Integrand size = 50, antiderivative size = 71 \[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^4\right )^m \Gamma \left (1+4 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-4 m}}{b c \log (F)} \]
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Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2219, 2212} \[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\frac {\left ((d+e x)^4\right )^m F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^{-4 m} \Gamma \left (4 m+1,-\frac {b c (d+e x) \log (F)}{e}\right )}{b c \log (F)} \]
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Rule 2212
Rule 2219
Rubi steps \begin{align*} \text {integral}& = (d+e x)^{-4 m} \left ((d+e x)^4\right )^m \int F^{c (a+b x)} (d+e x)^{4 m} \, dx \\ & = \frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^4\right )^m \Gamma \left (1+4 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-4 m}}{b c \log (F)} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^4\right )^m \Gamma \left (1+4 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-4 m}}{b c \log (F)} \]
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\[\int F^{c \left (b x +a \right )} \left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right )^{m}d x\]
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\[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\int { {\left (e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \]
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\[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\int F^{c \left (a + b x\right )} \left (\left (d + e x\right )^{4}\right )^{m}\, dx \]
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\[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\int { {\left (e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \]
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\[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\int { {\left (e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \]
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Timed out. \[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4\right )}^m \,d x \]
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