Integrand size = 33, antiderivative size = 295 \[ \int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 b^2 (9 A b+13 a B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]
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Time = 0.68 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3039, 4110, 4159, 4132, 3854, 3856, 2720, 4130, 2719} \[ \int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 b \left (22 a^2 B+27 a A b+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 a^3 B+21 a^2 A b+15 a b^2 B+5 A b^3\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 \left (7 a^3 B+21 a^2 A b+15 a b^2 B+5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 b^2 (13 a B+9 A b) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]
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Rule 2719
Rule 2720
Rule 3039
Rule 3854
Rule 3856
Rule 4110
Rule 4130
Rule 4132
Rule 4159
Rubi steps \begin{align*} \text {integral}& = \int \frac {(b+a \sec (c+d x))^3 (B+A \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}-\frac {2}{9} \int \frac {(b+a \sec (c+d x)) \left (-\frac {1}{2} b (9 A b+13 a B)-\frac {1}{2} \left (18 a A b+9 a^2 B+7 b^2 B\right ) \sec (c+d x)-\frac {3}{2} a (3 a A+b B) \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 b^2 (9 A b+13 a B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {4}{63} \int \frac {\frac {7}{4} b \left (27 a A b+22 a^2 B+7 b^2 B\right )+\frac {9}{4} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sec (c+d x)+\frac {21}{4} a^2 (3 a A+b B) \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 b^2 (9 A b+13 a B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {4}{63} \int \frac {\frac {7}{4} b \left (27 a A b+22 a^2 B+7 b^2 B\right )+\frac {21}{4} a^2 (3 a A+b B) \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx+\frac {1}{7} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 b^2 (9 A b+13 a B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{21} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{15} \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 b^2 (9 A b+13 a B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{21} \left (\left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{15} \left (\left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {2 \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \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 {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 b^2 (9 A b+13 a B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)} \\ \end{align*}
Time = 7.45 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {\sec (c+d x)} \left (168 \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+120 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\left (7 b \left (108 a A b+108 a^2 B+43 b^2 B\right ) \cos (c+d x)+5 \left (252 a^2 A b+78 A b^3+84 a^3 B+234 a b^2 B+18 b^2 (A b+3 a B) \cos (2 (c+d x))+7 b^3 B \cos (3 (c+d x))\right )\right ) \sin (2 (c+d x))\right )}{1260 d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(744\) vs. \(2(319)=638\).
Time = 18.82 (sec) , antiderivative size = 745, normalized size of antiderivative = 2.53
method | result | size |
default | \(\text {Expression too large to display}\) | \(745\) |
parts | \(\text {Expression too large to display}\) | \(971\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=-\frac {15 \, \sqrt {2} {\left (7 i \, B a^{3} + 21 i \, A a^{2} b + 15 i \, B a b^{2} + 5 i \, A b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 \, \sqrt {2} {\left (-7 i \, B a^{3} - 21 i \, A a^{2} b - 15 i \, B a b^{2} - 5 i \, A b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (-15 i \, A a^{3} - 27 i \, B a^{2} b - 27 i \, A a b^{2} - 7 i \, B b^{3}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (15 i \, A a^{3} + 27 i \, B a^{2} b + 27 i \, A a b^{2} + 7 i \, B b^{3}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (35 \, B b^{3} \cos \left (d x + c\right )^{4} + 45 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + 7 \, {\left (27 \, B a^{2} b + 27 \, A a b^{2} + 7 \, B b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (7 \, B a^{3} + 21 \, A a^{2} b + 15 \, B a b^{2} + 5 \, A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{315 \, d} \]
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\[ \int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{3}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \]
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\[ \int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
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\[ \int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]
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