Optimal. Leaf size=212 \[ -\frac{2 a^2 (15 A-5 B-7 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{4 a^2 (3 A+2 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 (5 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d \sqrt{\sec (c+d x)}}+\frac{4 a^2 (5 B+4 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 A \sin (c+d x) \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^2}{d} \]
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Rubi [A] time = 0.58893, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4221, 3043, 2976, 2968, 3023, 2748, 2641, 2639} \[ -\frac{2 a^2 (15 A-5 B-7 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{4 a^2 (3 A+2 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 (5 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d \sqrt{\sec (c+d x)}}+\frac{4 a^2 (5 B+4 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 A \sin (c+d x) \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^2}{d} \]
Antiderivative was successfully verified.
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Rule 4221
Rule 3043
Rule 2976
Rule 2968
Rule 3023
Rule 2748
Rule 2641
Rule 2639
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 A (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^2 \left (\frac{1}{2} a (4 A+B)-\frac{1}{2} a (5 A-C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{a}\\ &=-\frac{2 (5 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x)) \left (\frac{1}{4} a^2 (15 A+5 B+C)-\frac{1}{4} a^2 (15 A-5 B-7 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{5 a}\\ &=-\frac{2 (5 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a^3 (15 A+5 B+C)+\left (-\frac{1}{4} a^3 (15 A-5 B-7 C)+\frac{1}{4} a^3 (15 A+5 B+C)\right ) \cos (c+d x)-\frac{1}{4} a^3 (15 A-5 B-7 C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{5 a}\\ &=-\frac{2 a^2 (15 A-5 B-7 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 (5 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{5}{4} a^3 (3 A+2 B+C)+\frac{3}{4} a^3 (5 B+4 C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=-\frac{2 a^2 (15 A-5 B-7 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 (5 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{3} \left (2 a^2 (3 A+2 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (2 a^2 (5 B+4 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{4 a^2 (5 B+4 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+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 d}-\frac{2 a^2 (15 A-5 B-7 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 (5 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}\\ \end{align*}
Mathematica [A] time = 0.700849, size = 121, normalized size = 0.57 \[ \frac{a^2 \sqrt{\sec (c+d x)} \left (2 \sin (c+d x) (3 (10 A+C \cos (2 (c+d x))+C)+10 (B+2 C) \cos (c+d x))+40 (3 A+2 B+C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+24 (5 B+4 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{30 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.412, size = 595, normalized size = 2.8 \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{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \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 ({\left (C a^{2} \cos \left (d x + c\right )^{4} +{\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \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{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \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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