Optimal. Leaf size=274 \[ \frac{\sqrt{2} (A-B) \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{1}{2};\frac{1}{2},\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} B \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{3}{2};\frac{1}{2},\frac{1}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \]
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Rubi [A] time = 0.5435, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {2987, 2788, 140, 139, 138} \[ \frac{\sqrt{2} (A-B) \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{1}{2};\frac{1}{2},\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} B \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{3}{2};\frac{1}{2},\frac{1}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2987
Rule 2788
Rule 140
Rule 139
Rule 138
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt{c+d \sin (e+f x)}} \, dx &=(A-B) \int \frac{(a+a \sin (e+f x))^m}{\sqrt{c+d \sin (e+f x)}} \, dx+\frac{B \int \frac{(a+a \sin (e+f x))^{1+m}}{\sqrt{c+d \sin (e+f x)}} \, 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}}{\sqrt{a-a x} \sqrt{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{(a B \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m}}{\sqrt{a-a x} \sqrt{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 (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}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{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{\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}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{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{\left (a^2 (A-B) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\left (a B \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ &=\frac{\sqrt{2} (A-B) F_1\left (\frac{1}{2}+m;\frac{1}{2},\frac{1}{2};\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 \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}{f (1+2 m) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} B F_1\left (\frac{3}{2}+m;\frac{1}{2},\frac{1}{2};\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} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}{f (3+2 m) (a-a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [B] time = 6.31583, size = 672, normalized size = 2.45 \[ \frac{6 (c+d) \cot \left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^{\frac{1}{2}-m} \cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )^{m-\frac{1}{2}} (a (\sin (e+f x)+1))^m \left (\frac{(B c-A d) F_1\left (\frac{1}{2};\frac{1}{2}-m,\frac{1}{2};\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \left (2 d F_1\left (\frac{3}{2};\frac{1}{2}-m,\frac{3}{2};\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-(2 m-1) (c+d) F_1\left (\frac{3}{2};\frac{3}{2}-m,\frac{1}{2};\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )+3 (c+d) F_1\left (\frac{1}{2};\frac{1}{2}-m,\frac{1}{2};\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}-\frac{B (c+d \sin (e+f x)) F_1\left (\frac{1}{2};\frac{1}{2}-m,-\frac{1}{2};\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \left (-2 d F_1\left (\frac{3}{2};\frac{1}{2}-m,\frac{1}{2};\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-(2 m-1) (c+d) F_1\left (\frac{3}{2};\frac{3}{2}-m,-\frac{1}{2};\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )+3 (c+d) F_1\left (\frac{1}{2};\frac{1}{2}-m,-\frac{1}{2};\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}\right )}{d f \sqrt{c+d \sin (e+f x)}} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.343, 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 ){\frac{1}{\sqrt{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 (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt{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 (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt{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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin{\left (e + f x \right )}\right )}{\sqrt{c + d \sin{\left (e + f x \right )}}}\, dx \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 (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt{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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