Integrand size = 16, antiderivative size = 162 \[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {i 2^{-3-m} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-3-m} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )}{b} \]
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Time = 0.14 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3393, 3388, 2212} \[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\frac {i 2^{-m-3} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-m-3} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {(c+d x)^{m+1}}{2 d (m+1)} \]
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Rule 2212
Rule 3388
Rule 3393
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2} (c+d x)^m-\frac {1}{2} (c+d x)^m \cos (2 a+2 b x)\right ) \, dx \\ & = \frac {(c+d x)^{1+m}}{2 d (1+m)}-\frac {1}{2} \int (c+d x)^m \cos (2 a+2 b x) \, dx \\ & = \frac {(c+d x)^{1+m}}{2 d (1+m)}-\frac {1}{4} \int e^{-i (2 a+2 b x)} (c+d x)^m \, dx-\frac {1}{4} \int e^{i (2 a+2 b x)} (c+d x)^m \, dx \\ & = \frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {i 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 {i 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 = 0.50 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.93 \[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\frac {1}{8} (c+d x)^m \left (\frac {4 c+4 d x}{d+d m}+\frac {i 2^{-m} e^{2 i \left (a-\frac {b c}{d}\right )} \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-m} e^{-2 i \left (a-\frac {b c}{d}\right )} \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )}{b}\right ) \]
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\[\int \left (d x +c \right )^{m} \left (\sin ^{2}\left (b x +a \right )\right )d x\]
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none
Time = 0.10 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.84 \[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\frac {{\left (i \, d m + i \, d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 i \, b}{d}\right ) + 2 i \, b c - 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + {\left (-i \, d m - i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, b}{d}\right ) - 2 i \, b c + 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) + 4 \, {\left (b d x + b c\right )} {\left (d x + c\right )}^{m}}{8 \, {\left (b d m + b d\right )}} \]
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\[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\int \left (c + d x\right )^{m} \sin ^{2}{\left (a + b x \right )}\, dx \]
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\[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \sin \left (b x + a\right )^{2} \,d x } \]
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\[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \sin \left (b x + a\right )^{2} \,d x } \]
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Timed out. \[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^m \,d x \]
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