Optimal. Leaf size=100 \[ \frac{12 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d^3}-\frac{24 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{5 b d^2 \sqrt{\cos (a+b x)}}+\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}} \]
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Rubi [A] time = 0.104586, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2566, 2568, 2640, 2639} \[ \frac{12 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d^3}-\frac{24 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{5 b d^2 \sqrt{\cos (a+b x)}}+\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}} \]
Antiderivative was successfully verified.
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Rule 2566
Rule 2568
Rule 2640
Rule 2639
Rubi steps
\begin{align*} \int \frac{\sin ^4(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx &=\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}}-\frac{6 \int \sqrt{d \cos (a+b x)} \sin ^2(a+b x) \, dx}{d^2}\\ &=\frac{12 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b d^3}+\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}}-\frac{12 \int \sqrt{d \cos (a+b x)} \, dx}{5 d^2}\\ &=\frac{12 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b d^3}+\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}}-\frac{\left (12 \sqrt{d \cos (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \, dx}{5 d^2 \sqrt{\cos (a+b x)}}\\ &=-\frac{24 \sqrt{d \cos (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{5 b d^2 \sqrt{\cos (a+b x)}}+\frac{12 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b d^3}+\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}}\\ \end{align*}
Mathematica [C] time = 0.0594919, size = 60, normalized size = 0.6 \[ \frac{\sin ^5(a+b x) \sqrt [4]{\cos ^2(a+b x)} \, _2F_1\left (\frac{5}{4},\frac{5}{2};\frac{7}{2};\sin ^2(a+b x)\right )}{5 b d \sqrt{d \cos (a+b x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.068, size = 213, normalized size = 2.1 \begin{align*} -{\frac{8}{5\,db}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}d+ \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}d} \left ( -2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}\cos \left ( 1/2\,bx+a/2 \right ) +2\,\cos \left ( 1/2\,bx+a/2 \right ) \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+3\,\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) -3\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}\cos \left ( 1/2\,bx+a/2 \right ) \right ){\frac{1}{\sqrt{-d \left ( 2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}- \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2} \right ) }}} \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{d \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }}}} \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{\sin \left (b x + a\right )^{4}}{\left (d \cos \left (b x + a\right )\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{{\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sqrt{d \cos \left (b x + a\right )}}{d^{2} \cos \left (b x + a\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{\sin \left (b x + a\right )^{4}}{\left (d \cos \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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