Integrand size = 45, antiderivative size = 334 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\frac {2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{15015 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^3 (8368 A+9230 B+10439 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{45045 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+182 B+143 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{1287 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 a (5 A+13 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d \sec ^{\frac {11}{2}}(c+d x)} \]
2/143*a*(5*A+13*B)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(9/2)+2/ 13*A*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d/sec(d*x+c)^(11/2)+2/9009*a^3*(222 4*A+2522*B+2717*C)*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+2/ 15015*a^3*(8368*A+9230*B+10439*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d *x+c))^(1/2)+8/45045*a^3*(8368*A+9230*B+10439*C)*sin(d*x+c)/d/sec(d*x+c)^( 1/2)/(a+a*sec(d*x+c))^(1/2)+16/45045*a^3*(8368*A+9230*B+10439*C)*sin(d*x+c )*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/1287*a^2*(136*A+182*B+143*C) *sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(7/2)
Time = 5.98 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.48 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\frac {2 a^3 \left (3465 A+315 (38 A+13 B) \sec (c+d x)+35 (523 A+416 B+143 C) \sec ^2(c+d x)+5 (4184 A+4615 B+3718 C) \sec ^3(c+d x)+3 (8368 A+9230 B+10439 C) \sec ^4(c+d x)+4 (8368 A+9230 B+10439 C) \sec ^5(c+d x)+8 (8368 A+9230 B+10439 C) \sec ^6(c+d x)\right ) \sin (c+d x)}{45045 d \sec ^{\frac {11}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \]
Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x] ^2))/Sec[c + d*x]^(13/2),x]
(2*a^3*(3465*A + 315*(38*A + 13*B)*Sec[c + d*x] + 35*(523*A + 416*B + 143* C)*Sec[c + d*x]^2 + 5*(4184*A + 4615*B + 3718*C)*Sec[c + d*x]^3 + 3*(8368* A + 9230*B + 10439*C)*Sec[c + d*x]^4 + 4*(8368*A + 9230*B + 10439*C)*Sec[c + d*x]^5 + 8*(8368*A + 9230*B + 10439*C)*Sec[c + d*x]^6)*Sin[c + d*x])/(4 5045*d*Sec[c + d*x]^(11/2)*Sqrt[a*(1 + Sec[c + d*x])])
Time = 2.02 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.378, Rules used = {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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{13/2}}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^{5/2} (a (5 A+13 B)+a (6 A+13 C) \sec (c+d x))}{2 \sec ^{\frac {11}{2}}(c+d x)}dx}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^{5/2} (a (5 A+13 B)+a (6 A+13 C) \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)}dx}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+13 B)+a (6 A+13 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {2}{11} \int \frac {(\sec (c+d x) a+a)^{3/2} \left ((136 A+182 B+143 C) a^2+(96 A+78 B+143 C) \sec (c+d x) a^2\right )}{2 \sec ^{\frac {9}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{11} \int \frac {(\sec (c+d x) a+a)^{3/2} \left ((136 A+182 B+143 C) a^2+(96 A+78 B+143 C) \sec (c+d x) a^2\right )}{\sec ^{\frac {9}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{11} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((136 A+182 B+143 C) a^2+(96 A+78 B+143 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {2}{9} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((2224 A+2522 B+2717 C) a^3+3 (560 A+598 B+715 C) \sec (c+d x) a^3\right )}{2 \sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((2224 A+2522 B+2717 C) a^3+3 (560 A+598 B+715 C) \sec (c+d x) a^3\right )}{\sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((2224 A+2522 B+2717 C) a^3+3 (560 A+598 B+715 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4291 |
| \(\displaystyle \frac {\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{11} \left (\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{9} \left (\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {4}{5} \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )\right )\right )}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\) |
Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/S ec[c + d*x]^(13/2),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 + 182*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)*S qrt[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*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqr t[a + a*Sec[c + d*x]])))/5))/7)/9)/11)/(13*a)
3.7.4.3.1 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])
Time = 1.55 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.64
| method | result | size |
| default | \(\frac {2 a^{2} \left (3465 A \cos \left (d x +c \right )^{6}+11970 A \cos \left (d x +c \right )^{5}+4095 B \cos \left (d x +c \right )^{5}+18305 A \cos \left (d x +c \right )^{4}+14560 B \cos \left (d x +c \right )^{4}+5005 C \cos \left (d x +c \right )^{4}+20920 A \cos \left (d x +c \right )^{3}+23075 B \cos \left (d x +c \right )^{3}+18590 C \cos \left (d x +c \right )^{3}+25104 A \cos \left (d x +c \right )^{2}+27690 B \cos \left (d x +c \right )^{2}+31317 C \cos \left (d x +c \right )^{2}+33472 A \cos \left (d x +c \right )+36920 B \cos \left (d x +c \right )+41756 C \cos \left (d x +c \right )+66944 A +73840 B +83512 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{45045 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(214\) |
| parts | \(\frac {2 A \,a^{2} \left (3465 \cos \left (d x +c \right )^{6}+11970 \cos \left (d x +c \right )^{5}+18305 \cos \left (d x +c \right )^{4}+20920 \cos \left (d x +c \right )^{3}+25104 \cos \left (d x +c \right )^{2}+33472 \cos \left (d x +c \right )+66944\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{45045 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B \,a^{2} \left (63 \cos \left (d x +c \right )^{5}+224 \cos \left (d x +c \right )^{4}+355 \cos \left (d x +c \right )^{3}+426 \cos \left (d x +c \right )^{2}+568 \cos \left (d x +c \right )+1136\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{693 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 C \,a^{2} \left (35 \cos \left (d x +c \right )^{4}+130 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+292 \cos \left (d x +c \right )+584\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(287\) |
int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(13/ 2),x,method=_RETURNVERBOSE)
2/45045*a^2/d*(3465*A*cos(d*x+c)^6+11970*A*cos(d*x+c)^5+4095*B*cos(d*x+c)^ 5+18305*A*cos(d*x+c)^4+14560*B*cos(d*x+c)^4+5005*C*cos(d*x+c)^4+20920*A*co s(d*x+c)^3+23075*B*cos(d*x+c)^3+18590*C*cos(d*x+c)^3+25104*A*cos(d*x+c)^2+ 27690*B*cos(d*x+c)^2+31317*C*cos(d*x+c)^2+33472*A*cos(d*x+c)+36920*B*cos(d *x+c)+41756*C*cos(d*x+c)+66944*A+73840*B+83512*C)*(a*(1+sec(d*x+c)))^(1/2) /(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*tan(d*x+c)
Time = 0.27 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.59 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\frac {2 \, {\left (3465 \, A a^{2} \cos \left (d x + c\right )^{7} + 315 \, {\left (38 \, A + 13 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} + 35 \, {\left (523 \, A + 416 \, B + 143 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 5 \, {\left (4184 \, A + 4615 \, B + 3718 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 4 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(13/2),x, algorithm="fricas")
2/45045*(3465*A*a^2*cos(d*x + c)^7 + 315*(38*A + 13*B)*a^2*cos(d*x + c)^6 + 35*(523*A + 416*B + 143*C)*a^2*cos(d*x + c)^5 + 5*(4184*A + 4615*B + 371 8*C)*a^2*cos(d*x + c)^4 + 3*(8368*A + 9230*B + 10439*C)*a^2*cos(d*x + c)^3 + 4*(8368*A + 9230*B + 10439*C)*a^2*cos(d*x + c)^2 + 8*(8368*A + 9230*B + 10439*C)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d* x + c)/((d*cos(d*x + c) + d)*sqrt(cos(d*x + c)))
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1562 vs. \(2 (292) = 584\).
Time = 0.57 (sec) , antiderivative size = 1562, normalized size of antiderivative = 4.68 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\text {Too large to display} \]
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(13/2),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...
\[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {13}{2}}} \,d x } \]
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(13/2),x, algorithm="giac")
Time = 25.60 (sec) , antiderivative size = 458, normalized size of antiderivative = 1.37 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\frac {\sqrt {a-\frac {a}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {A\,a^2\,\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{208\,d}+\frac {a^2\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (7\,A+5\,B+2\,C\right )}{72\,d}+\frac {a^2\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (15\,A+13\,B+10\,C\right )}{56\,d}+\frac {a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (21\,A+23\,B+26\,C\right )}{4\,d}+\frac {a^2\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (51\,A+50\,B+48\,C\right )}{80\,d}+\frac {a^2\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (71\,A+76\,B+80\,C\right )}{48\,d}+\frac {a^2\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,\left (5\,A+2\,B\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{176\,d}\right )}{2\,\sqrt {-\frac {1}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {c}{4}+\frac {d\,x}{4}\right )}^2-1\right )} \]
int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/( 1/cos(c + d*x))^(13/2),x)
((a - a/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(sin((13*c)/2 + (13*d*x)/2)*1i + 2*sin((13*c)/4 + (13*d*x)/4)^2 - 1)*((A*a^2*sin((13*c)/2 + (13*d*x)/2)* (sin((13*c)/2 + (13*d*x)/2)*1i - 2*sin((13*c)/4 + (13*d*x)/4)^2 + 1))/(208 *d) + (a^2*sin((9*c)/2 + (9*d*x)/2)*(sin((13*c)/2 + (13*d*x)/2)*1i - 2*sin ((13*c)/4 + (13*d*x)/4)^2 + 1)*(7*A + 5*B + 2*C))/(72*d) + (a^2*sin((7*c)/ 2 + (7*d*x)/2)*(sin((13*c)/2 + (13*d*x)/2)*1i - 2*sin((13*c)/4 + (13*d*x)/ 4)^2 + 1)*(15*A + 13*B + 10*C))/(56*d) + (a^2*sin(c/2 + (d*x)/2)*(sin((13* c)/2 + (13*d*x)/2)*1i - 2*sin((13*c)/4 + (13*d*x)/4)^2 + 1)*(21*A + 23*B + 26*C))/(4*d) + (a^2*sin((5*c)/2 + (5*d*x)/2)*(sin((13*c)/2 + (13*d*x)/2)* 1i - 2*sin((13*c)/4 + (13*d*x)/4)^2 + 1)*(51*A + 50*B + 48*C))/(80*d) + (a ^2*sin((3*c)/2 + (3*d*x)/2)*(sin((13*c)/2 + (13*d*x)/2)*1i - 2*sin((13*c)/ 4 + (13*d*x)/4)^2 + 1)*(71*A + 76*B + 80*C))/(48*d) + (a^2*sin((11*c)/2 + (11*d*x)/2)*(5*A + 2*B)*(sin((13*c)/2 + (13*d*x)/2)*1i - 2*sin((13*c)/4 + (13*d*x)/4)^2 + 1))/(176*d)))/(2*(-1/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*( 2*sin(c/4 + (d*x)/4)^2 - 1))