Integrand size = 35, antiderivative size = 196 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {16 a^2 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^2 (17 A+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \]
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Time = 0.56 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4306, 3123, 3054, 3047, 3100, 2827, 2720, 2719} \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 a^2 (17 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {8 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{15 d}-\frac {16 a^2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]
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Rule 2719
Rule 2720
Rule 2827
Rule 3047
Rule 3054
Rule 3100
Rule 3123
Rule 4306
Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^2 \left (2 a A-\frac {1}{2} a (A-5 C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{5 a} \\ & = \frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{4} a^2 (17 A+15 C)-\frac {1}{4} a^2 (7 A-15 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a} \\ & = \frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} a^3 (17 A+15 C)+\left (-\frac {1}{4} a^3 (7 A-15 C)+\frac {1}{4} a^3 (17 A+15 C)\right ) \cos (c+d x)-\frac {1}{4} a^3 (7 A-15 C) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a} \\ & = \frac {2 a^2 (17 A+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{4} a^3 (A+3 C)-3 a^3 A \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a} \\ & = \frac {2 a^2 (17 A+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \left (8 a^2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (2 a^2 (A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {16 a^2 A \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 (A+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 {2 a^2 (17 A+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.42 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.49 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {4 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (12 A \left (1+e^{2 i (c+d x)}\right )+12 A \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+5 (A+3 C) e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\sqrt {\sec (c+d x)} (3 (16 A+5 C-5 C \cos (2 c)) \cos (d x) \csc (c)+30 C \cos (c) \sin (d x)+2 A (10+3 \sec (c+d x)) \tan (c+d x))\right )}{60 d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(755\) vs. \(2(224)=448\).
Time = 150.16 (sec) , antiderivative size = 756, normalized size of antiderivative = 3.86
method | result | size |
default | \(\text {Expression too large to display}\) | \(756\) |
parts | \(\text {Expression too large to display}\) | \(1043\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.14 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 12 i \, \sqrt {2} A a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 i \, \sqrt {2} A a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, {\left (8 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 10 \, A a^{2} \cos \left (d x + c\right ) + 3 \, A a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \]
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Timed out. \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]
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\[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \]
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\[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \]
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Timed out. \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2 \,d x \]
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