Optimal. Leaf size=270 \[ -\frac {a b^2 (7+2 m) \cos (e+f x) (d \sin (e+f x))^{1+m}}{d f (2+m) (3+m)}+\frac {a \left (3 b^2 (1+m)+a^2 (2+m)\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+m}}{d f (1+m) (2+m) \sqrt {\cos ^2(e+f x)}}+\frac {b \left (b^2 (2+m)+3 a^2 (3+m)\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+m}}{d^2 f (2+m) (3+m) \sqrt {\cos ^2(e+f x)}}-\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{1+m} (a+b \sin (e+f x))}{d f (3+m)} \]
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Rubi [A]
time = 0.27, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2872, 3102,
2827, 2722} \begin {gather*} \frac {b \left (3 a^2 (m+3)+b^2 (m+2)\right ) \cos (e+f x) (d \sin (e+f x))^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\sin ^2(e+f x)\right )}{d^2 f (m+2) (m+3) \sqrt {\cos ^2(e+f x)}}+\frac {a \left (a^2 (m+2)+3 b^2 (m+1)\right ) \cos (e+f x) (d \sin (e+f x))^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(e+f x)\right )}{d f (m+1) (m+2) \sqrt {\cos ^2(e+f x)}}-\frac {a b^2 (2 m+7) \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (m+2) (m+3)}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x)) (d \sin (e+f x))^{m+1}}{d f (m+3)} \end {gather*}
Antiderivative was successfully verified.
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Rule 2722
Rule 2827
Rule 2872
Rule 3102
Rubi steps
\begin {align*} \int (d \sin (e+f x))^m (a+b \sin (e+f x))^3 \, dx &=-\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{1+m} (a+b \sin (e+f x))}{d f (3+m)}+\frac {\int (d \sin (e+f x))^m \left (a d \left (b^2 (1+m)+a^2 (3+m)\right )+b d \left (b^2 (2+m)+3 a^2 (3+m)\right ) \sin (e+f x)+a b^2 d (7+2 m) \sin ^2(e+f x)\right ) \, dx}{d (3+m)}\\ &=-\frac {a b^2 (7+2 m) \cos (e+f x) (d \sin (e+f x))^{1+m}}{d f (2+m) (3+m)}-\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{1+m} (a+b \sin (e+f x))}{d f (3+m)}+\frac {\int (d \sin (e+f x))^m \left (a d^2 (3+m) \left (3 b^2 (1+m)+a^2 (2+m)\right )+b d^2 (2+m) \left (b^2 (2+m)+3 a^2 (3+m)\right ) \sin (e+f x)\right ) \, dx}{d^2 (2+m) (3+m)}\\ &=-\frac {a b^2 (7+2 m) \cos (e+f x) (d \sin (e+f x))^{1+m}}{d f (2+m) (3+m)}-\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{1+m} (a+b \sin (e+f x))}{d f (3+m)}+\left (a \left (a^2+\frac {3 b^2 (1+m)}{2+m}\right )\right ) \int (d \sin (e+f x))^m \, dx+\frac {\left (b \left (b^2 (2+m)+3 a^2 (3+m)\right )\right ) \int (d \sin (e+f x))^{1+m} \, dx}{d (3+m)}\\ &=-\frac {a b^2 (7+2 m) \cos (e+f x) (d \sin (e+f x))^{1+m}}{d f (2+m) (3+m)}+\frac {a \left (a^2+\frac {3 b^2 (1+m)}{2+m}\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+m}}{d f (1+m) \sqrt {\cos ^2(e+f x)}}+\frac {b \left (b^2 (2+m)+3 a^2 (3+m)\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+m}}{d^2 f (2+m) (3+m) \sqrt {\cos ^2(e+f x)}}-\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{1+m} (a+b \sin (e+f x))}{d f (3+m)}\\ \end {align*}
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Mathematica [A]
time = 0.52, size = 199, normalized size = 0.74 \begin {gather*} \frac {\cos (e+f x) \sin (e+f x) (d \sin (e+f x))^m \left (-\frac {a b^2 (7+2 m)}{2+m}+\frac {a (3+m) \left (3 b^2 (1+m)+a^2 (2+m)\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(e+f x)\right )}{(1+m) (2+m) \sqrt {\cos ^2(e+f x)}}+\frac {b \left (b^2 (2+m)+3 a^2 (3+m)\right ) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};\sin ^2(e+f x)\right ) \sin (e+f x)}{(2+m) \sqrt {\cos ^2(e+f x)}}-b^2 (a+b \sin (e+f x))\right )}{f (3+m)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 1.01, size = 0, normalized size = 0.00 \[\int \left (d \sin \left (f x +e \right )\right )^{m} \left (a +b \sin \left (f x +e \right )\right )^{3}\, dx\]
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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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 = -0.00 \begin {gather*} \int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,{\left (a+b\,\sin \left (e+f\,x\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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