Integrand size = 13, antiderivative size = 47 \[ \int F^{a+b (c+d x)^3} \, dx=-\frac {F^a (c+d x) \Gamma \left (\frac {1}{3},-b (c+d x)^3 \log (F)\right )}{3 d \sqrt [3]{-b (c+d x)^3 \log (F)}} \]
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Time = 0.00 (sec) , antiderivative size = 47, 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} \[ \int F^{a+b (c+d x)^3} \, dx=-\frac {F^a (c+d x) \Gamma \left (\frac {1}{3},-b (c+d x)^3 \log (F)\right )}{3 d \sqrt [3]{-b \log (F) (c+d x)^3}} \]
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Rule 2239
Rubi steps \begin{align*} \text {integral}& = -\frac {F^a (c+d x) \Gamma \left (\frac {1}{3},-b (c+d x)^3 \log (F)\right )}{3 d \sqrt [3]{-b (c+d x)^3 \log (F)}} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^3} \, dx=-\frac {F^a (c+d x) \Gamma \left (\frac {1}{3},-b (c+d x)^3 \log (F)\right )}{3 d \sqrt [3]{-b (c+d x)^3 \log (F)}} \]
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\[\int F^{a +b \left (d x +c \right )^{3}}d x\]
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none
Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.34 \[ \int F^{a+b (c+d x)^3} \, dx=\frac {\left (-b d^{3} \log \left (F\right )\right )^{\frac {2}{3}} F^{a} \Gamma \left (\frac {1}{3}, -{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right )}{3 \, b d^{3} \log \left (F\right )} \]
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\[ \int F^{a+b (c+d x)^3} \, dx=\int F^{a + b \left (c + d x\right )^{3}}\, dx \]
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\[ \int F^{a+b (c+d x)^3} \, dx=\int { F^{{\left (d x + c\right )}^{3} b + a} \,d x } \]
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\[ \int F^{a+b (c+d x)^3} \, dx=\int { F^{{\left (d x + c\right )}^{3} b + a} \,d x } \]
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Timed out. \[ \int F^{a+b (c+d x)^3} \, dx=\int F^{a+b\,{\left (c+d\,x\right )}^3} \,d x \]
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