Optimal. Leaf size=194 \[ \frac{2 b \left (15 a^2+7 b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 a \left (7 a^2+27 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a \left (7 a^2+27 b^2\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{2 b \left (15 a^2+7 b^2\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{40 a^2 b \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d}+\frac{2 a^2 \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x))}{9 d} \]
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Rubi [A] time = 0.286489, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4264, 3841, 4047, 3769, 3771, 2639, 4045, 2641} \[ \frac{2 b \left (15 a^2+7 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a \left (7 a^2+27 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a \left (7 a^2+27 b^2\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{2 b \left (15 a^2+7 b^2\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{40 a^2 b \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d}+\frac{2 a^2 \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x))}{9 d} \]
Antiderivative was successfully verified.
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Rule 4264
Rule 3841
Rule 4047
Rule 3769
Rule 3771
Rule 2639
Rule 4045
Rule 2641
Rubi steps
\begin{align*} \int \cos ^{\frac{9}{2}}(c+d x) (a+b \sec (c+d x))^3 \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+b \sec (c+d x))^3}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{9 d}+\frac{1}{9} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{10 a^2 b+\frac{1}{2} a \left (7 a^2+27 b^2\right ) \sec (c+d x)+\frac{1}{2} b \left (5 a^2+9 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{9 d}+\frac{1}{9} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{10 a^2 b+\frac{1}{2} b \left (5 a^2+9 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{7}{2}}(c+d x)} \, dx+\frac{1}{9} \left (a \left (7 a^2+27 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a \left (7 a^2+27 b^2\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{40 a^2 b \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 a^2 \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{9 d}+\frac{1}{7} \left (b \left (15 a^2+7 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx+\frac{1}{15} \left (a \left (7 a^2+27 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 b \left (15 a^2+7 b^2\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a \left (7 a^2+27 b^2\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{40 a^2 b \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 a^2 \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{9 d}+\frac{1}{15} \left (a \left (7 a^2+27 b^2\right )\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{21} \left (b \left (15 a^2+7 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 a \left (7 a^2+27 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 b \left (15 a^2+7 b^2\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a \left (7 a^2+27 b^2\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{40 a^2 b \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 a^2 \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{9 d}+\frac{1}{21} \left (b \left (15 a^2+7 b^2\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a \left (7 a^2+27 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 b \left (15 a^2+7 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 b \left (15 a^2+7 b^2\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a \left (7 a^2+27 b^2\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{40 a^2 b \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 a^2 \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{9 d}\\ \end{align*}
Mathematica [A] time = 0.942154, size = 137, normalized size = 0.71 \[ \frac{60 \left (15 a^2 b+7 b^3\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+84 \left (7 a^3+27 a b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\sin (c+d x) \sqrt{\cos (c+d x)} \left (7 a \left (43 a^2+108 b^2\right ) \cos (c+d x)+5 \left (54 a^2 b \cos (2 (c+d x))+234 a^2 b+7 a^3 \cos (3 (c+d x))+84 b^3\right )\right )}{630 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.598, size = 470, normalized size = 2.4 \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(-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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right )^{4} \sec \left (d x + c\right ) + a^{3} \cos \left (d x + c\right )^{4}\right )} \sqrt{\cos \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{\left (b \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac{9}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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