Integrand size = 45, antiderivative size = 334 \[ \int \cos ^{\frac {13}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {16 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {8 a^3 (8368 A+9230 B+10439 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{45045 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (8368 A+9230 B+10439 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (2224 A+2522 B+2717 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+182 B+143 C) \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{1287 d}+\frac {2 a (5 A+13 B) \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d}+\frac {2 A \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d} \]
2/143*a*(5*A+13*B)*cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/ 13*A*cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+2/15015*a^3*(83 68*A+9230*B+10439*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+ 2/9009*a^3*(2224*A+2522*B+2717*C)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d *x+c))^(1/2)+16/45045*a^3*(8368*A+9230*B+10439*C)*sin(d*x+c)/d/cos(d*x+c)^ (1/2)/(a+a*sec(d*x+c))^(1/2)+8/45045*a^3*(8368*A+9230*B+10439*C)*sin(d*x+c )*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/1287*a^2*(136*A+182*B+143*C) *cos(d*x+c)^(7/2)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
Time = 6.63 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.57 \[ \int \cos ^{\frac {13}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sqrt {\cos (c+d x)} (2798182 A+2980640 B+3233516 C+4 (453146 A+454285 B+445588 C) \cos (c+d x)+(746519 A+676000 B+581152 C) \cos (2 (c+d x))+287060 A \cos (3 (c+d x))+225550 B \cos (3 (c+d x))+148720 C \cos (3 (c+d x))+94010 A \cos (4 (c+d x))+58240 B \cos (4 (c+d x))+20020 C \cos (4 (c+d x))+23940 A \cos (5 (c+d x))+8190 B \cos (5 (c+d x))+3465 A \cos (6 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{720720 d} \]
Integrate[Cos[c + d*x]^(13/2)*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d* x] + C*Sec[c + d*x]^2),x]
(a^2*Sqrt[Cos[c + d*x]]*(2798182*A + 2980640*B + 3233516*C + 4*(453146*A + 454285*B + 445588*C)*Cos[c + d*x] + (746519*A + 676000*B + 581152*C)*Cos[ 2*(c + d*x)] + 287060*A*Cos[3*(c + d*x)] + 225550*B*Cos[3*(c + d*x)] + 148 720*C*Cos[3*(c + d*x)] + 94010*A*Cos[4*(c + d*x)] + 58240*B*Cos[4*(c + d*x )] + 20020*C*Cos[4*(c + d*x)] + 23940*A*Cos[5*(c + d*x)] + 8190*B*Cos[5*(c + d*x)] + 3465*A*Cos[6*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d* x)/2])/(720720*d)
Time = 2.30 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.08, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.422, Rules used = {3042, 4753, 3042, 4574, 27, 3042, 4505, 27, 3042, 4505, 27, 3042, 4503, 3042, 4292, 3042, 4292, 3042, 4291}
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.
\(\displaystyle \int \cos ^{\frac {13}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^{13/2} (a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 4753 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\sec (c+d x) a+a)^{5/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )}{\sec ^{\frac {13}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{13/2}}dx\) |
\(\Big \downarrow \) 4574 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\sec (c+d x) a+a)^{5/2} (a (5 A+13 B)+a (6 A+13 C) \sec (c+d x))}{2 \sec ^{\frac {11}{2}}(c+d x)}dx}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\sec (c+d x) a+a)^{5/2} (a (5 A+13 B)+a (6 A+13 C) \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)}dx}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+13 B)+a (6 A+13 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4505 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{11} \int \frac {(\sec (c+d x) a+a)^{3/2} \left ((136 A+182 B+143 C) a^2+(96 A+78 B+143 C) \sec (c+d x) a^2\right )}{2 \sec ^{\frac {9}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \int \frac {(\sec (c+d x) a+a)^{3/2} \left ((136 A+182 B+143 C) a^2+(96 A+78 B+143 C) \sec (c+d x) a^2\right )}{\sec ^{\frac {9}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((136 A+182 B+143 C) a^2+(96 A+78 B+143 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4505 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {2}{9} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((2224 A+2522 B+2717 C) a^3+3 (560 A+598 B+715 C) \sec (c+d x) a^3\right )}{2 \sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((2224 A+2522 B+2717 C) a^3+3 (560 A+598 B+715 C) \sec (c+d x) a^3\right )}{\sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((2224 A+2522 B+2717 C) a^3+3 (560 A+598 B+715 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4503 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4292 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4292 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4291 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{11} \left (\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{9} \left (\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {4}{5} \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )\right )\right )}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\right )\) |
Int[Cos[c + d*x]^(13/2)*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C *Sec[c + d*x]^2),x]
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Sec[c + d*x])^(5/2)*Sin [c + d*x])/(13*d*Sec[c + d*x]^(11/2)) + ((2*a^2*(5*A + 13*B)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2)) + ((2*a^3*(136*A + 1 82*B + 143*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/ 2)) + ((2*a^4*(2224*A + 2522*B + 2717*C)*Sin[c + d*x])/(7*d*Sec[c + d*x]^( 5/2)*Sqrt[a + a*Sec[c + d*x]]) + (3*a^3*(8368*A + 9230*B + 10439*C)*((2*a* Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (4*((2*a *Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a*Sq rt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])))/5))/7)/9)/ 11)/(13*a))
3.13.62.3.1 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[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] *(d_.)], x_Symbol] :> Simp[-2*a*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*S qrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n)) Int[Sqrt[a + b*Csc [e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, 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[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, 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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[ e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Time = 0.53 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.60
\[-\frac {2 a^{2} \left (\left (3465 \cos \left (d x +c \right )^{6}+11970 \cos \left (d x +c \right )^{5}+18305 \cos \left (d x +c \right )^{4}+20920 \cos \left (d x +c \right )^{3}+25104 \cos \left (d x +c \right )^{2}+33472 \cos \left (d x +c \right )+66944\right ) A +\left (4095 \cos \left (d x +c \right )^{5}+14560 \cos \left (d x +c \right )^{4}+23075 \cos \left (d x +c \right )^{3}+27690 \cos \left (d x +c \right )^{2}+36920 \cos \left (d x +c \right )+73840\right ) B +\left (5005 \cos \left (d x +c \right )^{4}+18590 \cos \left (d x +c \right )^{3}+31317 \cos \left (d x +c \right )^{2}+41756 \cos \left (d x +c \right )+83512\right ) C \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{45045 d}\]
-2/45045*a^2/d*((3465*cos(d*x+c)^6+11970*cos(d*x+c)^5+18305*cos(d*x+c)^4+2 0920*cos(d*x+c)^3+25104*cos(d*x+c)^2+33472*cos(d*x+c)+66944)*A+(4095*cos(d *x+c)^5+14560*cos(d*x+c)^4+23075*cos(d*x+c)^3+27690*cos(d*x+c)^2+36920*cos (d*x+c)+73840)*B+(5005*cos(d*x+c)^4+18590*cos(d*x+c)^3+31317*cos(d*x+c)^2+ 41756*cos(d*x+c)+83512)*C)*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)*(cot( d*x+c)-csc(d*x+c))
Time = 0.29 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.57 \[ \int \cos ^{\frac {13}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (3465 \, A a^{2} \cos \left (d x + c\right )^{6} + 315 \, {\left (38 \, A + 13 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 35 \, {\left (523 \, A + 416 \, B + 143 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (4184 \, A + 4615 \, B + 3718 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 4 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
integrate(cos(d*x+c)^(13/2)*(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*(3465*A*a^2*cos(d*x + c)^6 + 315*(38*A + 13*B)*a^2*cos(d*x + c)^5 + 35*(523*A + 416*B + 143*C)*a^2*cos(d*x + c)^4 + 5*(4184*A + 4615*B + 371 8*C)*a^2*cos(d*x + c)^3 + 3*(8368*A + 9230*B + 10439*C)*a^2*cos(d*x + c)^2 + 4*(8368*A + 9230*B + 10439*C)*a^2*cos(d*x + c) + 8*(8368*A + 9230*B + 1 0439*C)*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*si n(d*x + c)/(d*cos(d*x + c) + d)
Timed out. \[ \int \cos ^{\frac {13}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1135 vs. \(2 (292) = 584\).
Time = 0.52 (sec) , antiderivative size = 1135, normalized size of antiderivative = 3.40 \[ \int \cos ^{\frac {13}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)^(13/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d *x+c)^2),x, algorithm="maxima")
1/2882880*(sqrt(2)*(3783780*a^2*cos(12/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13/2*c) + 1066065*a^2*cos(10/13*ar ctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13/2 *c) + 459459*a^2*cos(8/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 1 3/2*c)))*sin(13/2*d*x + 13/2*c) + 193050*a^2*cos(6/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13/2*c) + 70070*a^2*co s(4/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d *x + 13/2*c) + 20475*a^2*cos(2/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2 *d*x + 13/2*c)))*sin(13/2*d*x + 13/2*c) - 3783780*a^2*cos(13/2*d*x + 13/2* c)*sin(12/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 10 66065*a^2*cos(13/2*d*x + 13/2*c)*sin(10/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 459459*a^2*cos(13/2*d*x + 13/2*c)*sin(8/13*arc tan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 193050*a^2*cos(13/ 2*d*x + 13/2*c)*sin(6/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13 /2*c))) - 70070*a^2*cos(13/2*d*x + 13/2*c)*sin(4/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 20475*a^2*cos(13/2*d*x + 13/2*c)*sin( 2/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 6930*a^2*s in(13/2*d*x + 13/2*c) + 20475*a^2*sin(11/13*arctan2(sin(13/2*d*x + 13/2*c) , cos(13/2*d*x + 13/2*c))) + 70070*a^2*sin(9/13*arctan2(sin(13/2*d*x + 13/ 2*c), cos(13/2*d*x + 13/2*c))) + 193050*a^2*sin(7/13*arctan2(sin(13/2*d...
\[ \int \cos ^{\frac {13}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {13}{2}} \,d x } \]
integrate(cos(d*x+c)^(13/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d *x+c)^2),x, algorithm="giac")
Timed out. \[ \int \cos ^{\frac {13}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{13/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]
int(cos(c + d*x)^(13/2)*(a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C /cos(c + d*x)^2),x)