\(\int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {13}{2}}(c+d x)} \, dx\) [604]

Optimal result

Integrand size = 45, antiderivative size = 334 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\frac {2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{15015 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^3 (8368 A+9230 B+10439 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{45045 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+182 B+143 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{1287 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 a (5 A+13 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d \sec ^{\frac {11}{2}}(c+d x)} \] Output:

2/9009*a^3*(2224*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+104 
39*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/1287*a^2*(136 
*A+182*B+143*C)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d/sec(d*x+c)^(7/2)+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)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 6.94 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.57 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\frac {a^2 (2798182 A+2980640 B+3233516 C+4 (453146 A+454285 B+445588 C) \cos (c+d x)+(746519 A+676000 B+581152 C) \cos (2 (c+d x))+287060 A \cos (3 (c+d x))+225550 B \cos (3 (c+d x))+148720 C \cos (3 (c+d x))+94010 A \cos (4 (c+d x))+58240 B \cos (4 (c+d x))+20020 C \cos (4 (c+d x))+23940 A \cos (5 (c+d x))+8190 B \cos (5 (c+d x))+3465 A \cos (6 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{720720 d \sqrt {\sec (c+d x)}} \] Input:

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]
 

Output:

(a^2*(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)] + 2870 
60*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*C 
os[6*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(720720*d*Sq 
rt[Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 2.08 (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

\(\displaystyle \frac {\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{11} \left (\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{9} \left (\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {4}{5} \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )\right )\right )}{13 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)}\)

Input:

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]
 

Output:

(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)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4291

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]
 

rule 4292

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]
 

rule 4503

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]
 

rule 4505

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]
 

rule 4574

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])
 

Maple [A] (verified)

Time = 4.02 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {2 a^{2} \left (\left (3465 \cos \left (d x +c \right )^{6}+11970 \cos \left (d x +c \right )^{5}+18305 \cos \left (d x +c \right )^{4}+20920 \cos \left (d x +c \right )^{3}+25104 \cos \left (d x +c \right )^{2}+33472 \cos \left (d x +c \right )+66944\right ) A +\left (4095 \cos \left (d x +c \right )^{5}+14560 \cos \left (d x +c \right )^{4}+23075 \cos \left (d x +c \right )^{3}+27690 \cos \left (d x +c \right )^{2}+36920 \cos \left (d x +c \right )+73840\right ) B +\left (5005 \cos \left (d x +c \right )^{4}+18590 \cos \left (d x +c \right )^{3}+31317 \cos \left (d x +c \right )^{2}+41756 \cos \left (d x +c \right )+83512\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{45045 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(202\)
parts	\(\frac {2 A \,a^{2} \left (3465 \cos \left (d x +c \right )^{6}+11970 \cos \left (d x +c \right )^{5}+18305 \cos \left (d x +c \right )^{4}+20920 \cos \left (d x +c \right )^{3}+25104 \cos \left (d x +c \right )^{2}+33472 \cos \left (d x +c \right )+66944\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{45045 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B \,a^{2} \left (63 \cos \left (d x +c \right )^{5}+224 \cos \left (d x +c \right )^{4}+355 \cos \left (d x +c \right )^{3}+426 \cos \left (d x +c \right )^{2}+568 \cos \left (d x +c \right )+1136\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{693 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 C \,a^{2} \left (35 \cos \left (d x +c \right )^{4}+130 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+292 \cos \left (d x +c \right )+584\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(287\)

Input:

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)
 

Output:

2/45045/d*a^2*((3465*cos(d*x+c)^6+11970*cos(d*x+c)^5+18305*cos(d*x+c)^4+20 
920*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+4 
1756*cos(d*x+c)+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)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.59 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\frac {2 \, {\left (3465 \, A a^{2} \cos \left (d x + c\right )^{7} + 315 \, {\left (38 \, A + 13 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} + 35 \, {\left (523 \, A + 416 \, B + 143 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 5 \, {\left (4184 \, A + 4615 \, B + 3718 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 4 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:

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")
 

Output:

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)))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

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)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1562 vs. \(2 (292) = 584\).

Time = 0.37 (sec) , antiderivative size = 1562, normalized size of antiderivative = 4.68 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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")
 

Output:

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...
 

Giac [A] (verification not implemented)

Time = 3.73 (sec) , antiderivative size = 448, normalized size of antiderivative = 1.34 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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")
 

Output:

8/45045*(45045*sqrt(2)*A*a^9*sgn(cos(d*x + c)) + 45045*sqrt(2)*B*a^9*sgn(c 
os(d*x + c)) + 45045*sqrt(2)*C*a^9*sgn(cos(d*x + c)) + (120120*sqrt(2)*A*a 
^9*sgn(cos(d*x + c)) + 150150*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 180180*sqr 
t(2)*C*a^9*sgn(cos(d*x + c)) + (294294*sqrt(2)*A*a^9*sgn(cos(d*x + c)) + 3 
00300*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 342342*sqrt(2)*C*a^9*sgn(cos(d*x + 
 c)) + (310596*sqrt(2)*A*a^9*sgn(cos(d*x + c)) + 356070*sqrt(2)*B*a^9*sgn( 
cos(d*x + c)) + 391248*sqrt(2)*C*a^9*sgn(cos(d*x + c)) + (212069*sqrt(2)*A 
*a^9*sgn(cos(d*x + c)) + 232375*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 265837*s 
qrt(2)*C*a^9*sgn(cos(d*x + c)) + 4*(19279*sqrt(2)*A*a^9*sgn(cos(d*x + c)) 
+ 21125*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 24167*sqrt(2)*C*a^9*sgn(cos(d*x 
+ c)) + 2*(1483*sqrt(2)*A*a^9*sgn(cos(d*x + c)) + 1625*sqrt(2)*B*a^9*sgn(c 
os(d*x + c)) + 1859*sqrt(2)*C*a^9*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^ 
2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2) 
*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a* 
tan(1/2*d*x + 1/2*c)^2 + a)^(13/2)*d)
 

Mupad [B] (verification not implemented)

Time = 23.06 (sec) , antiderivative size = 458, normalized size of antiderivative = 1.37 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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)
                                                                                    
                                                                                    
 

Output:

((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))
 

Reduce [F]

\[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx=\sqrt {a}\, a^{2} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{7}}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{6}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{6}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{5}}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{5}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{5}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}}d x \right ) b +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) c \right ) \] Input:

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)
 

Output:

sqrt(a)*a**2*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x) 
**7,x)*a + 2*int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)* 
*6,x)*a + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**6, 
x)*b + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**5,x)* 
a + 2*int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**5,x)*b 
 + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**5,x)*c + 
int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**4,x)*b + 2*i 
nt((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**4,x)*c + int( 
(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**3,x)*c)