Integrand size = 13, antiderivative size = 50 \[ \int F^{a+b (c+d x)^n} \, dx=-\frac {F^a (c+d x) \Gamma \left (\frac {1}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{-1/n}}{d n} \]
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Time = 0.00 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2239} \[ \int F^{a+b (c+d x)^n} \, dx=-\frac {F^a (c+d x) \left (-b \log (F) (c+d x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-b (c+d x)^n \log (F)\right )}{d n} \]
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Rule 2239
Rubi steps \begin{align*} \text {integral}& = -\frac {F^a (c+d x) \Gamma \left (\frac {1}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{-1/n}}{d n} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^n} \, dx=-\frac {F^a (c+d x) \Gamma \left (\frac {1}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{-1/n}}{d n} \]
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\[\int F^{a +b \left (d x +c \right )^{n}}d x\]
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\[ \int F^{a+b (c+d x)^n} \, dx=\int { F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]
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\[ \int F^{a+b (c+d x)^n} \, dx=\int F^{a + b \left (c + d x\right )^{n}}\, dx \]
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\[ \int F^{a+b (c+d x)^n} \, dx=\int { F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]
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\[ \int F^{a+b (c+d x)^n} \, dx=\int { F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]
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Timed out. \[ \int F^{a+b (c+d x)^n} \, dx=\int F^{a+b\,{\left (c+d\,x\right )}^n} \,d x \]
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