Optimal. Leaf size=607 \[ \frac {a^3 (c+d x)^{m+1}}{d (m+1)}-\frac {3 a^2 b e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i f (c+d x)}{d}\right )}{2 f}-\frac {3 a^2 b e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i f (c+d x)}{d}\right )}{2 f}+\frac {3 i a b^2 2^{-m-3} e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {3 i a b^2 2^{-m-3} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {3 a b^2 (c+d x)^{m+1}}{2 d (m+1)}-\frac {3 b^3 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i f (c+d x)}{d}\right )}{8 f}+\frac {b^3 3^{-m-1} e^{3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {3 i f (c+d x)}{d}\right )}{8 f}-\frac {3 b^3 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i f (c+d x)}{d}\right )}{8 f}+\frac {b^3 3^{-m-1} e^{-3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {3 i f (c+d x)}{d}\right )}{8 f} \]
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Rubi [A] time = 0.76, antiderivative size = 607, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3317, 3308, 2181, 3312, 3307} \[ -\frac {3 a^2 b e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {i f (c+d x)}{d}\right )}{2 f}-\frac {3 a^2 b e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {i f (c+d x)}{d}\right )}{2 f}+\frac {3 i a b^2 2^{-m-3} e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {3 i a b^2 2^{-m-3} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {3 b^3 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {i f (c+d x)}{d}\right )}{8 f}+\frac {b^3 3^{-m-1} e^{3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {3 i f (c+d x)}{d}\right )}{8 f}-\frac {3 b^3 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {i f (c+d x)}{d}\right )}{8 f}+\frac {b^3 3^{-m-1} e^{-3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {3 i f (c+d x)}{d}\right )}{8 f}+\frac {a^3 (c+d x)^{m+1}}{d (m+1)}+\frac {3 a b^2 (c+d x)^{m+1}}{2 d (m+1)} \]
Antiderivative was successfully verified.
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Rule 2181
Rule 3307
Rule 3308
Rule 3312
Rule 3317
Rubi steps
\begin {align*} \int (c+d x)^m (a+b \sin (e+f x))^3 \, dx &=\int \left (a^3 (c+d x)^m+3 a^2 b (c+d x)^m \sin (e+f x)+3 a b^2 (c+d x)^m \sin ^2(e+f x)+b^3 (c+d x)^m \sin ^3(e+f x)\right ) \, dx\\ &=\frac {a^3 (c+d x)^{1+m}}{d (1+m)}+\left (3 a^2 b\right ) \int (c+d x)^m \sin (e+f x) \, dx+\left (3 a b^2\right ) \int (c+d x)^m \sin ^2(e+f x) \, dx+b^3 \int (c+d x)^m \sin ^3(e+f x) \, dx\\ &=\frac {a^3 (c+d x)^{1+m}}{d (1+m)}+\frac {1}{2} \left (3 i a^2 b\right ) \int e^{-i (e+f x)} (c+d x)^m \, dx-\frac {1}{2} \left (3 i a^2 b\right ) \int e^{i (e+f x)} (c+d x)^m \, dx+\left (3 a b^2\right ) \int \left (\frac {1}{2} (c+d x)^m-\frac {1}{2} (c+d x)^m \cos (2 e+2 f x)\right ) \, dx+b^3 \int \left (\frac {3}{4} (c+d x)^m \sin (e+f x)-\frac {1}{4} (c+d x)^m \sin (3 e+3 f x)\right ) \, dx\\ &=\frac {a^3 (c+d x)^{1+m}}{d (1+m)}+\frac {3 a b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {3 a^2 b e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{2 f}-\frac {3 a^2 b e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{2 f}-\frac {1}{2} \left (3 a b^2\right ) \int (c+d x)^m \cos (2 e+2 f x) \, dx-\frac {1}{4} b^3 \int (c+d x)^m \sin (3 e+3 f x) \, dx+\frac {1}{4} \left (3 b^3\right ) \int (c+d x)^m \sin (e+f x) \, dx\\ &=\frac {a^3 (c+d x)^{1+m}}{d (1+m)}+\frac {3 a b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {3 a^2 b e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{2 f}-\frac {3 a^2 b e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{2 f}-\frac {1}{4} \left (3 a b^2\right ) \int e^{-i (2 e+2 f x)} (c+d x)^m \, dx-\frac {1}{4} \left (3 a b^2\right ) \int e^{i (2 e+2 f x)} (c+d x)^m \, dx-\frac {1}{8} \left (i b^3\right ) \int e^{-i (3 e+3 f x)} (c+d x)^m \, dx+\frac {1}{8} \left (i b^3\right ) \int e^{i (3 e+3 f x)} (c+d x)^m \, dx+\frac {1}{8} \left (3 i b^3\right ) \int e^{-i (e+f x)} (c+d x)^m \, dx-\frac {1}{8} \left (3 i b^3\right ) \int e^{i (e+f x)} (c+d x)^m \, dx\\ &=\frac {a^3 (c+d x)^{1+m}}{d (1+m)}+\frac {3 a b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {3 a^2 b e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{2 f}-\frac {3 b^3 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{8 f}-\frac {3 a^2 b e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{2 f}-\frac {3 b^3 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{8 f}+\frac {3 i 2^{-3-m} a b^2 e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {3 i 2^{-3-m} a b^2 e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {3^{-1-m} b^3 e^{3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 i f (c+d x)}{d}\right )}{8 f}+\frac {3^{-1-m} b^3 e^{-3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 i f (c+d x)}{d}\right )}{8 f}\\ \end {align*}
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Mathematica [A] time = 6.18, size = 415, normalized size = 0.68 \[ \frac {i (c+d x)^m \left (9 i b \left (4 a^2+b^2\right ) e^{i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i f (c+d x)}{d}\right )+9 i b \left (4 a^2+b^2\right ) e^{-i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i f (c+d x)}{d}\right )-\frac {12 i a f \left (2 a^2+3 b^2\right ) (c+d x)}{d (m+1)}+9 a b^2 2^{-m} e^{2 i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i f (c+d x)}{d}\right )-9 a b^2 2^{-m} e^{-2 i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )-i b^3 3^{-m} e^{3 i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {3 i f (c+d x)}{d}\right )-i b^3 3^{-m} e^{-3 i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {3 i f (c+d x)}{d}\right )\right )}{24 f} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 428, normalized size = 0.71 \[ \frac {{\left (b^{3} d m + b^{3} d\right )} e^{\left (-\frac {d m \log \left (\frac {3 i \, f}{d}\right ) + 3 i \, d e - 3 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {3 i \, d f x + 3 i \, c f}{d}\right ) + {\left (-9 i \, a b^{2} d m - 9 i \, a b^{2} d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) + 2 i \, d e - 2 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {2 i \, d f x + 2 i \, c f}{d}\right ) - 9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} d m + {\left (4 \, a^{2} b + b^{3}\right )} d\right )} e^{\left (-\frac {d m \log \left (\frac {i \, f}{d}\right ) + i \, d e - i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {i \, d f x + i \, c f}{d}\right ) - 9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} d m + {\left (4 \, a^{2} b + b^{3}\right )} d\right )} e^{\left (-\frac {d m \log \left (-\frac {i \, f}{d}\right ) - i \, d e + i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-i \, d f x - i \, c f}{d}\right ) + {\left (9 i \, a b^{2} d m + 9 i \, a b^{2} d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 i \, f}{d}\right ) - 2 i \, d e + 2 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-2 i \, d f x - 2 i \, c f}{d}\right ) + {\left (b^{3} d m + b^{3} d\right )} e^{\left (-\frac {d m \log \left (-\frac {3 i \, f}{d}\right ) - 3 i \, d e + 3 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-3 i \, d f x - 3 i \, c f}{d}\right ) + 12 \, {\left ({\left (2 \, a^{3} + 3 \, a b^{2}\right )} d f x + {\left (2 \, a^{3} + 3 \, a b^{2}\right )} c f\right )} {\left (d x + c\right )}^{m}}{24 \, {\left (d f m + d f\right )}} \]
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 \[ \int {\left (b \sin \left (f x + e\right ) + a\right )}^{3} {\left (d x + c\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.44, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{m} \left (a +b \sin \left (f x +e \right )\right )^{3}\, dx \]
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 \[ \frac {{\left (d x + c\right )}^{m + 1} a^{3}}{d {\left (m + 1\right )}} + \frac {6 \, a b^{2} e^{\left (m \log \left (d x + c\right ) + \log \left (d x + c\right )\right )} - 6 \, {\left (a b^{2} d m + a b^{2} d\right )} \int {\left (d x + c\right )}^{m} \cos \left (2 \, f x + 2 \, e\right )\,{d x} - {\left (b^{3} d m + b^{3} d\right )} \int {\left (d x + c\right )}^{m} \sin \left (3 \, f x + 3 \, e\right )\,{d x} + 3 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} d m + {\left (4 \, a^{2} b + b^{3}\right )} d\right )} \int {\left (d x + c\right )}^{m} \sin \left (f x + e\right )\,{d x}}{4 \, {\left (d m + d\right )}} \]
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 \[ \int {\left (a+b\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c+d\,x\right )}^m \,d x \]
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 \[ \int \left (a + b \sin {\left (e + f x \right )}\right )^{3} \left (c + d x\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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