 B \cos \left (d x +c \right )^{6}+66944 C \cos \left (d x +c \right )^{6}+41756 A \cos \left (d x +c \right )^{5}+36920 B \cos \left (d x +c \right )^{5}+33472 C \cos \left (d x +c \right )^{5}+31317 A \cos \left (d x +c \right )^{4}+27690 B \cos \left (d x +c \right )^{4}+25104 C \cos \left (d x +c \right )^{4}+18590 A \cos \left (d x +c \right )^{3}+23075 B \cos \left (d x +c \right )^{3}+20920 C \cos \left (d x +c \right )^{3}+5005 A \cos \left (d x +c \right )^{2}+14560 B \cos \left (d x +c \right )^{2}+18305 C \cos \left (d x +c \right )^{2}+4095 B \cos \left (d x +c \right )+11970 C \cos \left (d x +c \right )+3465 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{5}}{45045 d \left (\cos \left (d x +c \right )+1\right )}\]
2/45045*a^2/d*(83512*A*cos(d*x+c)^6+73840*B*cos(d*x+c)^6+66944*C*cos(d*x+c )^6+41756*A*cos(d*x+c)^5+36920*B*cos(d*x+c)^5+33472*C*cos(d*x+c)^5+31317*A *cos(d*x+c)^4+27690*B*cos(d*x+c)^4+25104*C*cos(d*x+c)^4+18590*A*cos(d*x+c) ^3+23075*B*cos(d*x+c)^3+20920*C*cos(d*x+c)^3+5005*A*cos(d*x+c)^2+14560*B*c os(d*x+c)^2+18305*C*cos(d*x+c)^2+4095*B*cos(d*x+c)+11970*C*cos(d*x+c)+3465 *C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^5
Time = 0.27 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.65 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (8 \, {\left (10439 \, A + 9230 \, B + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + 4 \, {\left (10439 \, A + 9230 \, B + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 3 \, {\left (10439 \, A + 9230 \, B + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (3718 \, A + 4615 \, B + 4184 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 35 \, {\left (143 \, A + 416 \, B + 523 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 315 \, {\left (13 \, B + 38 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3465 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}} \]
integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="fricas")
2/45045*(8*(10439*A + 9230*B + 8368*C)*a^2*cos(d*x + c)^6 + 4*(10439*A + 9 230*B + 8368*C)*a^2*cos(d*x + c)^5 + 3*(10439*A + 9230*B + 8368*C)*a^2*cos (d*x + c)^4 + 5*(3718*A + 4615*B + 4184*C)*a^2*cos(d*x + c)^3 + 35*(143*A + 416*B + 523*C)*a^2*cos(d*x + c)^2 + 315*(13*B + 38*C)*a^2*cos(d*x + c) + 3465*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d *x + c)^7 + d*cos(d*x + c)^6)
Timed out. \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="maxima")
\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="giac")
Time = 30.58 (sec) , antiderivative size = 1201, normalized size of antiderivative = 4.09 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*8i)/(13*d) - (a^2*(6*A + 5*B + 2*C)*16i)/(13*d) - (a^2*(1 0*A + 11*B + 10*C)*16i)/(13*d) + (a^2*(13*A + 15*B + 20*C)*16i)/(13*d) + ( a^2*(5*A + 2*B)*8i)/(13*d)) - (A*a^2*8i)/(13*d) + (a^2*(6*A + 5*B + 2*C)*1 6i)/(13*d) + (a^2*(10*A + 11*B + 10*C)*16i)/(13*d) - (a^2*(13*A + 15*B + 2 0*C)*16i)/(13*d) - (a^2*(5*A + 2*B)*8i)/(13*d)))/((exp(c*1i + d*x*1i) + 1) *(exp(c*2i + d*x*2i) + 1)^6) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*a^2*8i)/(9*d) - exp(c*1i + d*x*1i)*((a^2*(A - 8*B) *8i)/(9*d) - (a^2*(5*A + 9*B + 10*C)*16i)/(9*d) + (a^2*(5*A + 2*B)*8i)/(9* d) - (a^2*(13*B - 6*C)*64i)/(1287*d)) - (a^2*(5*A + 5*B + 2*C)*16i)/(9*d) + (a^2*(5*A + 10*B + 32*C)*8i)/(9*d) + (a^2*(B + 6*C)*64i)/(9*d)))/((exp(c *1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) - ((a + a/(exp(- c*1i - d*x *1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a^2*(403*B - 5 72*A + 1046*C)*16i)/(15015*d) - (a^2*(5*A + 2*B)*8i)/(5*d)) + (A*a^2*8i)/( 5*d) - (a^2*(4*A + 5*B + 2*C)*16i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp( c*2i + d*x*2i) + 1)^2) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x* 1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((C*a^2*128i)/(143*d) - (a^2*(A - 16*C)* 8i)/(11*d) - (a^2*(3*A + 4*B + 4*C)*40i)/(11*d) + (a^2*(11*A + 10*B + 20*C )*8i)/(11*d) + (a^2*(5*A + 2*B)*8i)/(11*d)) - (A*a^2*8i)/(11*d) - (C*a^2*1 28i)/(11*d) + (a^2*(5*A + 2*B - 16*C)*8i)/(11*d) + (a^2*(11*A + 10*B + ...