Optimal. Leaf size=100 \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m F_1\left (m+\frac{1}{2};\frac{1}{2},1;m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) (c-d) \sqrt{1-\sin (e+f x)}} \]
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Rubi [A] time = 0.128288, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2788, 137, 136} \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m F_1\left (m+\frac{1}{2};\frac{1}{2},1;m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) (c-d) \sqrt{1-\sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2788
Rule 137
Rule 136
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^m}{c+d \sin (e+f x)} \, dx &=\frac{\left (a^2 \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{\sqrt{a-a x} (c+d x)} \, dx,x,\sin (e+f x)\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=\frac{\left (a^2 \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}}{\sqrt{\frac{1}{2}-\frac{x}{2}} (c+d x)} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=\frac{\sqrt{2} F_1\left (\frac{1}{2}+m;\frac{1}{2},1;\frac{3}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{(c-d) f (1+2 m) \sqrt{1-\sin (e+f x)}}\\ \end{align*}
Mathematica [B] time = 1.17552, size = 363, normalized size = 3.63 \[ -\frac{6 (c+d) \cot \left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^{\frac{1}{2}-m} \cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )^{m-\frac{1}{2}} (a (\sin (e+f x)+1))^m F_1\left (\frac{1}{2};\frac{1}{2}-m,1;\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{f (c+d \sin (e+f x)) \left (\sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \left (4 d F_1\left (\frac{3}{2};\frac{1}{2}-m,2;\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-(2 m-1) (c+d) F_1\left (\frac{3}{2};\frac{3}{2}-m,1;\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )+3 (c+d) F_1\left (\frac{1}{2};\frac{1}{2}-m,1;\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.819, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}}{c+d\sin \left ( fx+e \right ) }}\, 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 \frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{d \sin \left (f x + e\right ) + c}\,{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 (\frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{d \sin \left (f x + e\right ) + c}, 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 \frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{d \sin \left (f x + e\right ) + c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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