Integrand size = 43, antiderivative size = 271 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4 a^3 (5 A-5 B-9 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^3 (5 A+20 B+21 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \]
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Time = 0.69 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4171, 4103, 4082, 3872, 3856, 2719, 2720} \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4 a^3 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {4 a^3 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a^3 (5 A-5 B-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{15 a d}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}} \]
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Rule 2719
Rule 2720
Rule 3856
Rule 3872
Rule 4082
Rule 4103
Rule 4171
Rubi steps \begin{align*} \text {integral}& = \frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \int \frac {(a+a \sec (c+d x))^3 \left (\frac {3}{2} a (2 A+B)-\frac {1}{2} a (5 A-3 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{3 a} \\ & = \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 {4 \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{4} a^2 (35 A+15 B-3 C)-\frac {3}{4} a^2 (5 A-5 B-9 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{15 a} \\ & = \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}+\frac {8 \int \frac {(a+a \sec (c+d x)) \left (\frac {3}{4} a^3 (20 A+5 B-6 C)+\frac {3}{4} a^3 (5 A+20 B+21 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{45 a} \\ & = \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}+\frac {16 \int \frac {\frac {9}{8} a^4 (5 A-5 B-9 C)+\frac {15}{8} a^4 (5 A+5 B+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{45 a} \\ & = \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}+\frac {1}{5} \left (2 a^3 (5 A-5 B-9 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (2 a^3 (5 A+5 B+3 C)\right ) \int \sqrt {\sec (c+d x)} \, dx \\ & = \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}+\frac {1}{5} \left (2 a^3 (5 A-5 B-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (2 a^3 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (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} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.47 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.17 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^3 e^{-i d x} \sec ^{\frac {5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (180 i A \cos (c+d x)-180 i B \cos (c+d x)-324 i C \cos (c+d x)+60 i A \cos (3 (c+d x))-60 i B \cos (3 (c+d x))-108 i C \cos (3 (c+d x))+80 (5 A+5 B+3 C) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-4 i (5 A-5 B-9 C) e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+30 A \sin (c+d x)+90 B \sin (c+d x)+132 C \sin (c+d x)+10 A \sin (2 (c+d x))+20 B \sin (2 (c+d x))+60 C \sin (2 (c+d x))+30 A \sin (3 (c+d x))+90 B \sin (3 (c+d x))+108 C \sin (3 (c+d x))+5 A \sin (4 (c+d x))\right )}{60 d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(1261\) vs. \(2(295)=590\).
Time = 6.21 (sec) , antiderivative size = 1262, normalized size of antiderivative = 4.66
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1262\) |
default | \(\text {Expression too large to display}\) | \(1328\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.00 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (5 \, A + 5 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (5 \, A + 5 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A - 5 \, B - 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A - 5 \, B - 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (5 \, A a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 15 \, B + 18 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 5 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 3 \, C a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \]
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Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
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