Integrand size = 20, antiderivative size = 299 \[ \int (c+d x)^m (a+a \sin (e+f x))^2 \, dx=\frac {3 a^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {a^2 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{f}-\frac {a^2 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{f}+\frac {i 2^{-3-m} a^2 e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {i 2^{-3-m} a^2 e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{f} \]
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Time = 0.23 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3399, 3393, 3388, 2212, 3389} \[ \int (c+d x)^m (a+a \sin (e+f x))^2 \, dx=-\frac {a^2 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i f (c+d x)}{d}\right )}{f}+\frac {i a^2 2^{-m-3} e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {a^2 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i f (c+d x)}{d}\right )}{f}-\frac {i a^2 2^{-m-3} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {3 a^2 (c+d x)^{m+1}}{2 d (m+1)} \]
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Rule 2212
Rule 3388
Rule 3389
Rule 3393
Rule 3399
Rubi steps \begin{align*} \text {integral}& = \left (4 a^2\right ) \int (c+d x)^m \sin ^4\left (\frac {1}{2} \left (e+\frac {\pi }{2}\right )+\frac {f x}{2}\right ) \, dx \\ & = \left (4 a^2\right ) \int \left (\frac {3}{8} (c+d x)^m-\frac {1}{8} (c+d x)^m \cos (2 e+2 f x)+\frac {1}{2} (c+d x)^m \sin (e+f x)\right ) \, dx \\ & = \frac {3 a^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {1}{2} a^2 \int (c+d x)^m \cos (2 e+2 f x) \, dx+\left (2 a^2\right ) \int (c+d x)^m \sin (e+f x) \, dx \\ & = \frac {3 a^2 (c+d x)^{1+m}}{2 d (1+m)}+\left (i a^2\right ) \int e^{-i (e+f x)} (c+d x)^m \, dx-\left (i a^2\right ) \int e^{i (e+f x)} (c+d x)^m \, dx-\frac {1}{4} a^2 \int e^{-i (2 e+2 f x)} (c+d x)^m \, dx-\frac {1}{4} a^2 \int e^{i (2 e+2 f x)} (c+d x)^m \, dx \\ & = \frac {3 a^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {a^2 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{f}-\frac {a^2 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{f}+\frac {i 2^{-3-m} a^2 e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {i 2^{-3-m} a^2 e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{f} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.87 \[ \int (c+d x)^m (a+a \sin (e+f x))^2 \, dx=\frac {1}{8} a^2 (c+d x)^m \left (\frac {12 (c+d x)}{d (1+m)}-\frac {8 e^{i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{f}-\frac {8 e^{-i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{f}+\frac {i 2^{-m} e^{2 i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {i 2^{-m} e^{-2 i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{f}\right ) \]
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\[\int \left (d x +c \right )^{m} \left (a +a \sin \left (f x +e \right )\right )^{2}d x\]
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none
Time = 0.12 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.90 \[ \int (c+d x)^m (a+a \sin (e+f x))^2 \, dx=-\frac {8 \, {\left (a^{2} d m + a^{2} d\right )} e^{\left (-\frac {d m \log \left (\frac {i \, f}{d}\right ) + i \, d e - i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {i \, d f x + i \, c f}{d}\right ) - {\left (i \, a^{2} d m + i \, a^{2} d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 i \, f}{d}\right ) - 2 i \, d e + 2 i \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) + 8 \, {\left (a^{2} d m + a^{2} d\right )} e^{\left (-\frac {d m \log \left (-\frac {i \, f}{d}\right ) - i \, d e + i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-i \, d f x - i \, c f}{d}\right ) - {\left (-i \, a^{2} d m - i \, a^{2} d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) + 2 i \, d e - 2 i \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) - 12 \, {\left (a^{2} d f x + a^{2} c f\right )} {\left (d x + c\right )}^{m}}{8 \, {\left (d f m + d f\right )}} \]
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\[ \int (c+d x)^m (a+a \sin (e+f x))^2 \, dx=a^{2} \left (\int 2 \left (c + d x\right )^{m} \sin {\left (e + f x \right )}\, dx + \int \left (c + d x\right )^{m} \sin ^{2}{\left (e + f x \right )}\, dx + \int \left (c + d x\right )^{m}\, dx\right ) \]
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\[ \int (c+d x)^m (a+a \sin (e+f x))^2 \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (d x + c\right )}^{m} \,d x } \]
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\[ \int (c+d x)^m (a+a \sin (e+f x))^2 \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (d x + c\right )}^{m} \,d x } \]
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Timed out. \[ \int (c+d x)^m (a+a \sin (e+f x))^2 \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,{\left (c+d\,x\right )}^m \,d x \]
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