Integrand size = 35, antiderivative size = 322 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\frac {32 a^3 (4184 A+4615 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{45045 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^3 (4184 A+4615 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45045 d \sqrt {a+a \cos (c+d x)}}+\frac {4 a^3 (4184 A+4615 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (4184 A+4615 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (280 A+299 B) \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (16 A+13 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {13}{2}}(c+d x) \sin (c+d x)}{13 d} \] Output:
32/45045*a^3*(4184*A+4615*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c) )^(1/2)+16/45045*a^3*(4184*A+4615*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*co s(d*x+c))^(1/2)+4/15015*a^3*(4184*A+4615*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/ (a+a*cos(d*x+c))^(1/2)+2/9009*a^3*(4184*A+4615*B)*sec(d*x+c)^(7/2)*sin(d*x +c)/d/(a+a*cos(d*x+c))^(1/2)+2/1287*a^3*(280*A+299*B)*sec(d*x+c)^(9/2)*sin (d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/143*a^2*(16*A+13*B)*(a+a*cos(d*x+c))^(1 /2)*sec(d*x+c)^(11/2)*sin(d*x+c)/d+2/13*a*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x +c)^(13/2)*sin(d*x+c)/d
Time = 1.45 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.53 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (171806 A+162955 B+35 (5552 A+5083 B) \cos (c+d x)+14 (15167 A+15925 B) \cos (2 (c+d x))+62760 A \cos (3 (c+d x))+69225 B \cos (3 (c+d x))+62760 A \cos (4 (c+d x))+69225 B \cos (4 (c+d x))+8368 A \cos (5 (c+d x))+9230 B \cos (5 (c+d x))+8368 A \cos (6 (c+d x))+9230 B \cos (6 (c+d x))) \sec ^{\frac {13}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{90090 d} \] Input:
Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(15 /2),x]
Output:
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(171806*A + 162955*B + 35*(5552*A + 5083*B )*Cos[c + d*x] + 14*(15167*A + 15925*B)*Cos[2*(c + d*x)] + 62760*A*Cos[3*( c + d*x)] + 69225*B*Cos[3*(c + d*x)] + 62760*A*Cos[4*(c + d*x)] + 69225*B* Cos[4*(c + d*x)] + 8368*A*Cos[5*(c + d*x)] + 9230*B*Cos[5*(c + d*x)] + 836 8*A*Cos[6*(c + d*x)] + 9230*B*Cos[6*(c + d*x)])*Sec[c + d*x]^(13/2)*Tan[(c + d*x)/2])/(90090*d)
Time = 1.92 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.06, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 3440, 3042, 3454, 27, 3042, 3454, 27, 3042, 3459, 3042, 3251, 3042, 3251, 3042, 3251, 3042, 3250}
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.
| \(\displaystyle \int \sec ^{\frac {15}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2} (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{15/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3440 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {15}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{15/2}}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{13} \int \frac {(\cos (c+d x) a+a)^{3/2} (a (16 A+13 B)+a (8 A+13 B) \cos (c+d x))}{2 \cos ^{\frac {13}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \int \frac {(\cos (c+d x) a+a)^{3/2} (a (16 A+13 B)+a (8 A+13 B) \cos (c+d x))}{\cos ^{\frac {13}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (16 A+13 B)+a (8 A+13 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{13/2}}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {2}{11} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((280 A+299 B) a^2+(216 A+247 B) \cos (c+d x) a^2\right )}{2 \cos ^{\frac {11}{2}}(c+d x)}dx+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {1}{11} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((280 A+299 B) a^2+(216 A+247 B) \cos (c+d x) a^2\right )}{\cos ^{\frac {11}{2}}(c+d x)}dx+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {1}{11} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((280 A+299 B) a^2+(216 A+247 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {1}{11} \left (\frac {1}{9} a^2 (4184 A+4615 B) \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {9}{2}}(c+d x)}dx+\frac {2 a^3 (280 A+299 B) \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {1}{11} \left (\frac {1}{9} a^2 (4184 A+4615 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx+\frac {2 a^3 (280 A+299 B) \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {1}{11} \left (\frac {1}{9} a^2 (4184 A+4615 B) \left (\frac {6}{7} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (280 A+299 B) \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {1}{11} \left (\frac {1}{9} a^2 (4184 A+4615 B) \left (\frac {6}{7} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (280 A+299 B) \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {1}{11} \left (\frac {1}{9} a^2 (4184 A+4615 B) \left (\frac {6}{7} \left (\frac {4}{5} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (280 A+299 B) \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {1}{11} \left (\frac {1}{9} a^2 (4184 A+4615 B) \left (\frac {6}{7} \left (\frac {4}{5} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (280 A+299 B) \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {1}{11} \left (\frac {1}{9} a^2 (4184 A+4615 B) \left (\frac {6}{7} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (280 A+299 B) \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {1}{11} \left (\frac {1}{9} a^2 (4184 A+4615 B) \left (\frac {6}{7} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (280 A+299 B) \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3250 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{13} \left (\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{11 d \cos ^{\frac {11}{2}}(c+d x)}+\frac {1}{11} \left (\frac {2 a^3 (280 A+299 B) \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {1}{9} a^2 (4184 A+4615 B) \left (\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {6}{7} \left (\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4}{5} \left (\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )\right )\right )\right )\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\) |
Input:
Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(15/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a*A*(a + a*Cos[c + d*x])^(3/2)*S in[c + d*x])/(13*d*Cos[c + d*x]^(13/2)) + ((2*a^2*(16*A + 13*B)*Sqrt[a + a *Cos[c + d*x]]*Sin[c + d*x])/(11*d*Cos[c + d*x]^(11/2)) + ((2*a^3*(280*A + 299*B)*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2)*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(4184*A + 4615*B)*((2*a*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)*Sqrt[a + a*Cos[c + d*x]]) + (6*((2*a*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*((2*a*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[ a + a*Cos[c + d*x]]) + (4*a*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])))/5))/7))/9)/11)/13)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq rt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x ] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p Int[(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g , m, n, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && !(IntegerQ[m] && I ntegerQ[n])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) *(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d)) Int[Sqrt[a + b*Sin[e + f*x] ]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x ] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]
Time = 3.70 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.53
\[\frac {2 \sin \left (d x +c \right ) \left (\left (66944 \cos \left (d x +c \right )^{6}+33472 \cos \left (d x +c \right )^{5}+25104 \cos \left (d x +c \right )^{4}+20920 \cos \left (d x +c \right )^{3}+18305 \cos \left (d x +c \right )^{2}+11970 \cos \left (d x +c \right )+3465\right ) A +\cos \left (d x +c \right ) \left (73840 \cos \left (d x +c \right )^{5}+36920 \cos \left (d x +c \right )^{4}+27690 \cos \left (d x +c \right )^{3}+23075 \cos \left (d x +c \right )^{2}+14560 \cos \left (d x +c \right )+4095\right ) B \right ) \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{\frac {15}{2}} a^{2}}{45045 d \left (\cos \left (d x +c \right )+1\right )}\]
Input:
int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x)
Output:
2/45045/d*sin(d*x+c)*((66944*cos(d*x+c)^6+33472*cos(d*x+c)^5+25104*cos(d*x +c)^4+20920*cos(d*x+c)^3+18305*cos(d*x+c)^2+11970*cos(d*x+c)+3465)*A+cos(d *x+c)*(73840*cos(d*x+c)^5+36920*cos(d*x+c)^4+27690*cos(d*x+c)^3+23075*cos( d*x+c)^2+14560*cos(d*x+c)+4095)*B)*(a*(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*sec (d*x+c)^(15/2)/(cos(d*x+c)+1)*a^2
Time = 0.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.55 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\frac {2 \, {\left (16 \, {\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} + 8 \, {\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 6 \, {\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 35 \, {\left (523 \, A + 416 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 315 \, {\left (38 \, A + 13 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3465 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x, alg orithm="fricas")
Output:
2/45045*(16*(4184*A + 4615*B)*a^2*cos(d*x + c)^6 + 8*(4184*A + 4615*B)*a^2 *cos(d*x + c)^5 + 6*(4184*A + 4615*B)*a^2*cos(d*x + c)^4 + 5*(4184*A + 461 5*B)*a^2*cos(d*x + c)^3 + 35*(523*A + 416*B)*a^2*cos(d*x + c)^2 + 315*(38* A + 13*B)*a^2*cos(d*x + c) + 3465*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^7 + d*cos(d*x + c)^6)*sqrt(cos(d*x + c)))
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**(15/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 763 vs. \(2 (280) = 560\).
Time = 0.20 (sec) , antiderivative size = 763, normalized size of antiderivative = 2.37 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Too large to display} \] Input:
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x, alg orithm="maxima")
Output:
8/45045*((45045*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 165165*s qrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 414414*sqrt(2)*a^(5/2 )*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 604890*sqrt(2)*a^(5/2)*sin(d*x + c )^7/(cos(d*x + c) + 1)^7 + 522665*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 289185*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 88980*sqrt(2)*a^(5/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 11864*sqr t(2)*a^(5/2)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15)*A*(sin(d*x + c)^2/(cos (d*x + c) + 1)^2 + 1)^5/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(-si n(d*x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^6/(cos(d *x + c) + 1)^6 + 5*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + sin(d*x + c)^10/( cos(d*x + c) + 1)^10 + 1)) + 65*(693*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 3003*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 69 30*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 10098*sqrt(2)*a^( 5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 9053*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 4875*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 1500*sqrt(2)*a^(5/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 200*sqrt(2)*a^(5/2)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^5/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(1 5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(5*sin(d*x + c)^2/(c...
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x, alg orithm="giac")
Output:
Timed out
Time = 30.29 (sec) , antiderivative size = 789, normalized size of antiderivative = 2.45 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Too large to display} \] Input:
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(15/2)*(a + a*cos(c + d*x))^(5/2 ),x)
Output:
((1/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a^2*(a + a*(e xp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(4184*A + 4615*B)*32i )/(45045*d) - (a^2*exp(c*5i + d*x*5i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp (c*1i + d*x*1i)/2))^(1/2)*(2*A + 5*B)*16i)/(5*d) + (a^2*exp(c*8i + d*x*8i) *(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(2*A + 5*B) *16i)/(5*d) + (a^2*exp(c*6i + d*x*6i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp (c*1i + d*x*1i)/2))^(1/2)*(116*A + 115*B)*16i)/(35*d) - (a^2*exp(c*7i + d* x*7i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(116*A + 115*B)*16i)/(35*d) + (a^2*exp(c*4i + d*x*4i)*(a + a*(exp(- c*1i - d*x*1 i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(1046*A + 1075*B)*16i)/(315*d) - (a^2* exp(c*9i + d*x*9i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)) ^(1/2)*(1046*A + 1075*B)*16i)/(315*d) + (a^2*exp(c*2i + d*x*2i)*(a + a*(ex p(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(4184*A + 4615*B)*16i) /(3465*d) - (a^2*exp(c*11i + d*x*11i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp (c*1i + d*x*1i)/2))^(1/2)*(4184*A + 4615*B)*16i)/(3465*d) - (a^2*exp(c*13i + d*x*13i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)* (4184*A + 4615*B)*32i)/(45045*d)))/(exp(c*1i + d*x*1i) + 6*exp(c*2i + d*x* 2i) + 6*exp(c*3i + d*x*3i) + 15*exp(c*4i + d*x*4i) + 15*exp(c*5i + d*x*5i) + 20*exp(c*6i + d*x*6i) + 20*exp(c*7i + d*x*7i) + 15*exp(c*8i + d*x*8i) + 15*exp(c*9i + d*x*9i) + 6*exp(c*10i + d*x*10i) + 6*exp(c*11i + d*x*11i...
\[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\sqrt {a}\, a^{2} \left (2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{7}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{7}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{7}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{7}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{7}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{7}d x \right ) a \right ) \] Input:
int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x)
Output:
sqrt(a)*a**2*(2*int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x) *sec(c + d*x)**7,x)*a + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos( c + d*x)*sec(c + d*x)**7,x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**7,x)*b + int(sqrt(sec(c + d*x))*sqrt(cos (c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**7,x)*a + 2*int(sqrt(sec(c + d *x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**7,x)*b + int(sqr t(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*sec(c + d*x)**7,x)*a)