Optimal. Leaf size=114 \[ \frac {c^2 2^{\frac {3}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac {1}{2} (2 m-1),\frac {1}{2} (2 m+1);\frac {1}{2} (2 m+3);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]
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Rubi [A] time = 0.19, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2745, 2689, 70, 69} \[ \frac {c^2 2^{\frac {3}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac {1}{2} (2 m-1),\frac {1}{2} (2 m+1);\frac {1}{2} (2 m+3);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 2689
Rule 2745
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx &=\left (\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))^{1-2 m} \, dx\\ &=\frac {\left (c^2 \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)^{1-2 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 {\left (2^{\frac {1}{2}-m} c^3 \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 )^{1-2 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 {2^{\frac {3}{2}-m} c^2 \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*}
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Mathematica [C] time = 7.18, size = 602, normalized size = 5.28 \[ -\frac {c 2^{2-m} (2 m-3) (\sin (e+f x)-1) \cos ^{3-2 m}\left (\frac {1}{4} (2 e+2 f x+\pi )\right ) (a (\sin (e+f x)+1))^m \left (F_1\left (\frac {1}{2}-m;-2 m,2;\frac {3}{2}-m;\tan ^2\left (\frac {1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{8} (2 e+2 f x-\pi )\right )\right )-F_1\left (\frac {1}{2}-m;-2 m,3;\frac {3}{2}-m;\tan ^2\left (\frac {1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{8} (2 e+2 f x-\pi )\right )\right )\right ) (c-c \sin (e+f x))^{-m} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{2 (m-1)}}{f (2 m-1) \left (2 \tan ^2\left (\frac {1}{8} (2 e+2 f x-\pi )\right ) \left (2 m F_1\left (\frac {3}{2}-m;1-2 m,2;\frac {5}{2}-m;\tan ^2\left (\frac {1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{8} (2 e+2 f x-\pi )\right )\right )-2 m F_1\left (\frac {3}{2}-m;1-2 m,3;\frac {5}{2}-m;\tan ^2\left (\frac {1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{8} (2 e+2 f x-\pi )\right )\right )+2 F_1\left (\frac {3}{2}-m;-2 m,3;\frac {5}{2}-m;\tan ^2\left (\frac {1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{8} (2 e+2 f x-\pi )\right )\right )-3 F_1\left (\frac {3}{2}-m;-2 m,4;\frac {5}{2}-m;\tan ^2\left (\frac {1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{8} (2 e+2 f x-\pi )\right )\right )\right )+(2 m-3) F_1\left (\frac {1}{2}-m;-2 m,2;\frac {3}{2}-m;\tan ^2\left (\frac {1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{8} (2 e+2 f x-\pi )\right )\right )+(3-2 m) F_1\left (\frac {1}{2}-m;-2 m,3;\frac {3}{2}-m;\tan ^2\left (\frac {1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{8} (2 e+2 f x-\pi )\right )\right )\right )} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.56, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{1-m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{1-m} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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