Optimal. Leaf size=196 \[ \frac{4 a^2 (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 a^2 (15 A+17 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{8 C \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{15 d}-\frac{16 a^2 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 C \sin (c+d x) \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d} \]
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Rubi [A] time = 0.389384, antiderivative size = 196, 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 = {4089, 4018, 3997, 3787, 3771, 2639, 2641} \[ \frac{2 a^2 (15 A+17 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{4 a^2 (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{8 C \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{15 d}-\frac{16 a^2 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 C \sin (c+d x) \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 4089
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 )}{\sqrt{\sec (c+d x)}} \, dx &=\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{2 \int \frac{(a+a \sec (c+d x))^2 \left (\frac{1}{2} a (5 A-C)+2 a C \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{5 a}\\ &=\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{8 C \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{4 \int \frac{(a+a \sec (c+d x)) \left (\frac{1}{4} a^2 (15 A-7 C)+\frac{1}{4} a^2 (15 A+17 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{15 a}\\ &=\frac{2 a^2 (15 A+17 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{8 C \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{8 \int \frac{-3 a^3 C+\frac{5}{4} a^3 (3 A+C) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a}\\ &=\frac{2 a^2 (15 A+17 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{8 C \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac{1}{5} \left (8 a^2 C\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (2 a^2 (3 A+C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 a^2 (15 A+17 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{8 C \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac{1}{5} \left (8 a^2 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (2 a^2 (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{16 a^2 C \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{4 a^2 (3 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 (15 A+17 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{8 C \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}\\ \end{align*}
Mathematica [C] time = 5.88315, size = 312, normalized size = 1.59 \[ \frac{a^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 \left (A+C \sec ^2(c+d x)\right ) \left (\frac{-3 \csc (c) \cos (d x) (5 A \cos (2 c)-5 A-16 C)+30 A \cos (c) \sin (d x)+2 C \tan (c+d x) (3 \sec (c+d x)+10)}{2 d \sec ^{\frac{7}{2}}(c+d x)}-\frac{2 i \sqrt{2} \cos ^4(c+d x) \left (5 \left (-1+e^{2 i c}\right ) (3 A+C) e^{i (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+12 \left (-1+e^{2 i c}\right ) C \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+12 C \sqrt{1+e^{2 i (c+d x)}}\right )}{\left (-1+e^{2 i c}\right ) d \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}}}\right )}{15 (A \cos (2 (c+d x))+A+2 C)} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 6.369, size = 756, normalized size = 3.9 \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}}{\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{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}}{\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 (C \sec \left (d x + c\right )^{2} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\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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