Integrand size = 17, antiderivative size = 67 \[ \int F^{c (a+b x)} (d+e x)^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} (d+e x)^m \Gamma \left (1+m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-m}}{b c \log (F)} \]
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Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2212} \[ \int F^{c (a+b x)} (d+e x)^m \, dx=\frac {(d+e x)^m F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^{-m} \Gamma \left (m+1,-\frac {b c (d+e x) \log (F)}{e}\right )}{b c \log (F)} \]
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Rule 2212
Rubi steps \begin{align*} \text {integral}& = \frac {F^{c \left (a-\frac {b d}{e}\right )} (d+e x)^m \Gamma \left (1+m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-m}}{b c \log (F)} \\ \end{align*}
Time = 0.42 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} (d+e x)^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} (d+e x)^m \Gamma \left (1+m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-m}}{b c \log (F)} \]
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\[\int F^{c \left (b x +a \right )} \left (e x +d \right )^{m}d x\]
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none
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.97 \[ \int F^{c (a+b x)} (d+e x)^m \, dx=\frac {e^{\left (-\frac {e m \log \left (-\frac {b c \log \left (F\right )}{e}\right ) + {\left (b c d - a c e\right )} \log \left (F\right )}{e}\right )} \Gamma \left (m + 1, -\frac {{\left (b c e x + b c d\right )} \log \left (F\right )}{e}\right )}{b c \log \left (F\right )} \]
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\[ \int F^{c (a+b x)} (d+e x)^m \, dx=\int F^{c \left (a + b x\right )} \left (d + e x\right )^{m}\, dx \]
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\[ \int F^{c (a+b x)} (d+e x)^m \, dx=\int { {\left (e x + d\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \]
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\[ \int F^{c (a+b x)} (d+e x)^m \, dx=\int { {\left (e x + d\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \]
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Timed out. \[ \int F^{c (a+b x)} (d+e x)^m \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (d+e\,x\right )}^m \,d x \]
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