Optimal. Leaf size=444 \[ \frac {a \left (2 a^2+3 b^2\right ) (e x)^{1+m}}{2 e (1+m)}+\frac {3 i b \left (4 a^2+b^2\right ) e^{i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-i d x^2\right )}{16 e}-\frac {3 i b \left (4 a^2+b^2\right ) e^{-i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},i d x^2\right )}{16 e}+\frac {3\ 2^{-\frac {7}{2}-\frac {m}{2}} a b^2 e^{2 i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-2 i d x^2\right )}{e}+\frac {3\ 2^{-\frac {7}{2}-\frac {m}{2}} a b^2 e^{-2 i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},2 i d x^2\right )}{e}-\frac {i 3^{-\frac {1}{2}-\frac {m}{2}} b^3 e^{3 i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-3 i d x^2\right )}{16 e}+\frac {i 3^{-\frac {1}{2}-\frac {m}{2}} b^3 e^{-3 i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},3 i d x^2\right )}{16 e} \]
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Rubi [A]
time = 0.31, antiderivative size = 444, 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 = {3484, 6, 3471,
2250, 3470} \begin {gather*} \frac {3 i b e^{i c} \left (4 a^2+b^2\right ) \left (-i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},-i d x^2\right )}{16 e}-\frac {3 i b e^{-i c} \left (4 a^2+b^2\right ) \left (i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},i d x^2\right )}{16 e}+\frac {3 a b^2 e^{2 i c} 2^{-\frac {m}{2}-\frac {7}{2}} \left (-i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},-2 i d x^2\right )}{e}+\frac {3 a b^2 e^{-2 i c} 2^{-\frac {m}{2}-\frac {7}{2}} \left (i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},2 i d x^2\right )}{e}-\frac {i b^3 e^{3 i c} 3^{-\frac {m}{2}-\frac {1}{2}} \left (-i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},-3 i d x^2\right )}{16 e}+\frac {i b^3 e^{-3 i c} 3^{-\frac {m}{2}-\frac {1}{2}} \left (i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},3 i d x^2\right )}{16 e}+\frac {a \left (2 a^2+3 b^2\right ) (e x)^{m+1}}{2 e (m+1)} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 2250
Rule 3470
Rule 3471
Rule 3484
Rubi steps
\begin {align*} \int (e x)^m \left (a+b \sin \left (c+d x^2\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^2\right )+3 a^2 b (e x)^m \sin \left (c+d x^2\right )+\frac {3}{4} b^3 (e x)^m \sin \left (c+d x^2\right )-\frac {1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^2\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^2\right )+3 a^2 b (e x)^m \sin \left (c+d x^2\right )+\frac {3}{4} b^3 (e x)^m \sin \left (c+d x^2\right )-\frac {1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^2\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^2\right )+\left (3 a^2 b+\frac {3 b^3}{4}\right ) (e x)^m \sin \left (c+d x^2\right )-\frac {1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^2\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^2\right ) \, dx-\frac {1}{4} b^3 \int (e x)^m \sin \left (3 c+3 d x^2\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^2\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^2} (e x)^m \, dx-\frac {1}{4} \left (3 a b^2\right ) \int e^{2 i c+2 i d x^2} (e x)^m \, dx-\frac {1}{8} \left (i b^3\right ) \int e^{-3 i c-3 i d x^2} (e x)^m \, dx+\frac {1}{8} \left (i b^3\right ) \int e^{3 i c+3 i d x^2} (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^2} (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^2} (e x)^m \, dx\\ &=\frac {a \left (2 a^2+3 b^2\right ) (e x)^{1+m}}{2 e (1+m)}+\frac {3 i b \left (4 a^2+b^2\right ) e^{i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-i d x^2\right )}{16 e}-\frac {3 i b \left (4 a^2+b^2\right ) e^{-i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},i d x^2\right )}{16 e}+\frac {3\ 2^{-\frac {7}{2}-\frac {m}{2}} a b^2 e^{2 i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-2 i d x^2\right )}{e}+\frac {3\ 2^{-\frac {7}{2}-\frac {m}{2}} a b^2 e^{-2 i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},2 i d x^2\right )}{e}-\frac {i 3^{-\frac {1}{2}-\frac {m}{2}} b^3 e^{3 i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-3 i d x^2\right )}{16 e}+\frac {i 3^{-\frac {1}{2}-\frac {m}{2}} b^3 e^{-3 i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},3 i d x^2\right )}{16 e}\\ \end {align*}
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Mathematica [A]
time = 8.33, size = 451, normalized size = 1.02 \begin {gather*} \frac {1}{16} i x (e x)^m \left (3 b \left (4 a^2+b^2\right ) e^{i c} \left (-i d x^2\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},-i d x^2\right )+\frac {-16 i a^3 d x^2-24 i a b^2 d x^2+12 i a^2 b e^{-i c} (1+m) \left (i d x^2\right )^{\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},i d x^2\right )+3 i b^3 e^{-i c} (1+m) \left (i d x^2\right )^{\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},i d x^2\right )+3\ 2^{\frac {1}{2}-\frac {m}{2}} a b^2 e^{2 i c} (1+m) \left (-i d x^2\right )^{\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},-2 i d x^2\right )-3\ 2^{\frac {1}{2}-\frac {m}{2}} a b^2 e^{-2 i c} (1+m) \left (i d x^2\right )^{\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},2 i d x^2\right )-i 3^{-\frac {1}{2}-\frac {m}{2}} b^3 e^{3 i c} (1+m) \left (-i d x^2\right )^{\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},-3 i d x^2\right )-i 3^{-\frac {1}{2}-\frac {m}{2}} b^3 e^{-3 i c} (1+m) \left (i d x^2\right )^{\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},3 i d x^2\right )}{d (1+m) x^2}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.18, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (a +b \sin \left (d \,x^{2}+c \right )\right )^{3}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.14, size = 302, normalized size = 0.68 \begin {gather*} \frac {24 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \left (x e\right )^{m} d x + {\left (b^{3} m + b^{3}\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (3 i \, d e^{\left (-2\right )}\right ) - 3 i \, c + 1\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 3 i \, d x^{2}\right ) - 9 i \, {\left (a b^{2} m + a b^{2}\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (2 i \, d e^{\left (-2\right )}\right ) - 2 i \, c + 1\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 2 i \, d x^{2}\right ) - 9 \, {\left (4 \, a^{2} b + b^{3} + {\left (4 \, a^{2} b + b^{3}\right )} m\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (i \, d e^{\left (-2\right )}\right ) - i \, c + 1\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, i \, d x^{2}\right ) - 9 \, {\left (4 \, a^{2} b + b^{3} + {\left (4 \, a^{2} b + b^{3}\right )} m\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-i \, d e^{\left (-2\right )}\right ) + i \, c + 1\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -i \, d x^{2}\right ) + 9 i \, {\left (a b^{2} m + a b^{2}\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-2 i \, d e^{\left (-2\right )}\right ) + 2 i \, c + 1\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -2 i \, d x^{2}\right ) + {\left (b^{3} m + b^{3}\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-3 i \, d e^{\left (-2\right )}\right ) + 3 i \, c + 1\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -3 i \, d x^{2}\right )}{48 \, {\left (d m + d\right )}} \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 (e x\right )^{m} \left (a + b \sin {\left (c + d x^{2} \right )}\right )^{3}\, 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 = -0.00 \begin {gather*} \int {\left (e\,x\right )}^m\,{\left (a+b\,\sin \left (d\,x^2+c\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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