Optimal. Leaf size=277 \[ \frac{\sqrt{2} B \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^{-m} \left (\frac{c+d \sin (e+f x)}{c-d}\right )^m F_1\left (m+\frac{3}{2};\frac{1}{2},m+1;m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) (c-d) \sqrt{1-\sin (e+f x)}}-\frac{a 2^{m+\frac{1}{2}} (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{m-1} \left (\frac{(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m} \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)} \]
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Rubi [A] time = 0.482663, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2987, 2788, 132, 140, 139, 138} \[ \frac{\sqrt{2} B \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^{-m} \left (\frac{c+d \sin (e+f x)}{c-d}\right )^m F_1\left (m+\frac{3}{2};\frac{1}{2},m+1;m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) (c-d) \sqrt{1-\sin (e+f x)}}-\frac{a 2^{m+\frac{1}{2}} (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{m-1} \left (\frac{(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m} \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)} \]
Antiderivative was successfully verified.
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Rule 2987
Rule 2788
Rule 132
Rule 140
Rule 139
Rule 138
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{-1-m} \, dx &=(A-B) \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx+\frac{B \int (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{-1-m} \, dx}{a}\\ &=\frac{\left (a^2 (A-B) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m} (c+d x)^{-1-m}}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}+\frac{(a B \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} (c+d x)^{-1-m}}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2^{\frac{1}{2}+m} a (A-B) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac{(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c+d) f}+\frac{\left (a B \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} (c+d x)^{-1-m}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2^{\frac{1}{2}+m} a (A-B) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac{(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c+d) f}+\frac{\left (a^2 B \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} (c+d \sin (e+f x))^{-m} \left (\frac{a (c+d \sin (e+f x))}{a c-a d}\right )^m\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} \left (\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}\right )^{-1-m}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} (a c-a d) f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2^{\frac{1}{2}+m} a (A-B) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac{(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c+d) f}+\frac{\sqrt{2} B F_1\left (\frac{3}{2}+m;\frac{1}{2},1+m;\frac{5}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) \sqrt{1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{-m} \left (\frac{c+d \sin (e+f x)}{c-d}\right )^m}{(c-d) f (3+2 m) (a-a \sin (e+f x))}\\ \end{align*}
Mathematica [B] time = 6.89663, size = 573, normalized size = 2.07 \[ \frac{2 \cot \left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^{\frac{1}{2}-m} \cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )^{m-\frac{1}{2}} (a (\sin (e+f x)+1))^m (c+d \sin (e+f x))^{-m} \left (-A \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d \sin (e+f x)}\right ) \left (\frac{(c+d) \cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m}-\frac{3 B (c+d)^2 F_1\left (\frac{1}{2};\frac{1}{2}-m,m;\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{d \left (3 (c+d) F_1\left (\frac{1}{2};\frac{1}{2}-m,m;\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \left ((2 m-1) (c+d) F_1\left (\frac{3}{2};\frac{3}{2}-m,m;\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-4 d m F_1\left (\frac{3}{2};\frac{1}{2}-m,m+1;\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )\right )}+\frac{B c \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d \sin (e+f x)}\right ) \left (\frac{(c+d) \cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m}}{d}\right )}{f (c+d)} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.44, 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+d\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 (d \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 (d \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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