Integrand size = 41, antiderivative size = 190 \[ \int \frac {(a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 (5 a A+3 b B+3 a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (7 A b+7 a B+5 b C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (b B+a C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (7 A b+7 a B+5 b C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 a A+3 b B+3 a C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]
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Time = 0.38 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4197, 3110, 3100, 2827, 2716, 2720, 2719} \[ \int \frac {(a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (7 a B+7 A b+5 b C)}{21 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (5 a A+3 a C+3 b B)}{5 d}+\frac {2 \sin (c+d x) (7 a B+7 A b+5 b C)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x) (5 a A+3 a C+3 b B)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (a C+b B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]
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Rule 2716
Rule 2719
Rule 2720
Rule 2827
Rule 3100
Rule 3110
Rule 4197
Rubi steps \begin{align*} \text {integral}& = \int \frac {(b+a \cos (c+d x)) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {2}{7} \int \frac {-\frac {7}{2} (b B+a C)-\frac {1}{2} (7 A b+7 a B+5 b C) \cos (c+d x)-\frac {7}{2} a A \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (b B+a C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {4}{35} \int \frac {-\frac {5}{4} (7 A b+7 a B+5 b C)-\frac {7}{4} (5 a A+3 b B+3 a C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (b B+a C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {1}{5} (-5 a A-3 b B-3 a C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx-\frac {1}{7} (-7 A b-7 a B-5 b C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (b B+a C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (7 A b+7 a B+5 b C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 a A+3 b B+3 a C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {1}{5} (5 a A+3 b B+3 a C) \int \sqrt {\cos (c+d x)} \, dx-\frac {1}{21} (-7 A b-7 a B-5 b C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {2 (5 a A+3 b B+3 a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (7 A b+7 a B+5 b C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (b B+a C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (7 A b+7 a B+5 b C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 a A+3 b B+3 a C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \\ \end{align*}
Time = 7.55 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-42 (5 a A+3 b B+3 a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 (7 A b+7 a B+5 b C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {(70 A b+70 a B+110 b C+21 (15 a A+13 b B+13 a C) \cos (c+d x)+10 (7 A b+7 a B+5 b C) \cos (2 (c+d x))+105 a A \cos (3 (c+d x))+63 b B \cos (3 (c+d x))+63 a C \cos (3 (c+d x))) \sin (c+d x)}{2 \cos ^{\frac {7}{2}}(c+d x)}}{105 d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(823\) vs. \(2(222)=444\).
Time = 4.70 (sec) , antiderivative size = 824, normalized size of antiderivative = 4.34
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.52 \[ \int \frac {(a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (7 i \, B a + i \, {\left (7 \, A + 5 \, C\right )} b\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-7 i \, B a - i \, {\left (7 \, A + 5 \, C\right )} b\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (i \, {\left (5 \, A + 3 \, C\right )} a + 3 i \, B b\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (-i \, {\left (5 \, A + 3 \, C\right )} a - 3 i \, B b\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (21 \, {\left ({\left (5 \, A + 3 \, C\right )} a + 3 \, B b\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (7 \, B a + {\left (7 \, A + 5 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} + 15 \, C b + 21 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{4}} \]
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\[ \int \frac {(a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \sec {\left (c + d x \right )}\right ) \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
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Timed out. \[ \int \frac {(a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
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Time = 21.02 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {6\,C\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+10\,B\,a\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {30\,C\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+70\,A\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+42\,B\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{105\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \]
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