Integrand size = 24, antiderivative size = 407 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=-\frac {e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i b (c+d x)}{d}\right )}{16 b}-\frac {e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i b (c+d x)}{d}\right )}{16 b}-\frac {3^{-1-m} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 i b (c+d x)}{d}\right )}{32 b}-\frac {3^{-1-m} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 i b (c+d x)}{d}\right )}{32 b}+\frac {5^{-1-m} e^{5 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {5 i b (c+d x)}{d}\right )}{32 b}+\frac {5^{-1-m} e^{-5 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {5 i b (c+d x)}{d}\right )}{32 b} \]
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Time = 0.46 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4491, 3389, 2212} \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=-\frac {e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i b (c+d x)}{d}\right )}{16 b}-\frac {3^{-m-1} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {3 i b (c+d x)}{d}\right )}{32 b}+\frac {5^{-m-1} e^{5 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {5 i b (c+d x)}{d}\right )}{32 b}-\frac {e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i b (c+d x)}{d}\right )}{16 b}-\frac {3^{-m-1} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {3 i b (c+d x)}{d}\right )}{32 b}+\frac {5^{-m-1} e^{-5 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {5 i b (c+d x)}{d}\right )}{32 b} \]
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Rule 2212
Rule 3389
Rule 4491
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{8} (c+d x)^m \sin (a+b x)+\frac {1}{16} (c+d x)^m \sin (3 a+3 b x)-\frac {1}{16} (c+d x)^m \sin (5 a+5 b x)\right ) \, dx \\ & = \frac {1}{16} \int (c+d x)^m \sin (3 a+3 b x) \, dx-\frac {1}{16} \int (c+d x)^m \sin (5 a+5 b x) \, dx+\frac {1}{8} \int (c+d x)^m \sin (a+b x) \, dx \\ & = \frac {1}{32} i \int e^{-i (3 a+3 b x)} (c+d x)^m \, dx-\frac {1}{32} i \int e^{i (3 a+3 b x)} (c+d x)^m \, dx-\frac {1}{32} i \int e^{-i (5 a+5 b x)} (c+d x)^m \, dx+\frac {1}{32} i \int e^{i (5 a+5 b x)} (c+d x)^m \, dx+\frac {1}{16} i \int e^{-i (a+b x)} (c+d x)^m \, dx-\frac {1}{16} i \int e^{i (a+b x)} (c+d x)^m \, dx \\ & = -\frac {e^{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 {i b (c+d x)}{d}\right )}{16 b}-\frac {e^{-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 {i b (c+d x)}{d}\right )}{16 b}-\frac {3^{-1-m} e^{3 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 {3 i b (c+d x)}{d}\right )}{32 b}-\frac {3^{-1-m} e^{-3 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 {3 i b (c+d x)}{d}\right )}{32 b}+\frac {5^{-1-m} e^{5 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 {5 i b (c+d x)}{d}\right )}{32 b}+\frac {5^{-1-m} e^{-5 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 {5 i b (c+d x)}{d}\right )}{32 b} \\ \end{align*}
Time = 1.04 (sec) , antiderivative size = 376, normalized size of antiderivative = 0.92 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {e^{-\frac {5 i (b c+a d)}{d}} (c+d x)^m \left (30 e^{\frac {4 i (b c+a d)}{d}} \left (-e^{2 i a} \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i b (c+d x)}{d}\right )-e^{\frac {2 i b c}{d}} \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i b (c+d x)}{d}\right )\right )-5\ 3^{-m} e^{\frac {2 i (b c+a d)}{d}} \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (e^{6 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {3 i b (c+d x)}{d}\right )+e^{\frac {6 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {3 i b (c+d x)}{d}\right )\right )+3\ 5^{-m} \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (e^{10 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {5 i b (c+d x)}{d}\right )+e^{\frac {10 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {5 i b (c+d x)}{d}\right )\right )\right )}{480 b} \]
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\[\int \left (d x +c \right )^{m} \cos \left (x b +a \right )^{2} \sin \left (x b +a \right )^{3}d x\]
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none
Time = 0.10 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.69 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=-\frac {30 \, e^{\left (-\frac {d m \log \left (\frac {i \, b}{d}\right ) - i \, b c + i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {i \, b d x + i \, b c}{d}\right ) + 5 \, e^{\left (-\frac {d m \log \left (-\frac {3 i \, b}{d}\right ) + 3 i \, b c - 3 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) - 3 \, e^{\left (-\frac {d m \log \left (-\frac {5 i \, b}{d}\right ) + 5 i \, b c - 5 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {5 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + 30 \, e^{\left (-\frac {d m \log \left (-\frac {i \, b}{d}\right ) + i \, b c - i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {-i \, b d x - i \, b c}{d}\right ) + 5 \, e^{\left (-\frac {d m \log \left (\frac {3 i \, b}{d}\right ) - 3 i \, b c + 3 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) - 3 \, e^{\left (-\frac {d m \log \left (\frac {5 i \, b}{d}\right ) - 5 i \, b c + 5 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {5 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )}{480 \, b} \]
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Exception generated. \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )^{3} \,d x } \]
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\[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )^{3} \,d x } \]
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Timed out. \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\int {\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^m \,d x \]
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