Integrand size = 19, antiderivative size = 69 \[ \int F^{c (a+b x)} (d+e x)^{-m} \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} (d+e x)^{-m} \Gamma \left (1-m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^m}{b c \log (F)} \]
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Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2212} \[ \int F^{c (a+b x)} (d+e x)^{-m} \, dx=\frac {(d+e x)^{-m} F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^m \Gamma \left (1-m,-\frac {b c (d+e x) \log (F)}{e}\right )}{b c \log (F)} \]
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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 = 0.05 (sec) , antiderivative size = 69, 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 = 67, 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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\[ \int F^{c (a+b x)} (d+e x)^{-m} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{m}} \,d x } \]
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\[ \int F^{c (a+b x)} (d+e x)^{-m} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{m}} \,d x } \]
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Timed out. \[ \int F^{c (a+b x)} (d+e x)^{-m} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (d+e\,x\right )}^m} \,d x \]
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