3.10.95 \(\int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [995]

3.10.95.1 Optimal result
3.10.95.2 Mathematica [A] (verified)
3.10.95.3 Rubi [A] (verified)
3.10.95.4 Maple [B] (verified)
3.10.95.5 Fricas [C] (verification not implemented)
3.10.95.6 Sympy [F]
3.10.95.7 Maxima [F]
3.10.95.8 Giac [F]
3.10.95.9 Mupad [F(-1)]

3.10.95.1 Optimal result

Integrand size = 43, antiderivative size = 242 \[ \int \frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \left (6 a A b+3 a^2 B+5 b^2 B+10 a b C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (14 a b B+7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a (4 A b+7 a B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (4 A b^2+14 a b B+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

output
2/35*a*(4*A*b+7*B*a)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/7*A*(a+b*sec(d*x+c))^ 
2*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/21*(4*A*b^2+14*B*a*b+a^2*(5*A+7*C))*sin( 
d*x+c)/d/sec(d*x+c)^(1/2)+2/5*(6*A*a*b+3*B*a^2+5*B*b^2+10*C*a*b)*(cos(1/2* 
d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2 
))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/21*(14*B*a*b+7*b^2*(A+3*C)+a^2*(5 
*A+7*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2 
*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d
 
3.10.95.2 Mathematica [A] (verified)

Time = 11.53 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.59 \[ \int \frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \cos ^4(c+d x) \left (\frac {2 \left (126 a A b+63 a^2 B+105 b^2 B+210 a b C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}+2 \left (25 a^2 A+35 A b^2+70 a b B+35 a^2 C+105 b^2 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}\right ) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{105 d (b+a \cos (c+d x))^2 (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {1}{5} a (2 A b+a B) \sin (c+d x)+\frac {1}{21} \left (13 a^2 A+14 A b^2+28 a b B+14 a^2 C\right ) \sin (2 (c+d x))+\frac {1}{5} a (2 A b+a B) \sin (3 (c+d x))+\frac {1}{14} a^2 A \sin (4 (c+d x))\right )}{d (b+a \cos (c+d x))^2 (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {7}{2}}(c+d x)} \]

input
Integrate[((a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)) 
/Sec[c + d*x]^(7/2),x]
 
output
(2*Cos[c + d*x]^4*((2*(126*a*A*b + 63*a^2*B + 105*b^2*B + 210*a*b*C)*Ellip 
ticE[(c + d*x)/2, 2])/(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 2*(25*a^2* 
A + 35*A*b^2 + 70*a*b*B + 35*a^2*C + 105*b^2*C)*Sqrt[Cos[c + d*x]]*Ellipti 
cF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])*(a + b*Sec[c + d*x])^2*(A + B*Sec[c 
 + d*x] + C*Sec[c + d*x]^2))/(105*d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B* 
Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((a + b*Sec[c + d*x])^2*(A + B*Sec[c 
 + d*x] + C*Sec[c + d*x]^2)*((a*(2*A*b + a*B)*Sin[c + d*x])/5 + ((13*a^2*A 
 + 14*A*b^2 + 28*a*b*B + 14*a^2*C)*Sin[2*(c + d*x)])/21 + (a*(2*A*b + a*B) 
*Sin[3*(c + d*x)])/5 + (a^2*A*Sin[4*(c + d*x)])/14))/(d*(b + a*Cos[c + d*x 
])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2))
 
3.10.95.3 Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4582, 27, 3042, 4562, 27, 3042, 4535, 3042, 4258, 3042, 3119, 4533, 3042, 4258, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4582

\(\displaystyle \frac {2}{7} \int \frac {(a+b \sec (c+d x)) \left (b (A+7 C) \sec ^2(c+d x)+(5 a A+7 b B+7 a C) \sec (c+d x)+4 A b+7 a B\right )}{2 \sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int \frac {(a+b \sec (c+d x)) \left (b (A+7 C) \sec ^2(c+d x)+(5 a A+7 b B+7 a C) \sec (c+d x)+4 A b+7 a B\right )}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(5 a A+7 b B+7 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 A b+7 a B\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4562

\(\displaystyle \frac {1}{7} \left (\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2}{5} \int -\frac {5 b^2 (A+7 C) \sec ^2(c+d x)+7 \left (3 B a^2+6 A b a+10 b C a+5 b^2 B\right ) \sec (c+d x)+5 \left ((5 A+7 C) a^2+14 b B a+4 A b^2\right )}{2 \sec ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {5 b^2 (A+7 C) \sec ^2(c+d x)+7 \left (3 B a^2+6 A b a+10 b C a+5 b^2 B\right ) \sec (c+d x)+5 \left ((5 A+7 C) a^2+14 b B a+4 A b^2\right )}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {5 b^2 (A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+7 \left (3 B a^2+6 A b a+10 b C a+5 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+5 \left ((5 A+7 C) a^2+14 b B a+4 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4535

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\int \frac {5 b^2 (A+7 C) \sec ^2(c+d x)+5 \left ((5 A+7 C) a^2+14 b B a+4 A b^2\right )}{\sec ^{\frac {3}{2}}(c+d x)}dx+7 \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (7 \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {5 b^2 (A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 \left ((5 A+7 C) a^2+14 b B a+4 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\int \frac {5 b^2 (A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 \left ((5 A+7 C) a^2+14 b B a+4 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+7 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right ) \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\int \frac {5 b^2 (A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 \left ((5 A+7 C) a^2+14 b B a+4 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+7 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\int \frac {5 b^2 (A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 \left ((5 A+7 C) a^2+14 b B a+4 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {14 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right )}{d}\right )+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4533

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {5}{3} \left (a^2 (5 A+7 C)+14 a b B+7 b^2 (A+3 C)\right ) \int \sqrt {\sec (c+d x)}dx+\frac {10 \sin (c+d x) \left (a^2 (5 A+7 C)+14 a b B+4 A b^2\right )}{3 d \sqrt {\sec (c+d x)}}+\frac {14 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right )}{d}\right )+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {5}{3} \left (a^2 (5 A+7 C)+14 a b B+7 b^2 (A+3 C)\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {10 \sin (c+d x) \left (a^2 (5 A+7 C)+14 a b B+4 A b^2\right )}{3 d \sqrt {\sec (c+d x)}}+\frac {14 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right )}{d}\right )+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {5}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^2 (5 A+7 C)+14 a b B+7 b^2 (A+3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {10 \sin (c+d x) \left (a^2 (5 A+7 C)+14 a b B+4 A b^2\right )}{3 d \sqrt {\sec (c+d x)}}+\frac {14 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right )}{d}\right )+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {5}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^2 (5 A+7 C)+14 a b B+7 b^2 (A+3 C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {10 \sin (c+d x) \left (a^2 (5 A+7 C)+14 a b B+4 A b^2\right )}{3 d \sqrt {\sec (c+d x)}}+\frac {14 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right )}{d}\right )+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

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

input
Int[((a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c 
 + d*x]^(7/2),x]
 
output
(2*A*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + ((2*a 
*(4*A*b + 7*a*B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ((14*(6*a*A*b + 
3*a^2*B + 5*b^2*B + 10*a*b*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] 
*Sqrt[Sec[c + d*x]])/d + (10*(14*a*b*B + 7*b^2*(A + 3*C) + a^2*(5*A + 7*C) 
)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + 
 (10*(4*A*b^2 + 14*a*b*B + a^2*(5*A + 7*C))*Sin[c + d*x])/(3*d*Sqrt[Sec[c 
+ d*x]]))/5)/7
 

3.10.95.3.1 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 4533
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + 
Simp[(C*m + A*(m + 1))/(b^2*m)   Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr 
eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
 

rule 4535
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

rule 4562
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 
_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si 
mp[1/(d*n)   Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* 
b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ 
{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
 

rule 4582
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/(d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d* 
Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs 
c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a 
, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
 
3.10.95.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(705\) vs. \(2(270)=540\).

Time = 7.49 (sec) , antiderivative size = 706, normalized size of antiderivative = 2.92

method result size
default \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (240 A \,a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-360 A \,a^{2}-336 a A b -168 B \,a^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (280 A \,a^{2}+336 a A b +140 A \,b^{2}+168 B \,a^{2}+280 B a b +140 C \,a^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-80 A \,a^{2}-84 a A b -70 A \,b^{2}-42 B \,a^{2}-140 B a b -70 C \,a^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+25 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, a^{2}+35 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, b^{2}-126 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, a b +70 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, a b -63 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, a^{2}-105 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, b^{2}+35 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, a^{2}+105 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, b^{2}-210 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, a b \right )}{105 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(706\)
parts \(\text {Expression too large to display}\) \(889\)

input
int((a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, 
method=_RETURNVERBOSE)
 
output
-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*A*a^2* 
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+(-360*A*a^2-336*A*a*b-168*B*a^2)*s 
in(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(280*A*a^2+336*A*a*b+140*A*b^2+168* 
B*a^2+280*B*a*b+140*C*a^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-80*A* 
a^2-84*A*a*b-70*A*b^2-42*B*a^2-140*B*a*b-70*C*a^2)*sin(1/2*d*x+1/2*c)^2*co 
s(1/2*d*x+1/2*c)+25*A*EllipticF(cos(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)*a^2+35*A*EllipticF(cos(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)*b^2-126*A*EllipticE(cos(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)*a*b+70*B*EllipticF(cos(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)*a*b-63*B*EllipticE(cos(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)*a^2-105*B*EllipticE(cos(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)*b^2+35*C*EllipticF(cos(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)*a^2+105*C*EllipticF(cos(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)* 
b^2-210*C*EllipticE(cos(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)*a*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*cos(1/2*d*x+1/2*c)^2-1)^(1/2)...
 
3.10.95.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, {\left (5 \, A + 7 \, C\right )} a^{2} + 14 i \, B a b + 7 i \, {\left (A + 3 \, C\right )} b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, {\left (5 \, A + 7 \, C\right )} a^{2} - 14 i \, B a b - 7 i \, {\left (A + 3 \, C\right )} b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (-3 i \, B a^{2} - 2 i \, {\left (3 \, A + 5 \, C\right )} a b - 5 i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (3 i \, B a^{2} + 2 i \, {\left (3 \, A + 5 \, C\right )} a b + 5 i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (15 \, A a^{2} \cos \left (d x + c\right )^{3} + 21 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left ({\left (5 \, A + 7 \, C\right )} a^{2} + 14 \, B a b + 7 \, A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d} \]

input
integrate((a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7 
/2),x, algorithm="fricas")
 
output
-1/105*(5*sqrt(2)*(I*(5*A + 7*C)*a^2 + 14*I*B*a*b + 7*I*(A + 3*C)*b^2)*wei 
erstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*(5* 
A + 7*C)*a^2 - 14*I*B*a*b - 7*I*(A + 3*C)*b^2)*weierstrassPInverse(-4, 0, 
cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(-3*I*B*a^2 - 2*I*(3*A + 5*C)* 
a*b - 5*I*B*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
 + c) + I*sin(d*x + c))) + 21*sqrt(2)*(3*I*B*a^2 + 2*I*(3*A + 5*C)*a*b + 5 
*I*B*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - 
 I*sin(d*x + c))) - 2*(15*A*a^2*cos(d*x + c)^3 + 21*(B*a^2 + 2*A*a*b)*cos( 
d*x + c)^2 + 5*((5*A + 7*C)*a^2 + 14*B*a*b + 7*A*b^2)*cos(d*x + c))*sin(d* 
x + c)/sqrt(cos(d*x + c)))/d
 
3.10.95.6 Sympy [F]

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

input
integrate((a+b*sec(d*x+c))**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)* 
*(7/2),x)
 
output
Integral((a + b*sec(c + d*x))**2*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)/ 
sec(c + d*x)**(7/2), x)
 
3.10.95.7 Maxima [F]

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

input
integrate((a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7 
/2),x, algorithm="maxima")
 
output
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^2/s 
ec(d*x + c)^(7/2), x)
 
3.10.95.8 Giac [F]

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

input
integrate((a+b*sec(d*x+c))^2*(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)*(b*sec(d*x + c) + a)^2/s 
ec(d*x + c)^(7/2), x)
 
3.10.95.9 Mupad [F(-1)]

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

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