Integrand size = 20, antiderivative size = 137 \[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=-\frac {2^{-3-m} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {2^{-3-m} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )}{b} \]
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Time = 0.18 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4491, 12, 3389, 2212} \[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=-\frac {2^{-m-3} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {2^{-m-3} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i b (c+d x)}{d}\right )}{b} \]
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Rule 12
Rule 2212
Rule 3389
Rule 4491
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{2} (c+d x)^m \sin (2 a+2 b x) \, dx \\ & = \frac {1}{2} \int (c+d x)^m \sin (2 a+2 b x) \, dx \\ & = \frac {1}{4} i \int e^{-i (2 a+2 b x)} (c+d x)^m \, dx-\frac {1}{4} i \int e^{i (2 a+2 b x)} (c+d x)^m \, dx \\ & = -\frac {2^{-3-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^{-3-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} \\ \end{align*}
Time = 0.41 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.01 \[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=-\frac {2^{-3-m} e^{-\frac {2 i (b c+a d)}{d}} (c+d x)^m \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (e^{4 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )+e^{\frac {4 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )\right )}{b} \]
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\[\int \left (d x +c \right )^{m} \cos \left (x b +a \right ) \sin \left (x b +a \right )d x\]
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none
Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.70 \[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=-\frac {e^{\left (-\frac {d m \log \left (-\frac {2 i \, b}{d}\right ) + 2 i \, b c - 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + e^{\left (-\frac {d m \log \left (\frac {2 i \, b}{d}\right ) - 2 i \, b c + 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )}{8 \, b} \]
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\[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=\int \left (c + d x\right )^{m} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}\, dx \]
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\[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right ) \sin \left (b x + a\right ) \,d x } \]
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\[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right ) \sin \left (b x + a\right ) \,d x } \]
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Timed out. \[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=\int \cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^m \,d x \]
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