Integrand size = 24, antiderivative size = 162 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {(c+d x)^{1+m}}{8 d (1+m)}+\frac {i 2^{-2 (3+m)} e^{4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {4 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-2 (3+m)} e^{-4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 i b (c+d x)}{d}\right )}{b} \]
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Time = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4491, 3388, 2212} \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {i 2^{-2 (m+3)} e^{4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {4 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-2 (m+3)} e^{-4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {4 i b (c+d x)}{d}\right )}{b}+\frac {(c+d x)^{m+1}}{8 d (m+1)} \]
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Rule 2212
Rule 3388
Rule 4491
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{8} (c+d x)^m-\frac {1}{8} (c+d x)^m \cos (4 a+4 b x)\right ) \, dx \\ & = \frac {(c+d x)^{1+m}}{8 d (1+m)}-\frac {1}{8} \int (c+d x)^m \cos (4 a+4 b x) \, dx \\ & = \frac {(c+d x)^{1+m}}{8 d (1+m)}-\frac {1}{16} \int e^{-i (4 a+4 b x)} (c+d x)^m \, dx-\frac {1}{16} \int e^{i (4 a+4 b x)} (c+d x)^m \, dx \\ & = \frac {(c+d x)^{1+m}}{8 d (1+m)}+\frac {i 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 {i 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 = 3.78 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.96 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=-\frac {(c+d x)^m \left (-8 b (c+d x)-i 4^{-m} d e^{4 i \left (a-\frac {b c}{d}\right )} (1+m) \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {4 i b (c+d x)}{d}\right )+i 4^{-m} d e^{-4 i \left (a-\frac {b c}{d}\right )} (1+m) \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 i b (c+d x)}{d}\right )\right )}{64 b d (1+m)} \]
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\[\int \left (d x +c \right )^{m} \cos \left (x b +a \right )^{2} \sin \left (x b +a \right )^{2}d x\]
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none
Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.84 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {{\left (i \, d m + i \, d\right )} e^{\left (-\frac {d m \log \left (-\frac {4 i \, b}{d}\right ) + 4 i \, b c - 4 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {4 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + {\left (-i \, d m - i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {4 i \, b}{d}\right ) - 4 i \, b c + 4 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {4 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) + 8 \, {\left (b d x + b c\right )} {\left (d x + c\right )}^{m}}{64 \, {\left (b d m + b d\right )}} \]
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Exception generated. \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )^{2} \,d x } \]
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\[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )^{2} \,d x } \]
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Timed out. \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\int {\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^m \,d x \]
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