Optimal. Leaf size=85 \[ \frac {2^{m+\frac {1}{2}} \cos (e+f x) (1-\sin (e+f x))^{-m-\frac {1}{2}} (a-a \sin (e+f x))^m F_1\left (\frac {1}{2};-n,\frac {1}{2}-m;\frac {3}{2};\sin (e+f x)+1,\frac {1}{2} (\sin (e+f x)+1)\right )}{f} \]
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Rubi [A] time = 0.11, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2787, 2785, 133} \[ \frac {2^{m+\frac {1}{2}} \cos (e+f x) (1-\sin (e+f x))^{-m-\frac {1}{2}} (a-a \sin (e+f x))^m F_1\left (\frac {1}{2};-n,\frac {1}{2}-m;\frac {3}{2};\sin (e+f x)+1,\frac {1}{2} (\sin (e+f x)+1)\right )}{f} \]
Antiderivative was successfully verified.
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Rule 133
Rule 2785
Rule 2787
Rubi steps
\begin {align*} \int (-\sin (e+f x))^n (a-a \sin (e+f x))^m \, dx &=\left ((1-\sin (e+f x))^{-m} (a-a \sin (e+f x))^m\right ) \int (1-\sin (e+f x))^m (-\sin (e+f x))^n \, dx\\ &=\frac {\left (\cos (e+f x) (1-\sin (e+f x))^{-\frac {1}{2}-m} (a-a \sin (e+f x))^m\right ) \operatorname {Subst}\left (\int \frac {(1-x)^n (2-x)^{-\frac {1}{2}+m}}{\sqrt {x}} \, dx,x,1+\sin (e+f x)\right )}{f \sqrt {1+\sin (e+f x)}}\\ &=\frac {2^{\frac {1}{2}+m} F_1\left (\frac {1}{2};-n,\frac {1}{2}-m;\frac {3}{2};1+\sin (e+f x),\frac {1}{2} (1+\sin (e+f x))\right ) \cos (e+f x) (1-\sin (e+f x))^{-\frac {1}{2}-m} (a-a \sin (e+f x))^m}{f}\\ \end {align*}
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Mathematica [B] time = 0.60, size = 301, normalized size = 3.54 \[ -\frac {(2 m+3) \cos (e+f x) (-\sin (e+f x))^n (a-a \sin (e+f x))^m F_1\left (m+\frac {1}{2};-n,m+n+1;m+\frac {3}{2};\cot ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )}{f (2 m+1) \left ((2 m+3) F_1\left (m+\frac {1}{2};-n,m+n+1;m+\frac {3}{2};\cot ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )-2 \tan ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \left (n F_1\left (m+\frac {3}{2};1-n,m+n+1;m+\frac {5}{2};\cot ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )+(m+n+1) F_1\left (m+\frac {3}{2};-n,m+n+2;m+\frac {5}{2};\cot ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )\right )\right )} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-a \sin \left (f x + e\right ) + a\right )}^{m} \left (-\sin \left (f x + e\right )\right )^{n}, x\right ) \]
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 \[ \int {\left (-a \sin \left (f x + e\right ) + a\right )}^{m} \left (-\sin \left (f x + e\right )\right )^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.17, size = 0, normalized size = 0.00 \[ \int \left (-\sin \left (f x +e \right )\right )^{n} \left (a -a \sin \left (f x +e \right )\right )^{m}\, dx \]
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 \[ \int {\left (-a \sin \left (f x + e\right ) + a\right )}^{m} \left (-\sin \left (f x + e\right )\right )^{n}\,{d x} \]
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 \[ \int {\left (-\sin \left (e+f\,x\right )\right )}^n\,{\left (a-a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (- \sin {\left (e + f x \right )}\right )^{n} \left (- a \left (\sin {\left (e + f x \right )} - 1\right )\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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