Integrand size = 35, antiderivative size = 270 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {16 a^2 (3 A+2 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {16 a^2 (3 A+2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {4 a^2 (7 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a^2 (21 A+19 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 C \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {8 C \sec ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d} \]
4/21*a^2*(7*A+5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/105*a^2*(21*A+19*C)*sec (d*x+c)^(5/2)*sin(d*x+c)/d+2/9*C*sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^2*sin(d *x+c)/d+8/63*C*sec(d*x+c)^(5/2)*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d+16/15*a^ 2*(3*A+2*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-16/15*a^2*(3*A+2*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*(7*A+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))*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 = 8.40 (sec) , antiderivative size = 821, normalized size of antiderivative = 3.04 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4(c+d x) \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{15 d (A+2 C+A \cos (2 c+2 d x))}+\frac {8 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4(c+d x) \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{45 d (A+2 C+A \cos (2 c+2 d x))}+\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{3 d (A+2 C+A \cos (2 c+2 d x)) \sec ^{\frac {7}{2}}(c+d x)}+\frac {10 C \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{21 d (A+2 C+A \cos (2 c+2 d x)) \sec ^{\frac {7}{2}}(c+d x)}+\frac {\sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (\frac {8 (3 A+2 C) \cos (d x) \csc (c)}{15 d}+\frac {C \sec (c) \sec ^4(c+d x) \sin (d x)}{9 d}+\frac {\sec (c) \sec ^3(c+d x) (7 C \sin (c)+18 C \sin (d x))}{63 d}+\frac {\sec (c) \sec ^2(c+d x) (90 C \sin (c)+63 A \sin (d x)+112 C \sin (d x))}{315 d}+\frac {\sec (c) \sec (c+d x) (63 A \sin (c)+112 C \sin (c)+210 A \sin (d x)+150 C \sin (d x))}{315 d}+\frac {2 (7 A+5 C) \tan (c)}{21 d}\right )}{(A+2 C+A \cos (2 c+2 d x)) \sec ^{\frac {7}{2}}(c+d x)} \]
(4*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^(( 2*I)*(c + d*x))]*Cos[c + d*x]^4*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2* I)*(c + d*x))])*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/(15*d*E^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])) + (8*Sqrt[2]*C*S qrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x) )]*Cos[c + d*x]^4*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x) *(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] )*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/(45* d*E^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])) + (2*A*Sqrt[Cos[c + d*x]]*Elli pticF[(c + d*x)/2, 2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*S ec[c + d*x]^2))/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2)) + (10*C*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/(21*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2)) + (Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2 *(A + C*Sec[c + d*x]^2)*((8*(3*A + 2*C)*Cos[d*x]*Csc[c])/(15*d) + (C*Sec[c ]*Sec[c + d*x]^4*Sin[d*x])/(9*d) + (Sec[c]*Sec[c + d*x]^3*(7*C*Sin[c] + 18 *C*Sin[d*x]))/(63*d) + (Sec[c]*Sec[c + d*x]^2*(90*C*Sin[c] + 63*A*Sin[d*x] + 112*C*Sin[d*x]))/(315*d) + (Sec[c]*Sec[c + d*x]*(63*A*Sin[c] + 112*C*Si n[c] + 210*A*Sin[d*x] + 150*C*Sin[d*x]))/(315*d) + (2*(7*A + 5*C)*Tan[c...
Time = 1.39 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4577, 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.
\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2 \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4577 |
\(\displaystyle \frac {2 \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^2 (3 a (3 A+C)+4 a C \sec (c+d x))dx}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^2 (3 a (3 A+C)+4 a C \sec (c+d x))dx}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (3 a (3 A+C)+4 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 4506 |
\(\displaystyle \frac {\frac {2}{7} \int \frac {3}{2} \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a) \left ((21 A+11 C) a^2+(21 A+19 C) \sec (c+d x) a^2\right )dx+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3}{7} \int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a) \left ((21 A+11 C) a^2+(21 A+19 C) \sec (c+d x) a^2\right )dx+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((21 A+11 C) a^2+(21 A+19 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 4485 |
\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \int \sec ^{\frac {3}{2}}(c+d x) \left (28 (3 A+2 C) a^3+15 (7 A+5 C) \sec (c+d x) a^3\right )dx+\frac {2 a^3 (21 A+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (28 (3 A+2 C) a^3+15 (7 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {2 a^3 (21 A+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (28 a^3 (3 A+2 C) \int \sec ^{\frac {3}{2}}(c+d x)dx+15 a^3 (7 A+5 C) \int \sec ^{\frac {5}{2}}(c+d x)dx\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (28 a^3 (3 A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+15 a^3 (7 A+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+5 C) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+28 a^3 (3 A+2 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+5 C) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+28 a^3 (3 A+2 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+5 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+28 a^3 (3 A+2 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+5 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+28 a^3 (3 A+2 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+5 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+28 a^3 (3 A+2 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {3}{7} \left (\frac {2 a^3 (21 A+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2}{5} \left (15 a^3 (7 A+5 C) \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )+28 a^3 (3 A+2 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {8 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\) |
(2*C*Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(9*d) + ((8*C *Sec[c + d*x]^(5/2)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(7*d) + (3*((2* a^3*(21*A + 19*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + (2*(28*a^3*(3*A + 2*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]]*Sin[c + d*x])/d) + 15*a^3*(7*A + 5*C)*((2*Sq rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2* Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d))))/5))/7)/(9*a)
3.3.14.3.1 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_)]^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*Csc[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n *Simp[A*b*(m + n + 1) + b*C*n + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 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. \(1140\) vs. \(2(290)=580\).
Time = 4.93 (sec) , antiderivative size = 1141, normalized size of antiderivative = 4.23
method | result | size |
default | \(\text {Expression too large to display}\) | \(1141\) |
parts | \(\text {Expression too large to display}\) | \(1403\) |
-a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*A*(-1/6* 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)/(c os(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(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)*E llipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+4*C*(-1/56*cos(1/2*d*x+1/2*c)*(-2*si n(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*co s(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*A/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)*sin(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))+2 *C*(-1/144*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)^5-7/180*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)^3-14/ 15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*s in(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(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))-7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.03 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (15 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 84 i \, \sqrt {2} {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 84 i \, \sqrt {2} {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (168 \, {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (7 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 7 \, {\left (9 \, A + 16 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 90 \, C a^{2} \cos \left (d x + c\right ) + 35 \, C a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{315 \, d \cos \left (d x + c\right )^{4}} \]
-2/315*(15*I*sqrt(2)*(7*A + 5*C)*a^2*cos(d*x + c)^4*weierstrassPInverse(-4 , 0, cos(d*x + c) + I*sin(d*x + c)) - 15*I*sqrt(2)*(7*A + 5*C)*a^2*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 84*I*s qrt(2)*(3*A + 2*C)*a^2*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPI nverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 84*I*sqrt(2)*(3*A + 2*C)*a ^2*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* x + c) - I*sin(d*x + c))) - (168*(3*A + 2*C)*a^2*cos(d*x + c)^4 + 30*(7*A + 5*C)*a^2*cos(d*x + c)^3 + 7*(9*A + 16*C)*a^2*cos(d*x + c)^2 + 90*C*a^2*c os(d*x + c) + 35*C*a^2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^4 )
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]