Optimal. Leaf size=145 \[ \frac{a^3 c^2 2^{m+\frac{1}{2}} (B (2-m)-A (m+3)) \cos ^5(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m-3} \, _2F_1\left (\frac{5}{2},\frac{1}{2}-m;\frac{7}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{5 f (m+3)}-\frac{a^2 B c^2 \cos ^5(e+f x) (a \sin (e+f x)+a)^{m-2}}{f (m+3)} \]
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Rubi [A] time = 0.334866, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {2967, 2860, 2689, 70, 69} \[ \frac{a^3 c^2 2^{m+\frac{1}{2}} (B (2-m)-A (m+3)) \cos ^5(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m-3} \, _2F_1\left (\frac{5}{2},\frac{1}{2}-m;\frac{7}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{5 f (m+3)}-\frac{a^2 B c^2 \cos ^5(e+f x) (a \sin (e+f x)+a)^{m-2}}{f (m+3)} \]
Antiderivative was successfully verified.
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Rule 2967
Rule 2860
Rule 2689
Rule 70
Rule 69
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx &=\left (a^2 c^2\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x))^{-2+m} (A+B \sin (e+f x)) \, dx\\ &=-\frac{a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)}+\left (a^2 c^2 \left (A-\frac{B (2-m)}{3+m}\right )\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x))^{-2+m} \, dx\\ &=-\frac{a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)}+\frac{\left (a^4 c^2 \left (A-\frac{B (2-m)}{3+m}\right ) \cos ^5(e+f x)\right ) \operatorname{Subst}\left (\int (a-a x)^{3/2} (a+a x)^{-\frac{1}{2}+m} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}\\ &=-\frac{a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)}+\frac{\left (2^{-\frac{1}{2}+m} a^4 c^2 \left (A-\frac{B (2-m)}{3+m}\right ) \cos ^5(e+f x) (a+a \sin (e+f x))^{-3+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} (a-a x)^{3/2} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2}}\\ &=-\frac{2^{\frac{1}{2}+m} a^3 c^2 \left (A-\frac{B (2-m)}{3+m}\right ) \cos ^5(e+f x) \, _2F_1\left (\frac{5}{2},\frac{1}{2}-m;\frac{7}{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))^{-3+m}}{5 f}-\frac{a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)}\\ \end{align*}
Mathematica [F] time = 180.043, size = 0, normalized size = 0. \[ \text{\$Aborted} \]
Verification is Not applicable to the result.
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Maple [F] time = 2.553, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \right ) \left ( c-c\sin \left ( fx+e \right ) \right ) ^{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{\left (B \sin \left (f x + e\right ) + A\right )}{\left (c \sin \left (f x + e\right ) - c\right )}^{2}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{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 ({\left (A - 2 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (A - B\right )} c^{2} +{\left (B c^{2} \cos \left (f x + e\right )^{2} + 2 \,{\left (A - B\right )} c^{2}\right )} \sin \left (f x + e\right )\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 (c \sin \left (f x + e\right ) - c\right )}^{2}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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