\(\int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [554]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F(-1)]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 43, antiderivative size = 271 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4 a^3 (7 A+9 B+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (13 A+21 B+35 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}-\frac {4 a^3 (41 A+42 B-35 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 (6 A+7 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (7 A+9 B+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}} \] Output:

4/5*a^3*(7*A+9*B+5*C)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2 
))*sec(d*x+c)^(1/2)/d+4/21*a^3*(13*A+21*B+35*C)*cos(d*x+c)^(1/2)*InverseJa 
cobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d-4/105*a^3*(41*A+42*B-35*C 
)*sec(d*x+c)^(1/2)*sin(d*x+c)/d+2/7*A*(a+a*sec(d*x+c))^3*sin(d*x+c)/d/sec( 
d*x+c)^(5/2)+2/35*(6*A+7*B)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d/sec(d*x+ 
c)^(3/2)+2/15*(7*A+9*B+5*C)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^( 
1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.67 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.98 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a^3 e^{-i d x} \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (2352 i A \cos (c+d x)+3024 i B \cos (c+d x)+1680 i C \cos (c+d x)+80 (13 A+21 B+35 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-112 i (7 A+9 B+5 C) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+126 A \sin (c+d x)+42 B \sin (c+d x)+840 C \sin (c+d x)+550 A \sin (2 (c+d x))+420 B \sin (2 (c+d x))+140 C \sin (2 (c+d x))+126 A \sin (3 (c+d x))+42 B \sin (3 (c+d x))+15 A \sin (4 (c+d x))\right )}{420 d} \] Input:

Integrate[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)) 
/Sec[c + d*x]^(7/2),x]
 

Output:

(a^3*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*((2352*I)*A*Cos[c + d*x] + 
 (3024*I)*B*Cos[c + d*x] + (1680*I)*C*Cos[c + d*x] + 80*(13*A + 21*B + 35* 
C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (112*I)*(7*A + 9*B + 5*C 
)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4 
, 7/4, -E^((2*I)*(c + d*x))] + 126*A*Sin[c + d*x] + 42*B*Sin[c + d*x] + 84 
0*C*Sin[c + d*x] + 550*A*Sin[2*(c + d*x)] + 420*B*Sin[2*(c + d*x)] + 140*C 
*Sin[2*(c + d*x)] + 126*A*Sin[3*(c + d*x)] + 42*B*Sin[3*(c + d*x)] + 15*A* 
Sin[4*(c + d*x)]))/(420*d*E^(I*d*x))
 

Rubi [A] (verified)

Time = 1.71 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.03, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4574, 27, 3042, 4505, 27, 3042, 4505, 3042, 4485, 27, 3042, 4274, 3042, 4258, 3042, 3119, 3120}

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.

\(\displaystyle \int \frac {(a \sec (c+d x)+a)^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\)

\(\Big \downarrow \) 4574

\(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^3 (a (6 A+7 B)-a (A-7 C) \sec (c+d x))}{2 \sec ^{\frac {5}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^3 (a (6 A+7 B)-a (A-7 C) \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a (6 A+7 B)-a (A-7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4505

\(\displaystyle \frac {\frac {2}{5} \int \frac {(\sec (c+d x) a+a)^2 \left (7 a^2 (7 A+9 B+5 C)-a^2 (11 A+7 B-35 C) \sec (c+d x)\right )}{2 \sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \int \frac {(\sec (c+d x) a+a)^2 \left (7 a^2 (7 A+9 B+5 C)-a^2 (11 A+7 B-35 C) \sec (c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (7 a^2 (7 A+9 B+5 C)-a^2 (11 A+7 B-35 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4505

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {(\sec (c+d x) a+a) \left (a^3 (106 A+147 B+140 C)-a^3 (41 A+42 B-35 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}}dx+\frac {14 (7 A+9 B+5 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^3 (106 A+147 B+140 C)-a^3 (41 A+42 B-35 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {14 (7 A+9 B+5 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4485

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (2 \int \frac {21 (7 A+9 B+5 C) a^4+5 (13 A+21 B+35 C) \sec (c+d x) a^4}{2 \sqrt {\sec (c+d x)}}dx-\frac {2 a^4 (41 A+42 B-35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (7 A+9 B+5 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (\int \frac {21 (7 A+9 B+5 C) a^4+5 (13 A+21 B+35 C) \sec (c+d x) a^4}{\sqrt {\sec (c+d x)}}dx-\frac {2 a^4 (41 A+42 B-35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (7 A+9 B+5 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (\int \frac {21 (7 A+9 B+5 C) a^4+5 (13 A+21 B+35 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^4 (41 A+42 B-35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (7 A+9 B+5 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (21 a^4 (7 A+9 B+5 C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+5 a^4 (13 A+21 B+35 C) \int \sqrt {\sec (c+d x)}dx-\frac {2 a^4 (41 A+42 B-35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (7 A+9 B+5 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (21 a^4 (7 A+9 B+5 C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 a^4 (13 A+21 B+35 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^4 (41 A+42 B-35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (7 A+9 B+5 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^4 (13 A+21 B+35 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+21 a^4 (7 A+9 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-\frac {2 a^4 (41 A+42 B-35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (7 A+9 B+5 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^4 (13 A+21 B+35 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+21 a^4 (7 A+9 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^4 (41 A+42 B-35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (7 A+9 B+5 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^4 (13 A+21 B+35 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^4 (41 A+42 B-35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {42 a^4 (7 A+9 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {14 (7 A+9 B+5 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {1}{5} \left (\frac {14 (7 A+9 B+5 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d \sqrt {\sec (c+d x)}}+\frac {2}{3} \left (-\frac {2 a^4 (41 A+42 B-35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 a^4 (13 A+21 B+35 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {42 a^4 (7 A+9 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 (6 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

Input:

Int[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c 
 + d*x]^(7/2),x]
 

Output:

(2*A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + ((2*( 
6*A + 7*B)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2 
)) + ((14*(7*A + 9*B + 5*C)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(3*d*Sq 
rt[Sec[c + d*x]]) + (2*((42*a^4*(7*A + 9*B + 5*C)*Sqrt[Cos[c + d*x]]*Ellip 
ticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (10*a^4*(13*A + 21*B + 35*C)* 
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (2*a^ 
4*(41*A + 42*B - 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))/3)/5)/(7*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 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4258
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4485
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ 
e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1)   Int[(d*Csc 
[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x 
], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[ 
n, -1]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(726\) vs. \(2(246)=492\).

Time = 18.07 (sec) , antiderivative size = 727, normalized size of antiderivative = 2.68

method result size
default \(\text {Expression too large to display}\) \(727\)
parts \(\text {Expression too large to display}\) \(1092\)

Input:

int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, 
method=_RETURNVERBOSE)
 

Output:

-4/105*a^3*(120*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos 
(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-12*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d 
*x+1/2*c)^2)^(1/2)*(36*A+7*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+14*( 
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(43*A+21*B+5*C)*sin(1/ 
2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1 
/2*c)^2)^(1/2)*(104*A+63*B+70*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+6 
5*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2* 
c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c), 
2^(1/2))-147*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2* 
d*x+1/2*c),2^(1/2))+105*B*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^( 
1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti 
cF(cos(1/2*d*x+1/2*c),2^(1/2))-189*B*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 
/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+175*C*(sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))* 
(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-105*C*(-2*sin(1/2*d*x 
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin( 
1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1 
/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.86 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (13 \, A + 21 \, B + 35 \, C\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (13 \, A + 21 \, B + 35 \, C\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 i \, \sqrt {2} {\left (7 \, A + 9 \, B + 5 \, C\right )} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} {\left (7 \, A + 9 \, B + 5 \, C\right )} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (15 \, A a^{3} \cos \left (d x + c\right )^{3} + 21 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 5 \, {\left (26 \, A + 21 \, B + 7 \, C\right )} a^{3} \cos \left (d x + c\right ) + 105 \, C a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{105 \, d} \] Input:

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7 
/2),x, algorithm="fricas")
 

Output:

-2/105*(5*I*sqrt(2)*(13*A + 21*B + 35*C)*a^3*weierstrassPInverse(-4, 0, co 
s(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(13*A + 21*B + 35*C)*a^3*weiers 
trassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 21*I*sqrt(2)*(7*A + 
9*B + 5*C)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + 
 c) + I*sin(d*x + c))) + 21*I*sqrt(2)*(7*A + 9*B + 5*C)*a^3*weierstrassZet 
a(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (15* 
A*a^3*cos(d*x + c)^3 + 21*(3*A + B)*a^3*cos(d*x + c)^2 + 5*(26*A + 21*B + 
7*C)*a^3*cos(d*x + c) + 105*C*a^3)*sin(d*x + c)/sqrt(cos(d*x + c)))/d
 

Sympy [F]

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

integrate((a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)* 
*(7/2),x)
 

Output:

a**3*(Integral(A/sec(c + d*x)**(7/2), x) + Integral(3*A/sec(c + d*x)**(5/2 
), x) + Integral(3*A/sec(c + d*x)**(3/2), x) + Integral(A/sqrt(sec(c + d*x 
)), x) + Integral(B/sec(c + d*x)**(5/2), x) + Integral(3*B/sec(c + d*x)**( 
3/2), x) + Integral(3*B/sqrt(sec(c + d*x)), x) + Integral(B*sqrt(sec(c + d 
*x)), x) + Integral(C/sec(c + d*x)**(3/2), x) + Integral(3*C/sqrt(sec(c + 
d*x)), x) + Integral(3*C*sqrt(sec(c + d*x)), x) + Integral(C*sec(c + d*x)* 
*(3/2), x))
                                                                                    
                                                                                    
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7 
/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7 
/2),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3/s 
ec(d*x + c)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \] Input:

int(((a + a/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/co 
s(c + d*x))^(7/2),x)
 

Output:

int(((a + a/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/co 
s(c + d*x))^(7/2), x)
 

Reduce [F]

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

int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x)
 

Output:

a**3*(int(sqrt(sec(c + d*x))/sec(c + d*x)**4,x)*a + 3*int(sqrt(sec(c + d*x 
))/sec(c + d*x)**3,x)*a + int(sqrt(sec(c + d*x))/sec(c + d*x)**3,x)*b + 3* 
int(sqrt(sec(c + d*x))/sec(c + d*x)**2,x)*a + 3*int(sqrt(sec(c + d*x))/sec 
(c + d*x)**2,x)*b + int(sqrt(sec(c + d*x))/sec(c + d*x)**2,x)*c + int(sqrt 
(sec(c + d*x))/sec(c + d*x),x)*a + 3*int(sqrt(sec(c + d*x))/sec(c + d*x),x 
)*b + 3*int(sqrt(sec(c + d*x))/sec(c + d*x),x)*c + int(sqrt(sec(c + d*x)), 
x)*b + 3*int(sqrt(sec(c + d*x)),x)*c + int(sqrt(sec(c + d*x))*sec(c + d*x) 
,x)*c)