Integrand size = 35, antiderivative size = 253 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {4 a^3 (9 A-5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (3 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {4 a^3 (21 A+5 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 (11 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:
-4/5*a^3*(9*A-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+4/3*a^3*(3*A+5*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2* d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d-4/15*a^3*(21*A+5*C)*sin(d*x+c)/d/sec (d*x+c)^(1/2)+2/5*(11*A+5*C)*(a^3+a^3*cos(d*x+c))*sec(d*x+c)^(1/2)*sin(d*x +c)/d+4/5*A*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d+2/5*A*( a+a*cos(d*x+c))^3*sec(d*x+c)^(5/2)*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.85 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.10 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {a^3 e^{-i d x} \csc (c) \sec (c) \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (4 (9 A-5 C) e^{-i (c-d x)} \left (-1+e^{4 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+\frac {1}{2} \sec ^2(c+d x) \sin (2 c) \left (-36 i (9 A-5 C) \cos (c+d x)-108 i A \cos (3 (c+d x))+60 i C \cos (3 (c+d x))+80 (3 A+5 C) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+132 A \sin (c+d x)+30 C \sin (c+d x)+60 A \sin (2 (c+d x))+10 C \sin (2 (c+d x))+108 A \sin (3 (c+d x))+30 C \sin (3 (c+d x))+5 C \sin (4 (c+d x))\right )\right )}{60 d} \] Input:
Integrate[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2) ,x]
Output:
(a^3*Csc[c]*Sec[c]*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*((4*(9*A - 5 *C)*(-1 + E^((4*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2 , 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c - d*x)) + (Sec[c + d*x]^2*Sin[2 *c]*((-36*I)*(9*A - 5*C)*Cos[c + d*x] - (108*I)*A*Cos[3*(c + d*x)] + (60*I )*C*Cos[3*(c + d*x)] + 80*(3*A + 5*C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d* x)/2, 2] + 132*A*Sin[c + d*x] + 30*C*Sin[c + d*x] + 60*A*Sin[2*(c + d*x)] + 10*C*Sin[2*(c + d*x)] + 108*A*Sin[3*(c + d*x)] + 30*C*Sin[3*(c + d*x)] + 5*C*Sin[4*(c + d*x)]))/2))/(60*d*E^(I*d*x))
Time = 1.69 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.94, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4709, 3042, 3523, 27, 3042, 3454, 27, 3042, 3454, 3042, 3447, 3042, 3502, 27, 3042, 3227, 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.
| \(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^3 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^{7/2} (a \cos (c+d x)+a)^3 \left (A+C \cos (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^3 \left (C \cos ^2(c+d x)+A\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^3 (6 a A-a (3 A-5 C) \cos (c+d x))}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^3 (6 a A-a (3 A-5 C) \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (6 a A-a (3 A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{3} \int \frac {3 (\cos (c+d x) a+a)^2 \left (a^2 (11 A+5 C)-a^2 (9 A-5 C) \cos (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^2 \left (a^2 (11 A+5 C)-a^2 (9 A-5 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a^2 (11 A+5 C)-a^2 (9 A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a) \left (2 a^3 (6 A+5 C)-a^3 (21 A+5 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 (11 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (2 a^3 (6 A+5 C)-a^3 (21 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (11 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {-\left ((21 A+5 C) \cos ^2(c+d x) a^4\right )+2 (6 A+5 C) a^4+\left (2 a^4 (6 A+5 C)-a^4 (21 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 (11 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {-\left ((21 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4\right )+2 (6 A+5 C) a^4+\left (2 a^4 (6 A+5 C)-a^4 (21 A+5 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (11 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (\frac {2}{3} \int \frac {5 a^4 (3 A+5 C)-3 a^4 (9 A-5 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 a^4 (21 A+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (11 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (\frac {1}{3} \int \frac {5 a^4 (3 A+5 C)-3 a^4 (9 A-5 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 a^4 (21 A+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (11 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (\frac {1}{3} \int \frac {5 a^4 (3 A+5 C)-3 a^4 (9 A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^4 (21 A+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (11 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (\frac {1}{3} \left (5 a^4 (3 A+5 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^4 (9 A-5 C) \int \sqrt {\cos (c+d x)}dx\right )-\frac {2 a^4 (21 A+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (11 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (\frac {1}{3} \left (5 a^4 (3 A+5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^4 (9 A-5 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 a^4 (21 A+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (11 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (\frac {1}{3} \left (5 a^4 (3 A+5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a^4 (9 A-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^4 (21 A+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (11 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 (11 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+2 \left (\frac {1}{3} \left (\frac {10 a^4 (3 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^4 (9 A-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^4 (21 A+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {4 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
Input:
Int[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((4*A*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)) + (2*(11*A + 5*C)*(a^4 + a^4*Cos[c + d*x])*S in[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + 2*(((-6*a^4*(9*A - 5*C)*EllipticE[(c + d*x)/2, 2])/d + (10*a^4*(3*A + 5*C)*EllipticF[(c + d*x)/2, 2])/d)/3 - ( 2*a^4*(21*A + 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/(5*a))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + 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/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a *d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* (c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
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. \(938\) vs. \(2(228)=456\).
Time = 586.82 (sec) , antiderivative size = 939, normalized size of antiderivative = 3.71
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(939\) |
| parts | \(\text {Expression too large to display}\) | \(1234\) |
Input:
int((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x,method=_RETUR NVERBOSE)
Output:
-4/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3/(8*sin (1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 2*d*x+1/2*c)^3*(-40*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+216*A*cos(1/ 2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-60*A*(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))*sin(1/2*d *x+1/2*c)^4-108*A*(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))*sin(1/2*d*x+1/2*c)^4+120*C*cos (1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-100*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))*sin(1 /2*d*x+1/2*c)^4+60*C*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-246*A* cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+60*A*(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))*sin (1/2*d*x+1/2*c)^2+108*A*(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))*sin(1/2*d*x+1/2*c)^2-90* C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+100*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))* sin(1/2*d*x+1/2*c)^2-60*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/ 2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+7 2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A-15*A*(sin(1/2*d*x+1/2*c)^2)...
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.00 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (9 \, A - 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (9 \, A - 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (5 \, C a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (18 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, A a^{3} \cos \left (d x + c\right ) + 3 \, A a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algori thm="fricas")
Output:
-2/15*(5*I*sqrt(2)*(3*A + 5*C)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(3*A + 5*C)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt( 2)*(9*A - 5*C)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInver se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(9*A - 5*C)*a^3*co s(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c ) - I*sin(d*x + c))) - (5*C*a^3*cos(d*x + c)^3 + 3*(18*A + 5*C)*a^3*cos(d* x + c)^2 + 15*A*a^3*cos(d*x + c) + 3*A*a^3)*sin(d*x + c)/sqrt(cos(d*x + c) ))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**3*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2),x)
Output:
Timed out
\[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algori thm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2) , x)
\[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algori thm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2) , x)
Timed out. \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3 \,d x \] Input:
int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^3,x )
Output:
int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^3, x)
\[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=a^{3} \left (3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )^{3}d x \right ) c +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{3}d x \right ) a +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{3}d x \right ) c +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) a \right ) \] Input:
int((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x)
Output:
a**3*(3*int(sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x)**3,x)*a + int(sqr t(sec(c + d*x))*cos(c + d*x)**5*sec(c + d*x)**3,x)*c + 3*int(sqrt(sec(c + d*x))*cos(c + d*x)**4*sec(c + d*x)**3,x)*c + int(sqrt(sec(c + d*x))*cos(c + d*x)**3*sec(c + d*x)**3,x)*a + 3*int(sqrt(sec(c + d*x))*cos(c + d*x)**3* sec(c + d*x)**3,x)*c + 3*int(sqrt(sec(c + d*x))*cos(c + d*x)**2*sec(c + d* x)**3,x)*a + int(sqrt(sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**3,x)*c + int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x)*a)