Optimal. Leaf size=127 \[ -\frac{a 2^{m+\frac{3}{2}} (A (m+3)+B m) \cos ^3(e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (\frac{3}{2},-m-\frac{1}{2};\frac{5}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 f (m+3)}-\frac{B \cos ^3(e+f x) (a \sin (e+f x)+a)^m}{f (m+3)} \]
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Rubi [A] time = 0.185491, antiderivative size = 127, 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 = {2860, 2689, 70, 69} \[ -\frac{a 2^{m+\frac{3}{2}} (A (m+3)+B m) \cos ^3(e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (\frac{3}{2},-m-\frac{1}{2};\frac{5}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 f (m+3)}-\frac{B \cos ^3(e+f x) (a \sin (e+f x)+a)^m}{f (m+3)} \]
Antiderivative was successfully verified.
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Rule 2860
Rule 2689
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx &=-\frac{B \cos ^3(e+f x) (a+a \sin (e+f x))^m}{f (3+m)}+\left (A+\frac{B m}{3+m}\right ) \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx\\ &=-\frac{B \cos ^3(e+f x) (a+a \sin (e+f x))^m}{f (3+m)}+\frac{\left (a^2 \left (A+\frac{B m}{3+m}\right ) \cos ^3(e+f x)\right ) \operatorname{Subst}\left (\int \sqrt{a-a x} (a+a x)^{\frac{1}{2}+m} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}}\\ &=-\frac{B \cos ^3(e+f x) (a+a \sin (e+f x))^m}{f (3+m)}+\frac{\left (2^{\frac{1}{2}+m} a^2 \left (A+\frac{B m}{3+m}\right ) \cos ^3(e+f x) (a+a \sin (e+f x))^{-1+m} \left (\frac{a+a \sin (e+f x)}{a}\right )^{-\frac{1}{2}-m}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{x}{2}\right )^{\frac{1}{2}+m} \sqrt{a-a x} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{3/2}}\\ &=-\frac{2^{\frac{3}{2}+m} a \left (A+\frac{B m}{3+m}\right ) \cos ^3(e+f x) \, _2F_1\left (\frac{3}{2},-\frac{1}{2}-m;\frac{5}{2};\frac{1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac{1}{2}-m} (a+a \sin (e+f x))^{-1+m}}{3 f}-\frac{B \cos ^3(e+f x) (a+a \sin (e+f x))^m}{f (3+m)}\\ \end{align*}
Mathematica [A] time = 0.318708, size = 111, normalized size = 0.87 \[ -\frac{\cos ^3(e+f x) (\sin (e+f x)+1)^{-m-\frac{3}{2}} (a (\sin (e+f x)+1))^m \left (2^{m+\frac{3}{2}} (A (m+3)+B m) \, _2F_1\left (\frac{3}{2},-m-\frac{1}{2};\frac{5}{2};\frac{1}{2} (1-\sin (e+f x))\right )+3 B (\sin (e+f x)+1)^{m+\frac{3}{2}}\right )}{3 f (m+3)} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.891, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \right ) \, 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{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\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 ({\left (B \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) + A \cos \left (f x + e\right )^{2}\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}, 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 (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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