Optimal. Leaf size=163 \[ \frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{f (2 m+1)}-\frac{B 2^{\frac{1}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac{1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac{1}{2} (2 m+1),\frac{1}{2} (2 m+1);\frac{1}{2} (2 m+3);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]
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Rubi [A] time = 0.30901, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2972, 2745, 2689, 70, 69} \[ \frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{f (2 m+1)}-\frac{B 2^{\frac{1}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac{1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac{1}{2} (2 m+1),\frac{1}{2} (2 m+1);\frac{1}{2} (2 m+3);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]
Antiderivative was successfully verified.
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Rule 2972
Rule 2745
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))^{-1-m} \, dx &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac{B \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m} \, dx}{c}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac{\left (B \cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 m}(e+f x) (c-c \sin (e+f x))^{-2 m} \, dx}{c}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac{\left (B c \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac{1}{2} (-1-2 m)+m} (c+c \sin (e+f x))^{\frac{1}{2} (-1-2 m)}\right ) \operatorname{Subst}\left (\int (c-c x)^{-2 m+\frac{1}{2} (-1+2 m)} (c+c x)^{\frac{1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac{\left (2^{-\frac{1}{2}-m} B c \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac{1}{2}+\frac{1}{2} (-1-2 m)} \left (\frac{c-c \sin (e+f x)}{c}\right )^{\frac{1}{2}+m} (c+c \sin (e+f x))^{\frac{1}{2} (-1-2 m)}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{-2 m+\frac{1}{2} (-1+2 m)} (c+c x)^{\frac{1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac{2^{\frac{1}{2}-m} B \cos (e+f x) \, _2F_1\left (\frac{1}{2} (1+2 m),\frac{1}{2} (1+2 m);\frac{1}{2} (3+2 m);\frac{1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac{1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}\\ \end{align*}
Mathematica [C] time = 11.3356, size = 675, normalized size = 4.14 \[ -\frac{2^{-m} (2 m-3) \cos ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ) \cot \left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ) \sin ^{-2 m}\left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m-1} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{-2 (-m-1)} \left (8 B (2 m+1) \tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ) F_1\left (\frac{1}{2}-m;-2 m,1;\frac{3}{2}-m;\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ),-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )-(A+B) \left ((2 m+1) \tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ) \, _2F_1\left (\frac{1}{2}-m,-2 m;\frac{3}{2}-m;\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )+(2 m-1) \, _2F_1\left (-m-\frac{1}{2},-2 m;\frac{1}{2}-m;\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )\right )\right )}{f \left (4 m^2-1\right ) \left ((2 m-3) \left ((A+B) \left (\cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right )+1\right ) \left (1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )^{2 m}-4 B \sin ^2\left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) F_1\left (\frac{1}{2}-m;-2 m,1;\frac{3}{2}-m;\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ),-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )\right )-64 B m \sin ^4\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ) F_1\left (\frac{3}{2}-m;1-2 m,1;\frac{5}{2}-m;\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ),-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )-32 B \sin ^4\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ) F_1\left (\frac{3}{2}-m;-2 m,2;\frac{5}{2}-m;\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ),-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.475, 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 ) ^{-1-m}\, 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}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 1}\,{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 \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 1}, 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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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