Optimal. Leaf size=198 \[ \frac{8 a^2 (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}-\frac{2 a^2 (A-5 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d}-\frac{2 (A-C) \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{3 d}+\frac{4 a^2 (A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt{\sec (c+d x)}} \]
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Rubi [A] time = 0.402225, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4087, 4018, 3997, 3787, 3771, 2639, 2641} \[ -\frac{2 a^2 (A-5 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d}-\frac{2 (A-C) \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{3 d}+\frac{8 a^2 (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a^2 (A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt{\sec (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 4087
Rule 4018
Rule 3997
Rule 3787
Rule 3771
Rule 2639
Rule 2641
Rubi steps
\begin{align*} \int \frac{(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 \int \frac{(a+a \sec (c+d x))^2 \left (2 a A-\frac{3}{2} a (A-C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{3 a}\\ &=\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{2 (A-C) \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\frac{4 \int \frac{(a+a \sec (c+d x)) \left (\frac{3}{4} a^2 (5 A-C)-\frac{3}{4} a^2 (A-5 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{9 a}\\ &=-\frac{2 a^2 (A-5 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{2 (A-C) \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\frac{8 \int \frac{\frac{9}{4} a^3 (A-C)+\frac{3}{2} a^3 (A+C) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{9 a}\\ &=-\frac{2 a^2 (A-5 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{2 (A-C) \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\left (2 a^2 (A-C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (4 a^2 (A+C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=-\frac{2 a^2 (A-5 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{2 (A-C) \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\left (2 a^2 (A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (4 a^2 (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^2 (A-C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{d}+\frac{8 a^2 (A+C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}-\frac{2 a^2 (A-5 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{2 (A-C) \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}\\ \end{align*}
Mathematica [C] time = 2.00566, size = 191, normalized size = 0.96 \[ \frac{a^2 e^{-i d x} \sec ^{\frac{3}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (-4 i (A-C) \left (1+e^{2 i (c+d x)}\right )^{3/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+16 (A+C) \cos ^{\frac{3}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+A \sin (c+d x)+A \sin (3 (c+d x))+12 i A \cos (2 (c+d x))+12 i A+4 C \sin (c+d x)+12 C \sin (2 (c+d x))-12 i C \cos (2 (c+d x))-12 i C\right )}{6 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 5.308, size = 651, normalized size = 3.3 \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 (C \sec \left (d x + c\right )^{2} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac{3}{2}}}\,{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{C a^{2} \sec \left (d x + c\right )^{4} + 2 \, C a^{2} \sec \left (d x + c\right )^{3} +{\left (A + C\right )} a^{2} \sec \left (d x + c\right )^{2} + 2 \, A a^{2} \sec \left (d x + c\right ) + A a^{2}}{\sec \left (d x + c\right )^{\frac{3}{2}}}, 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 (C \sec \left (d x + c\right )^{2} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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