Optimal. Leaf size=93 \[ \frac{g^3 2^{m+\frac{9}{4}} (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac{3}{4},-m-\frac{1}{4};\frac{1}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{3 a c^2 f (g \cos (e+f x))^{3/2}} \]
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Rubi [A] time = 0.280569, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2840, 2689, 70, 69} \[ \frac{g^3 2^{m+\frac{9}{4}} (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac{3}{4},-m-\frac{1}{4};\frac{1}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{3 a c^2 f (g \cos (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2840
Rule 2689
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \frac{(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx &=\frac{g^4 \int \frac{(a+a \sin (e+f x))^{2+m}}{(g \cos (e+f x))^{5/2}} \, dx}{a^2 c^2}\\ &=\frac{\left (g^3 (a-a \sin (e+f x))^{3/4} (a+a \sin (e+f x))^{3/4}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{4}+m}}{(a-a x)^{7/4}} \, dx,x,\sin (e+f x)\right )}{c^2 f (g \cos (e+f x))^{3/2}}\\ &=\frac{\left (2^{\frac{1}{4}+m} g^3 (a-a \sin (e+f x))^{3/4} (a+a \sin (e+f x))^{1+m} \left (\frac{a+a \sin (e+f x)}{a}\right )^{-\frac{1}{4}-m}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{x}{2}\right )^{\frac{1}{4}+m}}{(a-a x)^{7/4}} \, dx,x,\sin (e+f x)\right )}{c^2 f (g \cos (e+f x))^{3/2}}\\ &=\frac{2^{\frac{9}{4}+m} g^3 \, _2F_1\left (-\frac{3}{4},-\frac{1}{4}-m;\frac{1}{4};\frac{1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac{1}{4}-m} (a+a \sin (e+f x))^{1+m}}{3 a c^2 f (g \cos (e+f x))^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.202824, size = 96, normalized size = 1.03 \[ -\frac{g 2^{m+\frac{9}{4}} \sqrt{g \cos (e+f x)} (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a (\sin (e+f x)+1))^m \, _2F_1\left (-\frac{3}{4},-m-\frac{1}{4};\frac{1}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{3 c^2 f (\sin (e+f x)-1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.453, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}}{ \left ( c-c\sin \left ( fx+e \right ) \right ) ^{2}} \left ( g\cos \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}\, dx \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 (g \cos \left (f x + e\right )\right )^{\frac{3}{2}}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{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{\sqrt{g \cos \left (f x + e\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} g \cos \left (f x + e\right )}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{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 (g \cos \left (f x + e\right )\right )^{\frac{3}{2}}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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