Optimal. Leaf size=261 \[ -\frac{(13 A-3 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{6 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(49 A-9 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{10 a^3 d}-\frac{(13 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{(49 A-9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{(8 A-3 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.612115, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2960, 4019, 3787, 3771, 2641, 3768, 2639} \[ -\frac{(13 A-3 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{6 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(49 A-9 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{10 a^3 d}-\frac{(13 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{(49 A-9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{(8 A-3 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 2960
Rule 4019
Rule 3787
Rule 3771
Rule 2641
Rule 3768
Rule 2639
Rubi steps
\begin{align*} \int \frac{(A+B \cos (c+d x)) \sec ^{\frac{3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=\int \frac{\sec ^{\frac{7}{2}}(c+d x) (B+A \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\\ &=-\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^{\frac{5}{2}}(c+d x) \left (-\frac{5}{2} a (A-B)+\frac{1}{2} a (11 A-B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (-\frac{3}{2} a^2 (8 A-3 B)+\frac{1}{2} a^2 (41 A-6 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \sqrt{\sec (c+d x)} \left (-\frac{5}{4} a^3 (13 A-3 B)+\frac{3}{4} a^3 (49 A-9 B) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(49 A-9 B) \int \sec ^{\frac{3}{2}}(c+d x) \, dx}{20 a^3}-\frac{(13 A-3 B) \int \sqrt{\sec (c+d x)} \, dx}{12 a^3}\\ &=\frac{(49 A-9 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(49 A-9 B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}-\frac{\left ((13 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}\\ &=-\frac{(13 A-3 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{6 a^3 d}+\frac{(49 A-9 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\left ((49 A-9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=-\frac{(49 A-9 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{10 a^3 d}-\frac{(13 A-3 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{6 a^3 d}+\frac{(49 A-9 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}
Mathematica [C] time = 5.39708, size = 358, normalized size = 1.37 \[ -\frac{e^{-i d x} \cos \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \left (\cos \left (\frac{1}{2} (c+3 d x)\right )+i \sin \left (\frac{1}{2} (c+3 d x)\right )\right ) \left (-i (49 A-9 B) e^{-2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^5 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )+2 i ((1082 A-207 B) \cos (c+d x)+6 (87 A-17 B) \cos (2 (c+d x))+161 i A \sin (c+d x)+148 i A \sin (2 (c+d x))+41 i A \sin (3 (c+d x))+106 A \cos (3 (c+d x))+642 A-6 i B \sin (c+d x)-18 i B \sin (2 (c+d x))-6 i B \sin (3 (c+d x))-21 B \cos (3 (c+d x))-102 B)+160 (13 A-3 B) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-i \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 4.529, size = 685, normalized size = 2.6 \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 (\frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}}, 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 )} \sec \left (d x + c\right )^{\frac{3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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