Integrand size = 43, antiderivative size = 255 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 a^2 (5 A+4 B+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (14 A+7 B+6 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {4 a^2 (5 A+4 B+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (35 A+49 B+33 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac {2 (7 B+4 C) \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d} \]
2/105*a^2*(35*A+49*B+33*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/7*C*sec(d*x+c)^ (3/2)*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+2/35*(7*B+4*C)*sec(d*x+c)^(3/2)*(a^2 +a^2*sec(d*x+c))*sin(d*x+c)/d+4/5*a^2*(5*A+4*B+3*C)*sin(d*x+c)*sec(d*x+c)^ (1/2)/d-4/5*a^2*(5*A+4*B+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2 *c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2 )/d+4/21*a^2*(14*A+7*B+6*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/ d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.32 (sec) , antiderivative size = 555, normalized size of antiderivative = 2.18 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 e^{-i d x} \cos ^4(c+d x) \csc (c) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (7 \sqrt {2} (5 A+4 B+3 C) e^{2 i d x} \left (-1+e^{2 i c}\right ) \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )-\frac {e^{-i (c-d x)} \left (-1+e^{2 i c}\right ) \left (35 A \left (1+e^{2 i (c+d x)}\right )^2 \left (-1+6 e^{i (c+d x)}+e^{2 i (c+d x)}+6 e^{3 i (c+d x)}\right )+14 B \left (-5+9 e^{i (c+d x)}-5 e^{2 i (c+d x)}+36 e^{3 i (c+d x)}+5 e^{4 i (c+d x)}+39 e^{5 i (c+d x)}+5 e^{6 i (c+d x)}+12 e^{7 i (c+d x)}\right )+6 C \left (-10+7 e^{i (c+d x)}-20 e^{2 i (c+d x)}+63 e^{3 i (c+d x)}+20 e^{4 i (c+d x)}+77 e^{5 i (c+d x)}+10 e^{6 i (c+d x)}+21 e^{7 i (c+d x)}\right )\right ) \sqrt {\sec (c+d x)}}{2 \left (1+e^{2 i (c+d x)}\right )^3}+10 (14 A+7 B+6 C) e^{i d x} \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)} \sin (c)\right )}{105 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \]
Integrate[Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
(a^2*Cos[c + d*x]^4*Csc[c]*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(A + B* Sec[c + d*x] + C*Sec[c + d*x]^2)*(7*Sqrt[2]*(5*A + 4*B + 3*C)*E^((2*I)*d*x )*(-1 + E^((2*I)*c))*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[ 1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d *x))] - ((-1 + E^((2*I)*c))*(35*A*(1 + E^((2*I)*(c + d*x)))^2*(-1 + 6*E^(I *(c + d*x)) + E^((2*I)*(c + d*x)) + 6*E^((3*I)*(c + d*x))) + 14*B*(-5 + 9* E^(I*(c + d*x)) - 5*E^((2*I)*(c + d*x)) + 36*E^((3*I)*(c + d*x)) + 5*E^((4 *I)*(c + d*x)) + 39*E^((5*I)*(c + d*x)) + 5*E^((6*I)*(c + d*x)) + 12*E^((7 *I)*(c + d*x))) + 6*C*(-10 + 7*E^(I*(c + d*x)) - 20*E^((2*I)*(c + d*x)) + 63*E^((3*I)*(c + d*x)) + 20*E^((4*I)*(c + d*x)) + 77*E^((5*I)*(c + d*x)) + 10*E^((6*I)*(c + d*x)) + 21*E^((7*I)*(c + d*x))))*Sqrt[Sec[c + d*x]])/(2* E^(I*(c - d*x))*(1 + E^((2*I)*(c + d*x)))^3) + 10*(14*A + 7*B + 6*C)*E^(I* d*x)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]*Sin[c ]))/(105*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))
Time = 1.41 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4576, 27, 3042, 4506, 27, 3042, 4485, 3042, 4274, 3042, 4255, 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.
\(\displaystyle \int \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4576 |
\(\displaystyle \frac {2 \int \frac {1}{2} \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^2 (a (7 A+C)+a (7 B+4 C) \sec (c+d x))dx}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^2 (a (7 A+C)+a (7 B+4 C) \sec (c+d x))dx}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (7 A+C)+a (7 B+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 4506 |
\(\displaystyle \frac {\frac {2}{5} \int \frac {1}{2} \sqrt {\sec (c+d x)} (\sec (c+d x) a+a) \left ((35 A+7 B+9 C) a^2+(35 A+49 B+33 C) \sec (c+d x) a^2\right )dx+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{5} \int \sqrt {\sec (c+d x)} (\sec (c+d x) a+a) \left ((35 A+7 B+9 C) a^2+(35 A+49 B+33 C) \sec (c+d x) a^2\right )dx+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((35 A+7 B+9 C) a^2+(35 A+49 B+33 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 4485 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \sqrt {\sec (c+d x)} \left (5 (14 A+7 B+6 C) a^3+21 (5 A+4 B+3 C) \sec (c+d x) a^3\right )dx+\frac {2 a^3 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 (14 A+7 B+6 C) a^3+21 (5 A+4 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {2 a^3 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (21 a^3 (5 A+4 B+3 C) \int \sec ^{\frac {3}{2}}(c+d x)dx+5 a^3 (14 A+7 B+6 C) \int \sqrt {\sec (c+d x)}dx\right )+\frac {2 a^3 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (14 A+7 B+6 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+21 a^3 (5 A+4 B+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )+\frac {2 a^3 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (14 A+7 B+6 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+21 a^3 (5 A+4 B+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )\right )+\frac {2 a^3 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (14 A+7 B+6 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+21 a^3 (5 A+4 B+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {2 a^3 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (14 A+7 B+6 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+21 a^3 (5 A+4 B+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 a^3 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (14 A+7 B+6 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+21 a^3 (5 A+4 B+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 a^3 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (14 A+7 B+6 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+21 a^3 (5 A+4 B+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a^3 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {2 a^3 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2}{3} \left (\frac {10 a^3 (14 A+7 B+6 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+21 a^3 (5 A+4 B+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {2 (7 B+4 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d}\) |
(2*C*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(7*d) + ((2*( 7*B + 4*C)*Sec[c + d*x]^(3/2)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(5*d) + ((2*a^3*(35*A + 49*B + 33*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + ( 2*((10*a^3*(14*A + 7*B + 6*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] *Sqrt[Sec[c + d*x]])/d + 21*a^3*(5*A + 4*B + 3*C)*((-2*Sqrt[Cos[c + d*x]]* EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Si n[c + d*x])/d)))/3)/5)/(7*a)
3.6.42.3.1 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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
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[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m , n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(906\) vs. \(2(279)=558\).
Time = 6.49 (sec) , antiderivative size = 907, normalized size of antiderivative = 3.56
method | result | size |
default | \(\text {Expression too large to display}\) | \(907\) |
parts | \(\text {Expression too large to display}\) | \(1175\) |
int((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2),x, method=_RETURNVERBOSE)
-a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1 /2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+ 2*C*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/2*c)*(-2*sin(1/2* d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+5/21 *(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)))+8*(1/2*A+1/4*B)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(- 2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*s in(1/2*d*x+1/2*c)^2-(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)^2-1)^(1/2))+8/5*(1/4*B+1/2*C)/(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)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(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)^ 2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4 +12*(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)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c)*si n(1/2*d*x+1/2*c)^2-3*(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)^2-1)^(1/2))*(-2*sin(1/2*d*x+1/2*c)^...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.10 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (14 \, A + 7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (14 \, A + 7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (5 \, A + 4 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (5 \, A + 4 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (42 \, {\left (5 \, A + 4 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (7 \, A + 14 \, B + 12 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 21 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{105 \, d \cos \left (d x + c\right )^{3}} \]
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1 /2),x, algorithm="fricas")
-2/105*(5*I*sqrt(2)*(14*A + 7*B + 6*C)*a^2*cos(d*x + c)^3*weierstrassPInve rse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(14*A + 7*B + 6*C) *a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*I*sqrt(2)*(5*A + 4*B + 3*C)*a^2*cos(d*x + c)^3*weierstrassZeta(-4 , 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I*sqr t(2)*(5*A + 4*B + 3*C)*a^2*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstra ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (42*(5*A + 4*B + 3*C)* a^2*cos(d*x + c)^3 + 5*(7*A + 14*B + 12*C)*a^2*cos(d*x + c)^2 + 21*(B + 2* C)*a^2*cos(d*x + c) + 15*C*a^2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d* x + c)^3)
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )} \,d x } \]
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1 /2),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*s qrt(sec(d*x + c)), x)
\[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )} \,d x } \]
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1 /2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*s qrt(sec(d*x + c)), x)
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]