Integrand size = 43, antiderivative size = 342 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=-\frac {2 \left (18 a b B+3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 \left (10 a A b+5 a^2 B+7 b^2 B+14 a b C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 \left (18 a b B+3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 \left (10 a A b+5 a^2 B+7 b^2 B+14 a b C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 \left (4 A b^2+18 a b B+a^2 (7 A+9 C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a (4 A b+9 a B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d} \]
2/21*(10*A*a*b+5*B*a^2+7*B*b^2+14*C*a*b)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/4 5*(4*A*b^2+18*B*a*b+a^2*(7*A+9*C))*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/63*a*(4 *A*b+9*B*a)*sec(d*x+c)^(7/2)*sin(d*x+c)/d+2/9*A*(a+b*cos(d*x+c))^2*sec(d*x +c)^(9/2)*sin(d*x+c)/d+2/15*(18*B*a*b+3*b^2*(3*A+5*C)+a^2*(7*A+9*C))*sin(d *x+c)*sec(d*x+c)^(1/2)/d-2/15*(18*B*a*b+3*b^2*(3*A+5*C)+a^2*(7*A+9*C))*(co s(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*(10*A*a*b+5*B*a^2+7*B*b^ 2+14*C*a*b)*(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
Time = 8.03 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.04 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\frac {\frac {2 \left (-49 a^2 A-63 A b^2-126 a b B-63 a^2 C-105 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}+2 \left (50 a A b+25 a^2 B+35 b^2 B+70 a b C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{105 d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {2}{15} \left (7 a^2 A+9 A b^2+18 a b B+9 a^2 C+15 b^2 C\right ) \sin (c+d x)+\frac {2}{7} \sec ^3(c+d x) \left (2 a A b \sin (c+d x)+a^2 B \sin (c+d x)\right )+\frac {2}{45} \sec ^2(c+d x) \left (7 a^2 A \sin (c+d x)+9 A b^2 \sin (c+d x)+18 a b B \sin (c+d x)+9 a^2 C \sin (c+d x)\right )+\frac {2}{21} \sec (c+d x) \left (10 a A b \sin (c+d x)+5 a^2 B \sin (c+d x)+7 b^2 B \sin (c+d x)+14 a b C \sin (c+d x)\right )+\frac {2}{9} a^2 A \sec ^3(c+d x) \tan (c+d x)\right )}{d} \]
Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S ec[c + d*x]^(11/2),x]
((2*(-49*a^2*A - 63*A*b^2 - 126*a*b*B - 63*a^2*C - 105*b^2*C)*EllipticE[(c + d*x)/2, 2])/(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 2*(50*a*A*b + 25* a^2*B + 35*b^2*B + 70*a*b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]* Sqrt[Sec[c + d*x]])/(105*d) + (Sqrt[Sec[c + d*x]]*((2*(7*a^2*A + 9*A*b^2 + 18*a*b*B + 9*a^2*C + 15*b^2*C)*Sin[c + d*x])/15 + (2*Sec[c + d*x]^3*(2*a* A*b*Sin[c + d*x] + a^2*B*Sin[c + d*x]))/7 + (2*Sec[c + d*x]^2*(7*a^2*A*Sin [c + d*x] + 9*A*b^2*Sin[c + d*x] + 18*a*b*B*Sin[c + d*x] + 9*a^2*C*Sin[c + d*x]))/45 + (2*Sec[c + d*x]*(10*a*A*b*Sin[c + d*x] + 5*a^2*B*Sin[c + d*x] + 7*b^2*B*Sin[c + d*x] + 14*a*b*C*Sin[c + d*x]))/21 + (2*a^2*A*Sec[c + d* x]^3*Tan[c + d*x])/9))/d
Time = 1.60 (sec) , antiderivative size = 288, normalized size of antiderivative = 0.84, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4709, 3042, 3526, 27, 3042, 3510, 27, 3042, 3500, 27, 3042, 3227, 3042, 3116, 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.
| \(\displaystyle \int \sec ^{\frac {11}{2}}(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^{11/2} (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(a+b \cos (c+d x))^2 \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {11}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{9} \int \frac {(a+b \cos (c+d x)) \left (3 b (A+3 C) \cos ^2(c+d x)+(7 a A+9 b B+9 a C) \cos (c+d x)+4 A b+9 a B\right )}{2 \cos ^{\frac {9}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \int \frac {(a+b \cos (c+d x)) \left (3 b (A+3 C) \cos ^2(c+d x)+(7 a A+9 b B+9 a C) \cos (c+d x)+4 A b+9 a B\right )}{\cos ^{\frac {9}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (3 b (A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(7 a A+9 b B+9 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+4 A b+9 a B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {2}{7} \int -\frac {21 b^2 (A+3 C) \cos ^2(c+d x)+9 \left (5 B a^2+10 A b a+14 b C a+7 b^2 B\right ) \cos (c+d x)+7 \left ((7 A+9 C) a^2+18 b B a+4 A b^2\right )}{2 \cos ^{\frac {7}{2}}(c+d x)}dx\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {1}{7} \int \frac {21 b^2 (A+3 C) \cos ^2(c+d x)+9 \left (5 B a^2+10 A b a+14 b C a+7 b^2 B\right ) \cos (c+d x)+7 \left ((7 A+9 C) a^2+18 b B a+4 A b^2\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {1}{7} \int \frac {21 b^2 (A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+9 \left (5 B a^2+10 A b a+14 b C a+7 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+7 \left ((7 A+9 C) a^2+18 b B a+4 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {1}{7} \left (\frac {2}{5} \int \frac {3 \left (15 \left (5 B a^2+10 A b a+14 b C a+7 b^2 B\right )+7 \left ((7 A+9 C) a^2+18 b B a+3 b^2 (3 A+5 C)\right ) \cos (c+d x)\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {14 \sin (c+d x) \left (a^2 (7 A+9 C)+18 a b B+4 A b^2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} \int \frac {15 \left (5 B a^2+10 A b a+14 b C a+7 b^2 B\right )+7 \left ((7 A+9 C) a^2+18 b B a+3 b^2 (3 A+5 C)\right ) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {14 \sin (c+d x) \left (a^2 (7 A+9 C)+18 a b B+4 A b^2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} \int \frac {15 \left (5 B a^2+10 A b a+14 b C a+7 b^2 B\right )+7 \left ((7 A+9 C) a^2+18 b B a+3 b^2 (3 A+5 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {14 \sin (c+d x) \left (a^2 (7 A+9 C)+18 a b B+4 A b^2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} \left (15 \left (5 a^2 B+10 a A b+14 a b C+7 b^2 B\right ) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx+7 \left (a^2 (7 A+9 C)+18 a b B+3 b^2 (3 A+5 C)\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {14 \sin (c+d x) \left (a^2 (7 A+9 C)+18 a b B+4 A b^2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} \left (15 \left (5 a^2 B+10 a A b+14 a b C+7 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+7 \left (a^2 (7 A+9 C)+18 a b B+3 b^2 (3 A+5 C)\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )+\frac {14 \sin (c+d x) \left (a^2 (7 A+9 C)+18 a b B+4 A b^2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} \left (15 \left (5 a^2 B+10 a A b+14 a b C+7 b^2 B\right ) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 \left (a^2 (7 A+9 C)+18 a b B+3 b^2 (3 A+5 C)\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {14 \sin (c+d x) \left (a^2 (7 A+9 C)+18 a b B+4 A b^2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} \left (15 \left (5 a^2 B+10 a A b+14 a b C+7 b^2 B\right ) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 \left (a^2 (7 A+9 C)+18 a b B+3 b^2 (3 A+5 C)\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {14 \sin (c+d x) \left (a^2 (7 A+9 C)+18 a b B+4 A b^2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} \left (15 \left (5 a^2 B+10 a A b+14 a b C+7 b^2 B\right ) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 \left (a^2 (7 A+9 C)+18 a b B+3 b^2 (3 A+5 C)\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {14 \sin (c+d x) \left (a^2 (7 A+9 C)+18 a b B+4 A b^2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{9} \left (\frac {1}{7} \left (\frac {14 \sin (c+d x) \left (a^2 (7 A+9 C)+18 a b B+4 A b^2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {3}{5} \left (15 \left (5 a^2 B+10 a A b+14 a b C+7 b^2 B\right ) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 \left (a^2 (7 A+9 C)+18 a b B+3 b^2 (3 A+5 C)\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {2 a (9 a B+4 A b) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2)) + ((2*a*(4*A*b + 9*a*B)*Sin[c + d*x])/(7*d *Cos[c + d*x]^(7/2)) + ((14*(4*A*b^2 + 18*a*b*B + a^2*(7*A + 9*C))*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (3*(15*(10*a*A*b + 5*a^2*B + 7*b^2*B + 1 4*a*b*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[ c + d*x]^(3/2))) + 7*(18*a*b*B + 3*b^2*(3*A + 5*C) + a^2*(7*A + 9*C))*((-2 *EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))) /5)/7)/9)
3.15.58.3.1 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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
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]
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]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d , 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1168\) vs. \(2(362)=724\).
Time = 1463.60 (sec) , antiderivative size = 1169, normalized size of antiderivative = 3.42
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1169\) |
| parts | \(\text {Expression too large to display}\) | \(1422\) |
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x ,method=_RETURNVERBOSE)
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C*b^2/sin(1/ 2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2 *d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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)*EllipticE(cos(1/2*d*x+ 1/2*c),2^(1/2)))+2*A*a^2*(-1/144*cos(1/2*d*x+1/2*c)*(-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)^2-1/2)^5-7/180*cos(1/2* d*x+1/2*c)*(-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)^2-1/2)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos (1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(sin(1/2*d*x+1/2*c)^ 2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/ 2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-7/15*(sin(1/2* d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2* c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-El lipticE(cos(1/2*d*x+1/2*c),2^(1/2))))+2*b*(B*b+2*C*a)*(-1/6*cos(1/2*d*x+1/ 2*c)*(-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)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^ (1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1 /2*d*x+1/2*c),2^(1/2)))+2*a*(2*A*b+B*a)*(-1/56*cos(1/2*d*x+1/2*c)*(-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)^2-1/2)^4- 5/42*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.17 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=-\frac {15 \, \sqrt {2} {\left (5 i \, B a^{2} + 2 i \, {\left (5 \, A + 7 \, C\right )} a b + 7 i \, B b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 \, \sqrt {2} {\left (-5 i \, B a^{2} - 2 i \, {\left (5 \, A + 7 \, C\right )} a b - 7 i \, B b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (i \, {\left (7 \, A + 9 \, C\right )} a^{2} + 18 i \, B a b + 3 i \, {\left (3 \, A + 5 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (-i \, {\left (7 \, A + 9 \, C\right )} a^{2} - 18 i \, B a b - 3 i \, {\left (3 \, A + 5 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (21 \, {\left ({\left (7 \, A + 9 \, C\right )} a^{2} + 18 \, B a b + 3 \, {\left (3 \, A + 5 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (5 \, B a^{2} + 2 \, {\left (5 \, A + 7 \, C\right )} a b + 7 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 35 \, A a^{2} + 7 \, {\left ({\left (7 \, A + 9 \, C\right )} a^{2} + 18 \, B a b + 9 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 45 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{315 \, d \cos \left (d x + c\right )^{4}} \]
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1 1/2),x, algorithm="fricas")
-1/315*(15*sqrt(2)*(5*I*B*a^2 + 2*I*(5*A + 7*C)*a*b + 7*I*B*b^2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*sqrt( 2)*(-5*I*B*a^2 - 2*I*(5*A + 7*C)*a*b - 7*I*B*b^2)*cos(d*x + c)^4*weierstra ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(I*(7*A + 9* C)*a^2 + 18*I*B*a*b + 3*I*(3*A + 5*C)*b^2)*cos(d*x + c)^4*weierstrassZeta( -4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*sqr t(2)*(-I*(7*A + 9*C)*a^2 - 18*I*B*a*b - 3*I*(3*A + 5*C)*b^2)*cos(d*x + c)^ 4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d *x + c))) - 2*(21*((7*A + 9*C)*a^2 + 18*B*a*b + 3*(3*A + 5*C)*b^2)*cos(d*x + c)^4 + 15*(5*B*a^2 + 2*(5*A + 7*C)*a*b + 7*B*b^2)*cos(d*x + c)^3 + 35*A *a^2 + 7*((7*A + 9*C)*a^2 + 18*B*a*b + 9*A*b^2)*cos(d*x + c)^2 + 45*(B*a^2 + 2*A*a*b)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) ^4)
Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\text {Timed out} \]
\[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {11}{2}} \,d x } \]
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1 1/2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*s ec(d*x + c)^(11/2), x)
\[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {11}{2}} \,d x } \]
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1 1/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*s ec(d*x + c)^(11/2), x)
Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{11/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]
int((1/cos(c + d*x))^(11/2)*(a + b*cos(c + d*x))^2*(A + B*cos(c + d*x) + C *cos(c + d*x)^2),x)