Optimal. Leaf size=295 \[ \frac{2 b \left (22 a^2 B+27 a A b+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (21 a^2 A b+7 a^3 B+15 a b^2 B+5 A b^3\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \left (21 a^2 A b+7 a^3 B+15 a b^2 B+5 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 b^2 (13 a B+9 A b) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]
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Rubi [A] time = 0.578549, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {2960, 4025, 4074, 4047, 3769, 3771, 2641, 4045, 2639} \[ \frac{2 b \left (22 a^2 B+27 a A b+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (21 a^2 A b+7 a^3 B+15 a b^2 B+5 A b^3\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \left (21 a^2 A b+7 a^3 B+15 a b^2 B+5 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 b^2 (13 a B+9 A b) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 2960
Rule 4025
Rule 4074
Rule 4047
Rule 3769
Rule 3771
Rule 2641
Rule 4045
Rule 2639
Rubi steps
\begin{align*} \int \frac{(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt{\sec (c+d x)}} \, dx &=\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)} F\left (\left .\frac{1}{2} (c+d x)\right |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*}
Mathematica [A] time = 1.78156, size = 219, normalized size = 0.74 \[ \frac{\sqrt{\sec (c+d x)} \left (\sin (2 (c+d x)) \left (7 b \left (108 a^2 B+108 a A b+43 b^2 B\right ) \cos (c+d x)+5 \left (252 a^2 A b+84 a^3 B+18 b^2 (3 a B+A b) \cos (2 (c+d x))+234 a b^2 B+78 A b^3+7 b^3 B \cos (3 (c+d x))\right )\right )+120 \left (21 a^2 A b+7 a^3 B+15 a b^2 B+5 A b^3\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+168 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{1260 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 3.452, size = 745, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B b^{3} \cos \left (d x + c\right )^{4} + A a^{3} +{\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (B a^{2} b + A a b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )}{\sqrt{\sec \left (d x + c\right )}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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