Optimal. Leaf size=273 \[ -\frac{2^{-m-4} e^{2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 i b (c+d x)}{d}\right )}{b}-\frac{2^{-2 (m+3)} e^{4 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{4 i b (c+d x)}{d}\right )}{b}-\frac{2^{-m-4} e^{-2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i b (c+d x)}{d}\right )}{b}-\frac{2^{-2 (m+3)} e^{-4 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{4 i b (c+d x)}{d}\right )}{b} \]
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Rubi [A] time = 0.288119, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4406, 3308, 2181} \[ -\frac{2^{-m-4} e^{2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 i b (c+d x)}{d}\right )}{b}-\frac{2^{-2 (m+3)} e^{4 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{4 i b (c+d x)}{d}\right )}{b}-\frac{2^{-m-4} e^{-2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i b (c+d x)}{d}\right )}{b}-\frac{2^{-2 (m+3)} e^{-4 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{4 i b (c+d x)}{d}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 4406
Rule 3308
Rule 2181
Rubi steps
\begin{align*} \int (c+d x)^m \cos ^3(a+b x) \sin (a+b x) \, dx &=\int \left (\frac{1}{4} (c+d x)^m \sin (2 a+2 b x)+\frac{1}{8} (c+d x)^m \sin (4 a+4 b x)\right ) \, dx\\ &=\frac{1}{8} \int (c+d x)^m \sin (4 a+4 b x) \, dx+\frac{1}{4} \int (c+d x)^m \sin (2 a+2 b x) \, dx\\ &=\frac{1}{16} i \int e^{-i (4 a+4 b x)} (c+d x)^m \, dx-\frac{1}{16} i \int e^{i (4 a+4 b x)} (c+d x)^m \, dx+\frac{1}{8} i \int e^{-i (2 a+2 b x)} (c+d x)^m \, dx-\frac{1}{8} i \int e^{i (2 a+2 b x)} (c+d x)^m \, dx\\ &=-\frac{2^{-4-m} e^{2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac{2 i b (c+d x)}{d}\right )}{b}-\frac{2^{-4-m} e^{-2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac{2 i b (c+d x)}{d}\right )}{b}-\frac{4^{-3-m} e^{4 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac{4 i b (c+d x)}{d}\right )}{b}-\frac{4^{-3-m} e^{-4 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac{4 i b (c+d x)}{d}\right )}{b}\\ \end{align*}
Mathematica [A] time = 0.247976, size = 245, normalized size = 0.9 \[ -\frac{4^{-m-3} e^{-\frac{4 i (a d+b c)}{d}} (c+d x)^m \left (\frac{b^2 (c+d x)^2}{d^2}\right )^{-m} \left (2^{m+2} e^{2 i \left (a+\frac{3 b c}{d}\right )} \left (-\frac{i b (c+d x)}{d}\right )^m \text{Gamma}\left (m+1,\frac{2 i b (c+d x)}{d}\right )+2^{m+2} e^{2 i \left (3 a+\frac{b c}{d}\right )} \left (\frac{i b (c+d x)}{d}\right )^m \text{Gamma}\left (m+1,-\frac{2 i b (c+d x)}{d}\right )+e^{8 i a} \left (\frac{i b (c+d x)}{d}\right )^m \text{Gamma}\left (m+1,-\frac{4 i b (c+d x)}{d}\right )+e^{\frac{8 i b c}{d}} \left (-\frac{i b (c+d x)}{d}\right )^m \text{Gamma}\left (m+1,\frac{4 i b (c+d x)}{d}\right )\right )}{b} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.266, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{m} \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sin \left ( bx+a \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{3} \sin \left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.54418, size = 487, normalized size = 1.78 \begin{align*} -\frac{e^{\left (-\frac{d m \log \left (\frac{4 i \, b}{d}\right ) - 4 i \, b c + 4 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac{4 i \, b d x + 4 i \, b c}{d}\right ) + 4 \, e^{\left (-\frac{d m \log \left (\frac{2 i \, b}{d}\right ) - 2 i \, b c + 2 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac{2 i \, b d x + 2 i \, b c}{d}\right ) + 4 \, e^{\left (-\frac{d m \log \left (-\frac{2 i \, b}{d}\right ) + 2 i \, b c - 2 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac{-2 i \, b d x - 2 i \, b c}{d}\right ) + e^{\left (-\frac{d m \log \left (-\frac{4 i \, b}{d}\right ) + 4 i \, b c - 4 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac{-4 i \, b d x - 4 i \, b c}{d}\right )}{64 \, b} \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 (d x + c\right )}^{m} \cos \left (b x + a\right )^{3} \sin \left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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