Optimal. Leaf size=215 \[ -\frac{(5 A-2 B-C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a^2 d (\cos (c+d x)+1)}-\frac{(5 A-2 B-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(4 A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{a^2 d}-\frac{(4 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.481578, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4221, 3041, 2978, 2748, 2636, 2639, 2641} \[ -\frac{(5 A-2 B-C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a^2 d (\cos (c+d x)+1)}-\frac{(5 A-2 B-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(4 A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{a^2 d}-\frac{(4 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 4221
Rule 3041
Rule 2978
Rule 2748
Rule 2636
Rule 2639
Rule 2641
Rubi steps
\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx\\ &=-\frac{(A-B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a (7 A-B+C)-\frac{3}{2} a (A-B-C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(5 A-2 B-C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{3}{2} a^2 (4 A-B)-\frac{1}{2} a^2 (5 A-2 B-C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac{(5 A-2 B-C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\left ((4 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{2 a^2}-\frac{\left ((5 A-2 B-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}\\ &=-\frac{(5 A-2 B-C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}+\frac{(4 A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac{(5 A-2 B-C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{\left ((4 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}\\ &=-\frac{(4 A-B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}-\frac{(5 A-2 B-C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}+\frac{(4 A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac{(5 A-2 B-C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end{align*}
Mathematica [A] time = 3.19114, size = 172, normalized size = 0.8 \[ -\frac{2 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \left (2 (5 A-2 B-C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{1}{2} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (2 (19 A-4 B+C) \cos (c+d x)+3 (4 A-B) \cos (2 (c+d x))+24 A-3 B)+6 (4 A-B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 3.256, size = 563, 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 (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{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 \cos \left (d x + c\right )^{2} + 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 )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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