Optimal. Leaf size=121 \[ \frac{2 \left (3 a^2 B+6 a A b+b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (a^2 A-2 a b B-A b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^2 A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b^2 B \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]
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Rubi [A] time = 0.245916, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {2988, 3023, 2748, 2641, 2639} \[ \frac{2 \left (3 a^2 B+6 a A b+b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (a^2 A-2 a b B-A b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^2 A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b^2 B \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]
Antiderivative was successfully verified.
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Rule 2988
Rule 3023
Rule 2748
Rule 2641
Rule 2639
Rubi steps
\begin{align*} \int \frac{(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 a^2 A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-2 \int \frac{-\frac{1}{2} a (2 A b+a B)+\frac{1}{2} \left (a^2 A-A b^2-2 a b B\right ) \cos (c+d x)-\frac{1}{2} b^2 B \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a^2 A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b^2 B \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-\frac{4}{3} \int \frac{\frac{1}{4} \left (-b^2 B-3 a (2 A b+a B)\right )+\frac{3}{4} \left (a^2 A-A b^2-2 a b B\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a^2 A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b^2 B \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-\left (a^2 A-A b^2-2 a b B\right ) \int \sqrt{\cos (c+d x)} \, dx-\frac{1}{3} \left (-6 a A b-3 a^2 B-b^2 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 \left (a^2 A-A b^2-2 a b B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 \left (6 a A b+3 a^2 B+b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^2 A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b^2 B \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 0.605016, size = 102, normalized size = 0.84 \[ \frac{2 \left (\left (3 a^2 B+6 a A b+b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\left (-3 a^2 A+6 a b B+3 A b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{\sin (c+d x) \left (3 a^2 A+b^2 B \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}}\right )}{3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 3.398, size = 404, normalized size = 3.3 \begin{align*} -{\frac{2}{3\,d} \left ( 4\,{b}^{2}B\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+6\,Aab\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 ) +3\,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 ){a}^{2}-3\,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 ){b}^{2}-6\,A{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+3\,B{a}^{2}\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 ) +{b}^{2}B\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) -6\,B\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 ) ab-2\,B{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) \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 \frac{{\left (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \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{B b^{2} \cos \left (d x + c\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )}{\cos \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 (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \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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