Optimal. Leaf size=184 \[ \frac {(C-B (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {2^{\frac {1}{2}+m} \left (B m (2+m)+C \left (1+m+m^2\right )+A \left (2+3 m+m^2\right )\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)} \]
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Rubi [A]
time = 0.16, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3102, 2830,
2731, 2730} \begin {gather*} -\frac {2^{m+\frac {1}{2}} \left (A \left (m^2+3 m+2\right )+B m (m+2)+C \left (m^2+m+1\right )\right ) \cos (e+f x) (\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)}+\frac {(C-B (m+2)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1) (m+2)}-\frac {C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 2730
Rule 2731
Rule 2830
Rule 3102
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=-\frac {C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac {\int (a+a \sin (e+f x))^m (a (C (1+m)+A (2+m))-a (C-B (2+m)) \sin (e+f x)) \, dx}{a (2+m)}\\ &=\frac {(C-B (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac {\left (B m (2+m)+C \left (1+m+m^2\right )+A \left (2+3 m+m^2\right )\right ) \int (a+a \sin (e+f x))^m \, dx}{(1+m) (2+m)}\\ &=\frac {(C-B (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac {\left (\left (B m (2+m)+C \left (1+m+m^2\right )+A \left (2+3 m+m^2\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)}\\ &=\frac {(C-B (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {2^{\frac {1}{2}+m} \left (B m (2+m)+C \left (1+m+m^2\right )+A \left (2+3 m+m^2\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)}-\frac {C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 3.76, size = 525, normalized size = 2.85 \begin {gather*} -\frac {(a (1+\sin (e+f x)))^m \left (\frac {\sqrt [4]{-1} 4^{-m} B e^{-\frac {3}{2} i (e+f x)} \left (-(-1)^{3/4} e^{-\frac {1}{2} i (e+f x)} \left (i+e^{i (e+f x)}\right )\right )^{1+2 m} \left (e^{2 i (e+f x)} (-1+m) \, _2F_1\left (1,m;-m;-i e^{-i (e+f x)}\right )-(1+m) \, _2F_1\left (1,2+m;2-m;-i e^{-i (e+f x)}\right )\right )}{-1+m^2}+\frac {2^{-1-2 m} C e^{-3 i (e+f x)} \left (1-i e^{i (e+f x)}\right ) \left (-(-1)^{3/4} e^{-\frac {1}{2} i (e+f x)} \left (i+e^{i (e+f x)}\right )\right )^{2 m} \left (e^{4 i (e+f x)} (-2+m) \, _2F_1\left (1,-1+m;-1-m;-i e^{-i (e+f x)}\right )+(2+m) \, _2F_1\left (1,3+m;3-m;-i e^{-i (e+f x)}\right )\right )}{-4+m^2}+\frac {4 \sqrt {2} A \cos ^{1+2 m}\left (\frac {1}{4} (2 e-\pi +2 f x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1}{2}+m;\frac {3}{2}+m;\sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )\right ) \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{(1+2 m) \sqrt {1-\sin (e+f x)}}+\frac {2 \sqrt {2} C \cos ^{1+2 m}\left (\frac {1}{4} (2 e-\pi +2 f x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1}{2}+m;\frac {3}{2}+m;\sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )\right ) \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{(1+2 m) \sqrt {1-\sin (e+f x)}}\right ) \sin ^{-2 m}\left (\frac {1}{4} (2 e+\pi +2 f x)\right )}{2 f} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 1.21, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )} + C \sin ^{2}{\left (e + f x \right )}\right )\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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