Optimal. Leaf size=74 \[ \frac{2 A \sin (c+d x)}{b^2 d \sqrt{b \cos (c+d x)}}-\frac{2 (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{b^3 d \sqrt{\cos (c+d x)}} \]
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Rubi [A] time = 0.0706919, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {16, 3012, 2640, 2639} \[ \frac{2 A \sin (c+d x)}{b^2 d \sqrt{b \cos (c+d x)}}-\frac{2 (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{b^3 d \sqrt{\cos (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 16
Rule 3012
Rule 2640
Rule 2639
Rubi steps
\begin{align*} \int \frac{\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx &=\frac{\int \frac{A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx}{b}\\ &=\frac{2 A \sin (c+d x)}{b^2 d \sqrt{b \cos (c+d x)}}-\frac{(A-C) \int \sqrt{b \cos (c+d x)} \, dx}{b^3}\\ &=\frac{2 A \sin (c+d x)}{b^2 d \sqrt{b \cos (c+d x)}}-\frac{\left ((A-C) \sqrt{b \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{b^3 \sqrt{\cos (c+d x)}}\\ &=-\frac{2 (A-C) \sqrt{b \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sin (c+d x)}{b^2 d \sqrt{b \cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.115142, size = 57, normalized size = 0.77 \[ \frac{2 A \sin (c+d x)-2 (A-C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt{b \cos (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 3.707, size = 216, normalized size = 2.9 \begin{align*} -2\,{\frac{\sqrt{-2\,b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b} \left ( 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 ) -2\,A\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-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 ) \right ) }{{b}^{2}\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }d}} \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 \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac{5}{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{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt{b \cos \left (d x + c\right )}}{b^{3} \cos \left (d x + c\right )^{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} + A\right )} \cos \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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