Integrand size = 28, antiderivative size = 277 \[ \int \frac {(e+f x)^m \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {i e^{i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (-\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,-\frac {i d (e+f x)}{f}\right )}{2 a d}+\frac {i e^{-i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,\frac {i d (e+f x)}{f}\right )}{2 a d}+\frac {2^{-3-m} e^{2 i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (-\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,-\frac {2 i d (e+f x)}{f}\right )}{a d}+\frac {2^{-3-m} e^{-2 i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,\frac {2 i d (e+f x)}{f}\right )}{a d} \]
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Time = 0.21 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4619, 3388, 2212, 4491, 12, 3389} \[ \int \frac {(e+f x)^m \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {i e^{i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (-\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (m+1,-\frac {i d (e+f x)}{f}\right )}{2 a d}+\frac {2^{-m-3} e^{2 i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (-\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (m+1,-\frac {2 i d (e+f x)}{f}\right )}{a d}+\frac {i e^{-i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (m+1,\frac {i d (e+f x)}{f}\right )}{2 a d}+\frac {2^{-m-3} e^{-2 i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (m+1,\frac {2 i d (e+f x)}{f}\right )}{a d} \]
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Rule 12
Rule 2212
Rule 3388
Rule 3389
Rule 4491
Rule 4619
Rubi steps \begin{align*} \text {integral}& = \frac {\int (e+f x)^m \cos (c+d x) \, dx}{a}-\frac {\int (e+f x)^m \cos (c+d x) \sin (c+d x) \, dx}{a} \\ & = \frac {\int e^{-i (c+d x)} (e+f x)^m \, dx}{2 a}+\frac {\int e^{i (c+d x)} (e+f x)^m \, dx}{2 a}-\frac {\int \frac {1}{2} (e+f x)^m \sin (2 c+2 d x) \, dx}{a} \\ & = -\frac {i e^{i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (-\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,-\frac {i d (e+f x)}{f}\right )}{2 a d}+\frac {i e^{-i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,\frac {i d (e+f x)}{f}\right )}{2 a d}-\frac {\int (e+f x)^m \sin (2 c+2 d x) \, dx}{2 a} \\ & = -\frac {i e^{i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (-\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,-\frac {i d (e+f x)}{f}\right )}{2 a d}+\frac {i e^{-i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,\frac {i d (e+f x)}{f}\right )}{2 a d}-\frac {i \int e^{-i (2 c+2 d x)} (e+f x)^m \, dx}{4 a}+\frac {i \int e^{i (2 c+2 d x)} (e+f x)^m \, dx}{4 a} \\ & = -\frac {i e^{i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (-\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,-\frac {i d (e+f x)}{f}\right )}{2 a d}+\frac {i e^{-i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,\frac {i d (e+f x)}{f}\right )}{2 a d}+\frac {2^{-3-m} e^{2 i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (-\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,-\frac {2 i d (e+f x)}{f}\right )}{a d}+\frac {2^{-3-m} e^{-2 i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,\frac {2 i d (e+f x)}{f}\right )}{a d} \\ \end{align*}
Time = 8.21 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.91 \[ \int \frac {(e+f x)^m \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2^{-3-m} e^{-\frac {2 i (d e+c f)}{f}} (e+f x)^m \left (\frac {d^2 (e+f x)^2}{f^2}\right )^{-m} \left (-i 2^{2+m} e^{i \left (3 c+\frac {d e}{f}\right )} \left (\frac {i d (e+f x)}{f}\right )^m \Gamma \left (1+m,-\frac {i d (e+f x)}{f}\right )+i 2^{2+m} e^{i \left (c+\frac {3 d e}{f}\right )} \left (-\frac {i d (e+f x)}{f}\right )^m \Gamma \left (1+m,\frac {i d (e+f x)}{f}\right )+e^{4 i c} \left (\frac {i d (e+f x)}{f}\right )^m \Gamma \left (1+m,-\frac {2 i d (e+f x)}{f}\right )+e^{\frac {4 i d e}{f}} \left (-\frac {i d (e+f x)}{f}\right )^m \Gamma \left (1+m,\frac {2 i d (e+f x)}{f}\right )\right )}{a d} \]
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\[\int \frac {\left (f x +e \right )^{m} \left (\cos ^{3}\left (d x +c \right )\right )}{a +a \sin \left (d x +c \right )}d x\]
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Time = 0.11 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.68 \[ \int \frac {(e+f x)^m \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {4 i \, e^{\left (-\frac {f m \log \left (\frac {i \, d}{f}\right ) - i \, d e + i \, c f}{f}\right )} \Gamma \left (m + 1, \frac {i \, d f x + i \, d e}{f}\right ) + e^{\left (-\frac {f m \log \left (-\frac {2 i \, d}{f}\right ) + 2 i \, d e - 2 i \, c f}{f}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (i \, d f x + i \, d e\right )}}{f}\right ) - 4 i \, e^{\left (-\frac {f m \log \left (-\frac {i \, d}{f}\right ) + i \, d e - i \, c f}{f}\right )} \Gamma \left (m + 1, \frac {-i \, d f x - i \, d e}{f}\right ) + e^{\left (-\frac {f m \log \left (\frac {2 i \, d}{f}\right ) - 2 i \, d e + 2 i \, c f}{f}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, d f x - i \, d e\right )}}{f}\right )}{8 \, a d} \]
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Timed out. \[ \int \frac {(e+f x)^m \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {(e+f x)^m \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{m} \cos \left (d x + c\right )^{3}}{a \sin \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {(e+f x)^m \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{m} \cos \left (d x + c\right )^{3}}{a \sin \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {(e+f x)^m \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,{\left (e+f\,x\right )}^m}{a+a\,\sin \left (c+d\,x\right )} \,d x \]
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