Optimal. Leaf size=141 \[ \frac{2 a (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a (5 A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a C \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}} \]
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Rubi [A] time = 0.232775, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4221, 3034, 3023, 2748, 2641, 2639} \[ \frac{2 a (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a (5 A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a C \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 4221
Rule 3034
Rule 3023
Rule 2748
Rule 2641
Rule 2639
Rubi steps
\begin{align*} \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a C \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{5 a A}{2}+\frac{1}{2} a (5 A+3 C) \cos (c+d x)+\frac{5}{2} a C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a C \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{15} \left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{5}{4} a (3 A+C)+\frac{3}{4} a (5 A+3 C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a C \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{3} \left (a (3 A+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 (a (5 A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 a (5 A+3 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{2 a (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 C \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}\\ \end{align*}
Mathematica [C] time = 1.47329, size = 169, normalized size = 1.2 \[ \frac{a e^{-i d x} \sqrt{\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (-2 i (5 A+3 C) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )+\cos (c+d x) (6 i (5 A+3 C)+10 C \sin (c+d x)+3 C \sin (2 (c+d x)))+10 (3 A+C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{15 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.865, size = 345, normalized size = 2.5 \begin{align*} -{\frac{2\,a}{15\,d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( -24\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +44\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +15\,A\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -15\,A\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +5\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -9\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -16\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \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} + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )} \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 ({\left (C a \cos \left (d x + c\right )^{3} + C a \cos \left (d x + c\right )^{2} + A a \cos \left (d x + c\right ) + A a\right )} \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{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )} \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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