Optimal. Leaf size=449 \[ -\frac{15 a^3 e^{i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i f (c+d x)}{d}\right )}{8 f}+\frac{3 i a^3 2^{-m-3} e^{2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 i f (c+d x)}{d}\right )}{f}+\frac{a^3 3^{-m-1} e^{3 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{3 i f (c+d x)}{d}\right )}{8 f}-\frac{15 a^3 e^{-i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{i f (c+d x)}{d}\right )}{8 f}-\frac{3 i a^3 2^{-m-3} e^{-2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i f (c+d x)}{d}\right )}{f}+\frac{a^3 3^{-m-1} e^{-3 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{3 i f (c+d x)}{d}\right )}{8 f}+\frac{5 a^3 (c+d x)^{m+1}}{2 d (m+1)} \]
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Rubi [A] time = 0.605078, antiderivative size = 449, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3318, 3312, 3307, 2181, 3308} \[ -\frac{15 a^3 e^{i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i f (c+d x)}{d}\right )}{8 f}+\frac{3 i a^3 2^{-m-3} e^{2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 i f (c+d x)}{d}\right )}{f}+\frac{a^3 3^{-m-1} e^{3 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{3 i f (c+d x)}{d}\right )}{8 f}-\frac{15 a^3 e^{-i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{i f (c+d x)}{d}\right )}{8 f}-\frac{3 i a^3 2^{-m-3} e^{-2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i f (c+d x)}{d}\right )}{f}+\frac{a^3 3^{-m-1} e^{-3 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{3 i f (c+d x)}{d}\right )}{8 f}+\frac{5 a^3 (c+d x)^{m+1}}{2 d (m+1)} \]
Antiderivative was successfully verified.
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Rule 3318
Rule 3312
Rule 3307
Rule 2181
Rule 3308
Rubi steps
\begin{align*} \int (c+d x)^m (a+a \sin (e+f x))^3 \, dx &=\left (8 a^3\right ) \int (c+d x)^m \sin ^6\left (\frac{1}{2} \left (e+\frac{\pi }{2}\right )+\frac{f x}{2}\right ) \, dx\\ &=\left (8 a^3\right ) \int \left (\frac{5}{16} (c+d x)^m-\frac{3}{16} (c+d x)^m \cos (2 e+2 f x)+\frac{15}{32} (c+d x)^m \sin (e+f x)-\frac{1}{32} (c+d x)^m \sin (3 e+3 f x)\right ) \, dx\\ &=\frac{5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac{1}{4} a^3 \int (c+d x)^m \sin (3 e+3 f x) \, dx-\frac{1}{2} \left (3 a^3\right ) \int (c+d x)^m \cos (2 e+2 f x) \, dx+\frac{1}{4} \left (15 a^3\right ) \int (c+d x)^m \sin (e+f x) \, dx\\ &=\frac{5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac{1}{8} \left (i a^3\right ) \int e^{-i (3 e+3 f x)} (c+d x)^m \, dx+\frac{1}{8} \left (i a^3\right ) \int e^{i (3 e+3 f x)} (c+d x)^m \, dx+\frac{1}{8} \left (15 i a^3\right ) \int e^{-i (e+f x)} (c+d x)^m \, dx-\frac{1}{8} \left (15 i a^3\right ) \int e^{i (e+f x)} (c+d x)^m \, dx-\frac{1}{4} \left (3 a^3\right ) \int e^{-i (2 e+2 f x)} (c+d x)^m \, dx-\frac{1}{4} \left (3 a^3\right ) \int e^{i (2 e+2 f x)} (c+d x)^m \, dx\\ &=\frac{5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac{15 a^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{15 a^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^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 (1+m,-\frac{2 i f (c+d x)}{d}\right )}{f}-\frac{3 i 2^{-3-m} a^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 (1+m,\frac{2 i f (c+d x)}{d}\right )}{f}+\frac{3^{-1-m} a^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} a^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*}
Mathematica [A] time = 0.844823, size = 376, normalized size = 0.84 \[ \frac{1}{24} a^3 (c+d x)^m \left (-\frac{45 e^{i \left (e-\frac{c f}{d}\right )} \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i f (c+d x)}{d}\right )}{f}+\frac{9 i 2^{-m} e^{2 i \left (e-\frac{c f}{d}\right )} \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 i f (c+d x)}{d}\right )}{f}+\frac{3^{-m} e^{3 i \left (e-\frac{c f}{d}\right )} \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{3 i f (c+d x)}{d}\right )}{f}-\frac{45 e^{-i \left (e-\frac{c f}{d}\right )} \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{i f (c+d x)}{d}\right )}{f}-\frac{9 i 2^{-m} e^{-2 i \left (e-\frac{c f}{d}\right )} \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i f (c+d x)}{d}\right )}{f}+\frac{3^{-m} e^{-3 i \left (e-\frac{c f}{d}\right )} \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{3 i f (c+d x)}{d}\right )}{f}+\frac{60 (c+d x)}{d (m+1)}\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 0.279, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{m} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.1351, size = 934, normalized size = 2.08 \begin{align*} \frac{{\left (a^{3} d m + a^{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^{3} d m - 9 i \, a^{3} 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 ) - 45 \,{\left (a^{3} d m + a^{3} 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 ) - 45 \,{\left (a^{3} d m + a^{3} 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^{3} d m + 9 i \, a^{3} 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 (a^{3} d m + a^{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 ) + 60 \,{\left (a^{3} d f x + a^{3} c f\right )}{\left (d x + c\right )}^{m}}{24 \,{\left (d f m + d f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{3}{\left (d x + c\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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