Integrand size = 21, antiderivative size = 139 \[ \int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx=-\frac {e^{-i d} F^{a c} (f x)^m \Gamma (1+m,x (i e-b c \log (F))) (x (i e-b c \log (F)))^{-m}}{2 (e+i b c \log (F))}-\frac {e^{i d} F^{a c} (f x)^m \Gamma (1+m,-x (i e+b c \log (F))) (-x (i e+b c \log (F)))^{-m}}{2 (e-i b c \log (F))} \]
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\[ \int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx=\int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int F^{a c+b c x} (f x)^m \sin (d+e x) \, dx \\ \end{align*}
Time = 0.59 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.03 \[ \int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx=\frac {1}{2} F^{a c} (f x)^m (x (-i e-b c \log (F)))^{-m} \left (-i x \Gamma (1+m,i e x-b c x \log (F)) (i x (e+i b c \log (F)))^{-1-m} (-i e x-b c x \log (F))^m (\cos (d)-i \sin (d))-\frac {\Gamma (1+m,-i e x-b c x \log (F)) (\cos (d)+i \sin (d))}{e-i b c \log (F)}\right ) \]
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\[\int F^{c \left (x b +a \right )} \left (f x \right )^{m} \sin \left (e x +d \right )d x\]
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Time = 0.10 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.94 \[ \int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx=\frac {{\left (i \, b c \log \left (F\right ) - e\right )} e^{\left (a c \log \left (F\right ) - m \log \left (-\frac {b c \log \left (F\right ) - i \, e}{f}\right ) - i \, d\right )} \Gamma \left (m + 1, -b c x \log \left (F\right ) + i \, e x\right ) + {\left (-i \, b c \log \left (F\right ) - e\right )} e^{\left (a c \log \left (F\right ) - m \log \left (-\frac {b c \log \left (F\right ) + i \, e}{f}\right ) + i \, d\right )} \Gamma \left (m + 1, -b c x \log \left (F\right ) - i \, e x\right )}{2 \, {\left (b^{2} c^{2} \log \left (F\right )^{2} + e^{2}\right )}} \]
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\[ \int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx=\int F^{c \left (a + b x\right )} \left (f x\right )^{m} \sin {\left (d + e x \right )}\, dx \]
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\[ \int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx=\int { \left (f x\right )^{m} F^{{\left (b x + a\right )} c} \sin \left (e x + d\right ) \,d x } \]
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\[ \int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx=\int { \left (f x\right )^{m} F^{{\left (b x + a\right )} c} \sin \left (e x + d\right ) \,d x } \]
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Timed out. \[ \int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx=\int F^{c\,\left (a+b\,x\right )}\,\sin \left (d+e\,x\right )\,{\left (f\,x\right )}^m \,d x \]
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