Integrand size = 19, antiderivative size = 49 \[ \int F^{a+b (c+d x)^3} (c+d x) \, dx=-\frac {F^a (c+d x)^2 \Gamma \left (\frac {2}{3},-b (c+d x)^3 \log (F)\right )}{3 d \left (-b (c+d x)^3 \log (F)\right )^{2/3}} \]
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Time = 0.02 (sec) , antiderivative size = 49, 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 = {2250} \[ \int F^{a+b (c+d x)^3} (c+d x) \, dx=-\frac {F^a (c+d x)^2 \Gamma \left (\frac {2}{3},-b (c+d x)^3 \log (F)\right )}{3 d \left (-b \log (F) (c+d x)^3\right )^{2/3}} \]
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Rule 2250
Rubi steps \begin{align*} \text {integral}& = -\frac {F^a (c+d x)^2 \Gamma \left (\frac {2}{3},-b (c+d x)^3 \log (F)\right )}{3 d \left (-b (c+d x)^3 \log (F)\right )^{2/3}} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^3} (c+d x) \, dx=-\frac {F^a (c+d x)^2 \Gamma \left (\frac {2}{3},-b (c+d x)^3 \log (F)\right )}{3 d \left (-b (c+d x)^3 \log (F)\right )^{2/3}} \]
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\[\int F^{a +b \left (d x +c \right )^{3}} \left (d x +c \right )d x\]
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none
Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.29 \[ \int F^{a+b (c+d x)^3} (c+d x) \, dx=\frac {\left (-b d^{3} \log \left (F\right )\right )^{\frac {1}{3}} F^{a} \Gamma \left (\frac {2}{3}, -{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right )}{3 \, b d^{2} \log \left (F\right )} \]
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\[ \int F^{a+b (c+d x)^3} (c+d x) \, dx=\int F^{a + b \left (c + d x\right )^{3}} \left (c + d x\right )\, dx \]
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\[ \int F^{a+b (c+d x)^3} (c+d x) \, dx=\int { {\left (d x + c\right )} F^{{\left (d x + c\right )}^{3} b + a} \,d x } \]
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\[ \int F^{a+b (c+d x)^3} (c+d x) \, dx=\int { {\left (d x + c\right )} F^{{\left (d x + c\right )}^{3} b + a} \,d x } \]
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Timed out. \[ \int F^{a+b (c+d x)^3} (c+d x) \, dx=\int F^{a+b\,{\left (c+d\,x\right )}^3}\,\left (c+d\,x\right ) \,d x \]
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