Integrand size = 43, antiderivative size = 226 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a^3 (9 A+5 B-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (3 A+5 (B+C)) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a^3 (6 A-5 B-20 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (B+2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}+\frac {2 (3 A-15 B-35 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d} \]
4/5*a^3*(9*A+5*B-5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli pticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/3*a^3*(3*A+5*B+5*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))/d+2/ 3*C*(a+a*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2*(B+2*C)*(a^2+a^2*co s(d*x+c))^2*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)+4/15*a^3*(6*A-5*B-20*C)*sin(d* x+c)*cos(d*x+c)^(1/2)/d+2/15*(3*A-15*B-35*C)*(a^3+a^3*cos(d*x+c))*sin(d*x+ c)*cos(d*x+c)^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 13.47 (sec) , antiderivative size = 1672, normalized size of antiderivative = 7.40 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \]
Integrate[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
(Cos[c + d*x]^(11/2)*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Se c[c + d*x] + C*Sec[c + d*x]^2)*(-1/20*((18*A + 5*B - 25*C + 18*A*Cos[2*c] + 15*B*Cos[2*c] + 5*C*Cos[2*c])*Csc[c]*Sec[c])/d + ((3*A + B)*Cos[d*x]*Sin [c])/(6*d) + (A*Cos[2*d*x]*Sin[2*c])/(20*d) + ((3*A + B)*Cos[c]*Sin[d*x])/ (6*d) + (C*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(6*d) + (Sec[c]*Sec[c + d*x]*(C *Sin[c] + 3*B*Sin[d*x] + 9*C*Sin[d*x]))/(6*d) + (A*Cos[2*c]*Sin[2*d*x])/(2 0*d)))/(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]) - (A*Cos[c + d*x] ^5*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2 ]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[ c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]* Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[ d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x ])*Sqrt[1 + Cot[c]^2]) - (5*B*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4 , 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*(a + a*Se c[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot [c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c] *Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (5*C* Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTa n[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c ...
Time = 1.64 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.05, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.488, Rules used = {3042, 4600, 3042, 3522, 27, 3042, 3454, 27, 3042, 3455, 27, 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.
| \(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^{5/2} (a \sec (c+d x)+a)^3 \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4600 |
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^3 \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^3 (3 a (B+2 C)+a (3 A-5 C) \cos (c+d x))}{2 \cos ^{\frac {3}{2}}(c+d x)}dx}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^3 (3 a (B+2 C)+a (3 A-5 C) \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)}dx}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (3 a (B+2 C)+a (3 A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^2 \left ((3 A+15 B+25 C) a^2+(3 A-15 B-35 C) \cos (c+d x) a^2\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^2 \left ((3 A+15 B+25 C) a^2+(3 A-15 B-35 C) \cos (c+d x) a^2\right )}{\sqrt {\cos (c+d x)}}dx+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left ((3 A+15 B+25 C) a^2+(3 A-15 B-35 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {2}{5} \int \frac {3 (\cos (c+d x) a+a) \left ((3 A+10 B+15 C) a^3+(6 A-5 B-20 C) \cos (c+d x) a^3\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {6}{5} \int \frac {(\cos (c+d x) a+a) \left ((3 A+10 B+15 C) a^3+(6 A-5 B-20 C) \cos (c+d x) a^3\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((3 A+10 B+15 C) a^3+(6 A-5 B-20 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {6}{5} \int \frac {(6 A-5 B-20 C) \cos ^2(c+d x) a^4+(3 A+10 B+15 C) a^4+\left ((6 A-5 B-20 C) a^4+(3 A+10 B+15 C) a^4\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6}{5} \int \frac {(6 A-5 B-20 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+(3 A+10 B+15 C) a^4+\left ((6 A-5 B-20 C) a^4+(3 A+10 B+15 C) a^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {2}{3} \int \frac {5 (3 A+5 (B+C)) a^4+3 (9 A+5 B-5 C) \cos (c+d x) a^4}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 a^4 (6 A-5 B-20 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \int \frac {5 (3 A+5 (B+C)) a^4+3 (9 A+5 B-5 C) \cos (c+d x) a^4}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^4 (6 A-5 B-20 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \int \frac {5 (3 A+5 (B+C)) a^4+3 (9 A+5 B-5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^4 (6 A-5 B-20 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \left (5 a^4 (3 A+5 (B+C)) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^4 (9 A+5 B-5 C) \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 a^4 (6 A-5 B-20 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \left (5 a^4 (3 A+5 (B+C)) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^4 (9 A+5 B-5 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 a^4 (6 A-5 B-20 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \left (5 a^4 (3 A+5 (B+C)) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^4 (9 A+5 B-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a^4 (6 A-5 B-20 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {2 (3 A-15 B-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6}{5} \left (\frac {2 a^4 (6 A-5 B-20 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {1}{3} \left (\frac {10 a^4 (3 A+5 (B+C)) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^4 (9 A+5 B-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {6 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
(2*C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((6*( B + 2*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (2*(3*A - 15*B - 35*C)*Sqrt[Cos[c + d*x]]*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(5*d) + (6*(((6*a^4*(9*A + 5*B - 5*C)*EllipticE[(c + d*x)/2, 2])/d + (10*a^4*(3*A + 5*(B + C))*EllipticF[(c + d*x)/2, 2])/d)/3 + (2*a^4*(6*A - 5*B - 20*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/5)/(3*a)
3.13.5.3.1 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[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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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_.) + (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/(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 - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(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, B, 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[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x _)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) *(x_)]^2), x_Symbol] :> Simp[d^(m + 2) Int[(b + a*Cos[e + f*x])^m*(d*Cos[ e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr eeQ[{a, b, d, e, f, A, B, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(949\) vs. \(2(261)=522\).
Time = 142.26 (sec) , antiderivative size = 950, normalized size of antiderivative = 4.20
int(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, method=_RETURNVERBOSE)
-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/(4*sin (1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)^3*(24*A*sin (1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)-96*A*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x +1/2*c)-20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+78*A*sin(1/2*d*x+1/2* c)^4*cos(1/2*d*x+1/2*c)-30*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x +1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^ 2+54*A*(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)^2+50*B*cos(1/2*d*x+1/2 *c)*sin(1/2*d*x+1/2*c)^4-50*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1 /2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c) ^2+30*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elli pticE(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)^4-50*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c )^2-30*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)^2-18*A*sin(1/2*d*x+1 /2*c)^2*cos(1/2*d*x+1/2*c)+15*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))-27*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))-20*B*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+25*B*(sin(1...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.18 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 5 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 5 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B - 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B - 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, A a^{3} \cos \left (d x + c\right )^{3} + 5 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 5 \, C a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \]
integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="fricas")
-2/15*(5*I*sqrt(2)*(3*A + 5*B + 5*C)*a^3*cos(d*x + c)^2*weierstrassPInvers e(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(3*A + 5*B + 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*B - 5*C)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)* (9*A + 5*B - 5*C)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPIn verse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (3*A*a^3*cos(d*x + c)^3 + 5 *(3*A + B)*a^3*cos(d*x + c)^2 + 15*(B + 3*C)*a^3*cos(d*x + c) + 5*C*a^3)*s qrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="maxima")
\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3*c os(d*x + c)^(5/2), x)
Time = 20.04 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.58 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (C\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+3\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {B\,a^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {6\,A\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,B\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,B\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^3\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{d}-\frac {2\,A\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(2*(C*a^3*ellipticE(c/2 + (d*x)/2, 2) + 3*C*a^3*ellipticF(c/2 + (d*x)/2, 2 )))/d + (B*a^3*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (6*A*a^3*ellipticE(c/2 + (d*x)/2, 2))/d + (4*A*a^3*e llipticF(c/2 + (d*x)/2, 2))/d + (6*B*a^3*ellipticE(c/2 + (d*x)/2, 2))/d + (6*B*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*a^3*cos(c + d*x)^(1/2)*sin( c + d*x))/d - (2*A*a^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4 ], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) + (2*B*a^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(s in(c + d*x)^2)^(1/2)) + (6*C*a^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*C*a^3* sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x )^(3/2)*(sin(c + d*x)^2)^(1/2))