Integrand size = 22, antiderivative size = 273 \[ \int (c+d x)^m \cos ^3(a+b x) \sin (a+b x) \, dx=-\frac {2^{-4-m} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {2^{-4-m} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {2^{-2 (3+m)} e^{4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {4 i b (c+d x)}{d}\right )}{b}-\frac {2^{-2 (3+m)} e^{-4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 i b (c+d x)}{d}\right )}{b} \]
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Time = 0.35 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4491, 3389, 2212} \[ \int (c+d x)^m \cos ^3(a+b x) \sin (a+b x) \, dx=-\frac {2^{-m-4} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {2^{-2 (m+3)} e^{4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {4 i b (c+d x)}{d}\right )}{b}-\frac {2^{-m-4} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {2^{-2 (m+3)} e^{-4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {4 i b (c+d x)}{d}\right )}{b} \]
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Rule 2212
Rule 3389
Rule 4491
Rubi steps \begin{align*} \text {integral}& = \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*}
Time = 0.28 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.90 \[ \int (c+d x)^m \cos ^3(a+b x) \sin (a+b x) \, dx=-\frac {4^{-3-m} e^{-\frac {4 i (b c+a d)}{d}} (c+d x)^m \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (2^{2+m} e^{2 i \left (3 a+\frac {b c}{d}\right )} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )+2^{2+m} e^{2 i \left (a+\frac {3 b c}{d}\right )} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )+e^{8 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {4 i b (c+d x)}{d}\right )+e^{\frac {8 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {4 i b (c+d x)}{d}\right )\right )}{b} \]
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\[\int \left (d x +c \right )^{m} \cos \left (x b +a \right )^{3} \sin \left (x b +a \right )d x\]
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none
Time = 0.10 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.69 \[ \int (c+d x)^m \cos ^3(a+b x) \sin (a+b x) \, dx=-\frac {4 \, e^{\left (-\frac {d m \log \left (-\frac {2 i \, b}{d}\right ) + 2 i \, b c - 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + e^{\left (-\frac {d m \log \left (-\frac {4 i \, b}{d}\right ) + 4 i \, b c - 4 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {4 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + 4 \, e^{\left (-\frac {d m \log \left (\frac {2 i \, b}{d}\right ) - 2 i \, b c + 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) + e^{\left (-\frac {d m \log \left (\frac {4 i \, b}{d}\right ) - 4 i \, b c + 4 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {4 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )}{64 \, b} \]
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Exception generated. \[ \int (c+d x)^m \cos ^3(a+b x) \sin (a+b x) \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int (c+d x)^m \cos ^3(a+b x) \sin (a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{3} \sin \left (b x + a\right ) \,d x } \]
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\[ \int (c+d x)^m \cos ^3(a+b x) \sin (a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{3} \sin \left (b x + a\right ) \,d x } \]
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Timed out. \[ \int (c+d x)^m \cos ^3(a+b x) \sin (a+b x) \, dx=\int {\cos \left (a+b\,x\right )}^3\,\sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^m \,d x \]
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