Integrand size = 43, antiderivative size = 271 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4 a^3 (5 A-5 B-9 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^3 (5 A+20 B+21 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \] Output:
4/5*a^3*(5*A-5*B-9*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*(5*A+5*B+3*C)*cos(d*x+c)^(1/2)*InverseJacobi AM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+4/15*a^3*(5*A+20*B+21*C)*sec( d*x+c)^(1/2)*sin(d*x+c)/d+2/3*A*(a+a*sec(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c) ^(1/2)-2/15*(5*A-3*C)*sec(d*x+c)^(1/2)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a /d-2/15*(5*A-5*B-9*C)*sec(d*x+c)^(1/2)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.38 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.17 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^3 e^{-i d x} \sec ^{\frac {5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (180 i A \cos (c+d x)-180 i B \cos (c+d x)-324 i C \cos (c+d x)+60 i A \cos (3 (c+d x))-60 i B \cos (3 (c+d x))-108 i C \cos (3 (c+d x))+80 (5 A+5 B+3 C) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-4 i (5 A-5 B-9 C) e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+30 A \sin (c+d x)+90 B \sin (c+d x)+132 C \sin (c+d x)+10 A \sin (2 (c+d x))+20 B \sin (2 (c+d x))+60 C \sin (2 (c+d x))+30 A \sin (3 (c+d x))+90 B \sin (3 (c+d x))+108 C \sin (3 (c+d x))+5 A \sin (4 (c+d x))\right )}{60 d} \] Input:
Integrate[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)) /Sec[c + d*x]^(3/2),x]
Output:
(a^3*Sec[c + d*x]^(5/2)*(Cos[d*x] + I*Sin[d*x])*((180*I)*A*Cos[c + d*x] - (180*I)*B*Cos[c + d*x] - (324*I)*C*Cos[c + d*x] + (60*I)*A*Cos[3*(c + d*x) ] - (60*I)*B*Cos[3*(c + d*x)] - (108*I)*C*Cos[3*(c + d*x)] + 80*(5*A + 5*B + 3*C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] - ((4*I)*(5*A - 5*B - 9*C)*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^ ((2*I)*(c + d*x))])/E^(I*(c + d*x)) + 30*A*Sin[c + d*x] + 90*B*Sin[c + d*x ] + 132*C*Sin[c + d*x] + 10*A*Sin[2*(c + d*x)] + 20*B*Sin[2*(c + d*x)] + 6 0*C*Sin[2*(c + d*x)] + 30*A*Sin[3*(c + d*x)] + 90*B*Sin[3*(c + d*x)] + 108 *C*Sin[3*(c + d*x)] + 5*A*Sin[4*(c + d*x)]))/(60*d*E^(I*d*x))
Time = 1.73 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.02, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.442, Rules used = {3042, 4574, 27, 3042, 4506, 27, 3042, 4506, 27, 3042, 4485, 27, 3042, 4274, 3042, 4258, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^3 (3 a (2 A+B)-a (5 A-3 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^3 (3 a (2 A+B)-a (5 A-3 C) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (3 a (2 A+B)-a (5 A-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {2}{5} \int \frac {(\sec (c+d x) a+a)^2 \left (a^2 (35 A+15 B-3 C)-3 a^2 (5 A-5 B-9 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {(\sec (c+d x) a+a)^2 \left (a^2 (35 A+15 B-3 C)-3 a^2 (5 A-5 B-9 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}}dx-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a^2 (35 A+15 B-3 C)-3 a^2 (5 A-5 B-9 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {3 (\sec (c+d x) a+a) \left ((20 A+5 B-6 C) a^3+(5 A+20 B+21 C) \sec (c+d x) a^3\right )}{\sqrt {\sec (c+d x)}}dx-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \left (2 \int \frac {(\sec (c+d x) a+a) \left ((20 A+5 B-6 C) a^3+(5 A+20 B+21 C) \sec (c+d x) a^3\right )}{\sqrt {\sec (c+d x)}}dx-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (2 \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((20 A+5 B-6 C) a^3+(5 A+20 B+21 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {\frac {1}{5} \left (2 \left (2 \int \frac {3 (5 A-5 B-9 C) a^4+5 (5 A+5 B+3 C) \sec (c+d x) a^4}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 a^4 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \left (2 \left (\int \frac {3 (5 A-5 B-9 C) a^4+5 (5 A+5 B+3 C) \sec (c+d x) a^4}{\sqrt {\sec (c+d x)}}dx+\frac {2 a^4 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (2 \left (\int \frac {3 (5 A-5 B-9 C) a^4+5 (5 A+5 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^4 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {1}{5} \left (2 \left (3 a^4 (5 A-5 B-9 C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+5 a^4 (5 A+5 B+3 C) \int \sqrt {\sec (c+d x)}dx+\frac {2 a^4 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (2 \left (3 a^4 (5 A-5 B-9 C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 a^4 (5 A+5 B+3 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^4 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {1}{5} \left (2 \left (5 a^4 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^4 (5 A-5 B-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {2 a^4 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (2 \left (5 a^4 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^4 (5 A-5 B-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^4 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {1}{5} \left (2 \left (5 a^4 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^4 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {6 a^4 (5 A-5 B-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {1}{5} \left (2 \left (\frac {2 a^4 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 a^4 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^4 (5 A-5 B-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}}\) |
Input:
Int[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(3/2),x]
Output:
(2*A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ((-2* (5*A - 3*C)*Sqrt[Sec[c + d*x]]*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(5 *d) + ((-2*(5*A - 5*B - 9*C)*Sqrt[Sec[c + d*x]]*(a^4 + a^4*Sec[c + d*x])*S in[c + d*x])/d + 2*((6*a^4*(5*A - 5*B - 9*C)*Sqrt[Cos[c + d*x]]*EllipticE[ (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (10*a^4*(5*A + 5*B + 3*C)*Sqrt[Cos [c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*a^4*(5*A + 20*B + 21*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))/5)/(3*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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[ e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(1261\) vs. \(2(246)=492\).
Time = 11.46 (sec) , antiderivative size = 1262, normalized size of antiderivative = 4.66
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1262\) |
| default | \(\text {Expression too large to display}\) | \(1328\) |
Input:
int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, method=_RETURNVERBOSE)
Output:
2*(3*A*a^3+B*a^3)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)* (sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(c os(1/2*d*x+1/2*c),2^(1/2))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^ (1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d-2/3*(B*a^3+3*C *a^3)*(-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El lipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*((2*cos(1/2*d*x+1/2*c) ^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2 *c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d-2*(A*a^ 3+3*B*a^3+3*C*a^3)*(-2*(-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)*sin(1/2*d*x+1/2*c)^2+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2* sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+s in(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^( 1/2)/d-2*(3*A*a^3+3*B*a^3+C*a^3)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1 /2*c)^2)^(1/2)*(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))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d -2/3*a^3*A*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*c...
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.00 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (5 \, A + 5 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (5 \, A + 5 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A - 5 \, B - 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A - 5 \, B - 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (5 \, A a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 15 \, B + 18 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 5 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 3 \, C a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3 /2),x, algorithm="fricas")
Output:
-2/15*(5*I*sqrt(2)*(5*A + 5*B + 3*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)*(5*A + 5*B + 3*C)*a^ 3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(5*A - 5*B - 9*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)* (5*A - 5*B - 9*C)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPIn verse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (5*A*a^3*cos(d*x + c)^3 + 3 *(5*A + 15*B + 18*C)*a^3*cos(d*x + c)^2 + 5*(B + 3*C)*a^3*cos(d*x + c) + 3 *C*a^3)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)* *(3/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3 /2),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3/s ec(d*x + c)^(3/2), x)
\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3 /2),x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3/s ec(d*x + c)^(3/2), x)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int(((a + a/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/co s(c + d*x))^(3/2),x)
Output:
int(((a + a/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/co s(c + d*x))^(3/2), x)
\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=a^{3} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x \right ) a +3 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) b +3 \left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a +3 \left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) b +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) b +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) c \right ) \] Input:
int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x)
Output:
a**3*(int(sqrt(sec(c + d*x))/sec(c + d*x)**2,x)*a + 3*int(sqrt(sec(c + d*x ))/sec(c + d*x),x)*a + int(sqrt(sec(c + d*x))/sec(c + d*x),x)*b + 3*int(sq rt(sec(c + d*x)),x)*a + 3*int(sqrt(sec(c + d*x)),x)*b + int(sqrt(sec(c + d *x)),x)*c + int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x)*c + int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*b + 3*int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)* c + int(sqrt(sec(c + d*x))*sec(c + d*x),x)*a + 3*int(sqrt(sec(c + d*x))*se c(c + d*x),x)*b + 3*int(sqrt(sec(c + d*x))*sec(c + d*x),x)*c)