Optimal. Leaf size=351 \[ -\frac{2^{m+\frac{1}{2}} \cos (e+f x) \left (A (m+3) \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right )+B \left (c^2 m \left (m^2+5 m+6\right )+2 c d \left (m^3+4 m^2+4 m+3\right )+d^2 m \left (m^2+3 m+5\right )\right )\right ) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2) (m+3)}+\frac{\cos (e+f x) \left (d (A d (m+3)+B (2 c+d m))-2 (m+2) \left (A c d (m+3)+B \left (c^2+c d m+d^2\right )\right )\right ) (a \sin (e+f x)+a)^m}{f (m+1) (m+2) (m+3)}-\frac{d \cos (e+f x) (A d (m+3)+B (2 c+d m)) (a \sin (e+f x)+a)^{m+1}}{a f (m+2) (m+3)}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2}{f (m+3)} \]
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Rubi [A] time = 0.986798, antiderivative size = 351, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {2983, 2968, 3023, 2751, 2652, 2651} \[ -\frac{2^{m+\frac{1}{2}} \cos (e+f x) \left (A (m+3) \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right )+B \left (c^2 m \left (m^2+5 m+6\right )+2 c d \left (m^3+4 m^2+4 m+3\right )+d^2 m \left (m^2+3 m+5\right )\right )\right ) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2) (m+3)}+\frac{\cos (e+f x) \left (d (A d (m+3)+B (2 c+d m))-2 (m+2) \left (A c d (m+3)+B \left (c^2+c d m+d^2\right )\right )\right ) (a \sin (e+f x)+a)^m}{f (m+1) (m+2) (m+3)}-\frac{d \cos (e+f x) (A d (m+3)+B (2 c+d m)) (a \sin (e+f x)+a)^{m+1}}{a f (m+2) (m+3)}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2}{f (m+3)} \]
Antiderivative was successfully verified.
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Rule 2983
Rule 2968
Rule 3023
Rule 2751
Rule 2652
Rule 2651
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac{\int (a+a \sin (e+f x))^m (c+d \sin (e+f x)) (a (A c (3+m)+B (2 d+c m))+a (A d (3+m)+B (2 c+d m)) \sin (e+f x)) \, dx}{a (3+m)}\\ &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac{\int (a+a \sin (e+f x))^m \left (a c (A c (3+m)+B (2 d+c m))+(a d (A c (3+m)+B (2 d+c m))+a c (A d (3+m)+B (2 c+d m))) \sin (e+f x)+a d (A d (3+m)+B (2 c+d m)) \sin ^2(e+f x)\right ) \, dx}{a (3+m)}\\ &=-\frac{d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac{\int (a+a \sin (e+f x))^m \left (a^2 (c (2+m) (A c (3+m)+B (2 d+c m))+d (1+m) (A d (3+m)+B (2 c+d m)))-a^2 \left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \sin (e+f x)\right ) \, dx}{a^2 (2+m) (3+m)}\\ &=\frac{\left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac{d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac{\left (A (3+m) \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right )+B \left (d^2 m \left (5+3 m+m^2\right )+c^2 m \left (6+5 m+m^2\right )+2 c d \left (3+4 m+4 m^2+m^3\right )\right )\right ) \int (a+a \sin (e+f x))^m \, dx}{(1+m) (2+m) (3+m)}\\ &=\frac{\left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac{d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac{\left (\left (A (3+m) \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right )+B \left (d^2 m \left (5+3 m+m^2\right )+c^2 m \left (6+5 m+m^2\right )+2 c d \left (3+4 m+4 m^2+m^3\right )\right )\right ) (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{(1+m) (2+m) (3+m)}\\ &=\frac{\left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac{2^{\frac{1}{2}+m} \left (A (3+m) \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right )+B \left (d^2 m \left (5+3 m+m^2\right )+c^2 m \left (6+5 m+m^2\right )+2 c d \left (3+4 m+4 m^2+m^3\right )\right )\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\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}{f (1+m) (2+m) (3+m)}-\frac{d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}\\ \end{align*}
Mathematica [A] time = 7.64892, size = 300, normalized size = 0.85 \[ -\frac{\csc ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^m (a (\sin (e+f x)+1))^m \left (-\frac{2}{7} (A-B) (c-d)^2 \tan ^7\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac{7}{2},m+4;\frac{9}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )-\frac{2}{5} (c-d) (A (3 c+d)-B (c+3 d)) \tan ^5\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac{5}{2},m+4;\frac{7}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )-\frac{2}{3} (c+d) (3 A c-A d+B c-3 B d) \tan ^3\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac{3}{2},m+4;\frac{5}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )-2 (A+B) (c+d)^2 \tan \left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac{1}{2},m+4;\frac{3}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )\right )}{f} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 2.762, 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 ) ^{2}\, 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 (d \sin \left (f x + e\right ) + c\right )}^{2}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{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 (A c^{2} + 2 \, B c d + A d^{2} -{\left (2 \, B c d + A d^{2}\right )} \cos \left (f x + e\right )^{2} -{\left (B d^{2} \cos \left (f x + e\right )^{2} - B c^{2} - 2 \, A c d - B d^{2}\right )} \sin \left (f x + e\right )\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}, 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{\left (B \sin \left (f x + e\right ) + A\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{2}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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