Optimal. Leaf size=442 \[ \frac {i b e^{i c} \left (4 a^2+b^2\right ) \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},-i d x^3\right )}{8 e}-\frac {i b e^{-i c} \left (4 a^2+b^2\right ) \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},i d x^3\right )}{8 e}+\frac {a \left (2 a^2+3 b^2\right ) (e x)^{m+1}}{2 e (m+1)}+\frac {a b^2 e^{2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},-2 i d x^3\right )}{e}+\frac {a b^2 e^{-2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},2 i d x^3\right )}{e}-\frac {i b^3 e^{3 i c} 3^{-\frac {m}{3}-\frac {4}{3}} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},-3 i d x^3\right )}{8 e}+\frac {i b^3 e^{-3 i c} 3^{-\frac {m}{3}-\frac {4}{3}} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},3 i d x^3\right )}{8 e} \]
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Rubi [A] time = 0.41, antiderivative size = 442, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3403, 6, 3390, 2218, 3389} \[ \frac {i b e^{i c} \left (4 a^2+b^2\right ) \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{3},-i d x^3\right )}{8 e}-\frac {i b e^{-i c} \left (4 a^2+b^2\right ) \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{3},i d x^3\right )}{8 e}+\frac {a b^2 e^{2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{3},-2 i d x^3\right )}{e}+\frac {a b^2 e^{-2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{3},2 i d x^3\right )}{e}-\frac {i b^3 e^{3 i c} 3^{-\frac {m}{3}-\frac {4}{3}} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{3},-3 i d x^3\right )}{8 e}+\frac {i b^3 e^{-3 i c} 3^{-\frac {m}{3}-\frac {4}{3}} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{3},3 i d x^3\right )}{8 e}+\frac {a \left (2 a^2+3 b^2\right ) (e x)^{m+1}}{2 e (m+1)} \]
Antiderivative was successfully verified.
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Rule 6
Rule 2218
Rule 3389
Rule 3390
Rule 3403
Rubi steps
\begin {align*} \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx &=\int \left (a^3 (e x)^m+\frac {3}{2} a b^2 (e x)^m-\frac {3}{2} a b^2 (e x)^m \cos \left (2 c+2 d x^3\right )+3 a^2 b (e x)^m \sin \left (c+d x^3\right )+\frac {3}{4} b^3 (e x)^m \sin \left (c+d x^3\right )-\frac {1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^3\right )\right ) \, dx\\ &=\int \left (\left (a^3+\frac {3 a b^2}{2}\right ) (e x)^m-\frac {3}{2} a b^2 (e x)^m \cos \left (2 c+2 d x^3\right )+3 a^2 b (e x)^m \sin \left (c+d x^3\right )+\frac {3}{4} b^3 (e x)^m \sin \left (c+d x^3\right )-\frac {1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^3\right )\right ) \, dx\\ &=\int \left (\left (a^3+\frac {3 a b^2}{2}\right ) (e x)^m-\frac {3}{2} a b^2 (e x)^m \cos \left (2 c+2 d x^3\right )+\left (3 a^2 b+\frac {3 b^3}{4}\right ) (e x)^m \sin \left (c+d x^3\right )-\frac {1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^3\right )\right ) \, dx\\ &=\frac {a \left (2 a^2+3 b^2\right ) (e x)^{1+m}}{2 e (1+m)}-\frac {1}{2} \left (3 a b^2\right ) \int (e x)^m \cos \left (2 c+2 d x^3\right ) \, dx-\frac {1}{4} b^3 \int (e x)^m \sin \left (3 c+3 d x^3\right ) \, dx+\frac {1}{4} \left (3 b \left (4 a^2+b^2\right )\right ) \int (e x)^m \sin \left (c+d x^3\right ) \, dx\\ &=\frac {a \left (2 a^2+3 b^2\right ) (e x)^{1+m}}{2 e (1+m)}-\frac {1}{4} \left (3 a b^2\right ) \int e^{-2 i c-2 i d x^3} (e x)^m \, dx-\frac {1}{4} \left (3 a b^2\right ) \int e^{2 i c+2 i d x^3} (e x)^m \, dx-\frac {1}{8} \left (i b^3\right ) \int e^{-3 i c-3 i d x^3} (e x)^m \, dx+\frac {1}{8} \left (i b^3\right ) \int e^{3 i c+3 i d x^3} (e x)^m \, dx+\frac {1}{8} \left (3 i b \left (4 a^2+b^2\right )\right ) \int e^{-i c-i d x^3} (e x)^m \, dx-\frac {1}{8} \left (3 i b \left (4 a^2+b^2\right )\right ) \int e^{i c+i d x^3} (e x)^m \, dx\\ &=\frac {a \left (2 a^2+3 b^2\right ) (e x)^{1+m}}{2 e (1+m)}+\frac {i b \left (4 a^2+b^2\right ) e^{i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-i d x^3\right )}{8 e}-\frac {i b \left (4 a^2+b^2\right ) e^{-i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},i d x^3\right )}{8 e}+\frac {2^{-\frac {7}{3}-\frac {m}{3}} a b^2 e^{2 i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-2 i d x^3\right )}{e}+\frac {2^{-\frac {7}{3}-\frac {m}{3}} a b^2 e^{-2 i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},2 i d x^3\right )}{e}-\frac {i 3^{-\frac {4}{3}-\frac {m}{3}} b^3 e^{3 i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-3 i d x^3\right )}{8 e}+\frac {i 3^{-\frac {4}{3}-\frac {m}{3}} b^3 e^{-3 i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},3 i d x^3\right )}{8 e}\\ \end {align*}
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Mathematica [A] time = 12.02, size = 373, normalized size = 0.84 \[ \frac {1}{24} i x (e x)^m \left (3 b e^{i c} \left (4 a^2+b^2\right ) \left (-i d x^3\right )^{-\frac {m}{3}-\frac {1}{3}} \Gamma \left (\frac {m+1}{3},-i d x^3\right )-3 b e^{-i c} \left (4 a^2+b^2\right ) \left (i d x^3\right )^{-\frac {m}{3}-\frac {1}{3}} \Gamma \left (\frac {m+1}{3},i d x^3\right )-\frac {12 i a \left (2 a^2+3 b^2\right )}{m+1}-3 i a b^2 e^{2 i c} 2^{\frac {2}{3}-\frac {m}{3}} \left (-i d x^3\right )^{-\frac {m}{3}-\frac {1}{3}} \Gamma \left (\frac {m+1}{3},-2 i d x^3\right )-3 i a b^2 e^{-2 i c} 2^{\frac {2}{3}-\frac {m}{3}} \left (i d x^3\right )^{-\frac {m}{3}-\frac {1}{3}} \Gamma \left (\frac {m+1}{3},2 i d x^3\right )-b^3 e^{3 i c} 3^{-\frac {m}{3}-\frac {1}{3}} \left (-i d x^3\right )^{-\frac {m}{3}-\frac {1}{3}} \Gamma \left (\frac {m+1}{3},-3 i d x^3\right )+b^3 e^{-3 i c} 3^{-\frac {m}{3}-\frac {1}{3}} \left (i d x^3\right )^{-\frac {m}{3}-\frac {1}{3}} \Gamma \left (\frac {m+1}{3},3 i d x^3\right )\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.94, size = 345, normalized size = 0.78 \[ \frac {36 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \left (e x\right )^{m} d x + {\left (b^{3} e^{2} m + b^{3} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {3 i \, d}{e^{3}}\right ) - 3 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, 3 i \, d x^{3}\right ) + {\left (-9 i \, a b^{2} e^{2} m - 9 i \, a b^{2} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {2 i \, d}{e^{3}}\right ) - 2 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, 2 i \, d x^{3}\right ) - 9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} e^{2} m + {\left (4 \, a^{2} b + b^{3}\right )} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {i \, d}{e^{3}}\right ) - i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, i \, d x^{3}\right ) - 9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} e^{2} m + {\left (4 \, a^{2} b + b^{3}\right )} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {i \, d}{e^{3}}\right ) + i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -i \, d x^{3}\right ) + {\left (9 i \, a b^{2} e^{2} m + 9 i \, a b^{2} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {2 i \, d}{e^{3}}\right ) + 2 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -2 i \, d x^{3}\right ) + {\left (b^{3} e^{2} m + b^{3} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {3 i \, d}{e^{3}}\right ) + 3 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -3 i \, d x^{3}\right )}{72 \, {\left (d m + d\right )}} \]
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 \[ \int {\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{3} \left (e x\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.62, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (a +b \sin \left (d \,x^{3}+c \right )\right )^{3}\, dx \]
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 \[ \frac {\left (e x\right )^{m + 1} a^{3}}{e {\left (m + 1\right )}} + \frac {\frac {{\left ({\left (4 \, a^{2} b + b^{3}\right )} e^{m} m \cos \relax (c) + {\left (4 \, a^{2} b + b^{3}\right )} e^{m} \cos \relax (c)\right )} d x^{4} x^{m} \Gamma \left (\frac {1}{6} \, m + \frac {2}{3}\right ) \,_1F_2\left (\begin {matrix} \frac {1}{6} \, m + \frac {2}{3} \\ \frac {3}{2},\frac {1}{6} \, m + \frac {5}{3} \end {matrix} ; -\frac {1}{4} \, d^{2} x^{6} \right )}{2 \, \Gamma \left (\frac {1}{6} \, m + \frac {5}{3}\right )} + 12 \, a b^{2} e^{m} x x^{m} + \frac {{\left ({\left (4 \, a^{2} b + b^{3}\right )} e^{m} m \sin \relax (c) + {\left (4 \, a^{2} b + b^{3}\right )} e^{m} \sin \relax (c)\right )} x x^{m} \Gamma \left (\frac {1}{6} \, m + \frac {1}{6}\right ) \,_1F_2\left (\begin {matrix} \frac {1}{6} \, m + \frac {1}{6} \\ \frac {1}{2},\frac {1}{6} \, m + \frac {7}{6} \end {matrix} ; -\frac {1}{4} \, d^{2} x^{6} \right )}{2 \, \Gamma \left (\frac {1}{6} \, m + \frac {7}{6}\right )} - 12 \, {\left (a b^{2} e^{m} m + a b^{2} e^{m}\right )} \int x^{m} \cos \left (2 \, d x^{3} + 2 \, c\right )\,{d x} - 2 \, {\left (b^{3} e^{m} m + b^{3} e^{m}\right )} \int x^{m} \sin \left (3 \, d x^{3} + 3 \, c\right )\,{d x} + 3 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} e^{m} m + {\left (4 \, a^{2} b + b^{3}\right )} e^{m}\right )} \int x^{m} \sin \left (d x^{3} + c\right )\,{d x}}{8 \, {\left (m + 1\right )}} \]
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 \[ \int {\left (e\,x\right )}^m\,{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^3 \,d x \]
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 \[ \int \left (e x\right )^{m} \left (a + b \sin {\left (c + d x^{3} \right )}\right )^{3}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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