Integrand size = 22, antiderivative size = 32 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {e^{x^3}}{3}-\frac {2 e^{5 x^3}}{15}+\frac {e^{9 x^3}}{27} \]
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Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6847, 2281, 200} \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {e^{x^3}}{3}-\frac {2 e^{5 x^3}}{15}+\frac {e^{9 x^3}}{27} \]
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Rule 200
Rule 2281
Rule 6847
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int e^x \left (1-e^{4 x}\right )^2 \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (1-x^4\right )^2 \, dx,x,e^{x^3}\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (1-2 x^4+x^8\right ) \, dx,x,e^{x^3}\right ) \\ & = \frac {e^{x^3}}{3}-\frac {2 e^{5 x^3}}{15}+\frac {e^{9 x^3}}{27} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {1}{135} e^{x^3} \left (45-18 e^{4 x^3}+5 e^{8 x^3}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {{\mathrm e}^{x^{3}}}{3}-\frac {2 \,{\mathrm e}^{5 x^{3}}}{15}+\frac {{\mathrm e}^{9 x^{3}}}{27}\) | \(24\) |
default | \(\frac {{\mathrm e}^{x^{3}}}{3}-\frac {2 \,{\mathrm e}^{5 x^{3}}}{15}+\frac {{\mathrm e}^{9 x^{3}}}{27}\) | \(24\) |
risch | \(\frac {{\mathrm e}^{x^{3}}}{3}-\frac {2 \,{\mathrm e}^{5 x^{3}}}{15}+\frac {{\mathrm e}^{9 x^{3}}}{27}\) | \(24\) |
meijerg | \(-\frac {32}{135}+\frac {{\mathrm e}^{x^{3}}}{3}-\frac {2 \,{\mathrm e}^{5 x^{3}}}{15}+\frac {{\mathrm e}^{9 x^{3}}}{27}\) | \(25\) |
parallelrisch | \(\frac {{\mathrm e}^{x^{3}}}{3}+\frac {{\mathrm e}^{8 x^{3}} {\mathrm e}^{x^{3}}}{27}-\frac {2 \,{\mathrm e}^{x^{3}} {\mathrm e}^{4 x^{3}}}{15}\) | \(34\) |
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Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {1}{27} \, e^{\left (9 \, x^{3}\right )} - \frac {2}{15} \, e^{\left (5 \, x^{3}\right )} + \frac {1}{3} \, e^{\left (x^{3}\right )} \]
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Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {e^{9 x^{3}}}{27} - \frac {2 e^{5 x^{3}}}{15} + \frac {e^{x^{3}}}{3} \]
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Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {1}{27} \, e^{\left (9 \, x^{3}\right )} - \frac {2}{15} \, e^{\left (5 \, x^{3}\right )} + \frac {1}{3} \, e^{\left (x^{3}\right )} \]
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Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {1}{27} \, e^{\left (9 \, x^{3}\right )} - \frac {2}{15} \, e^{\left (5 \, x^{3}\right )} + \frac {1}{3} \, e^{\left (x^{3}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {{\mathrm {e}}^{x^3}\,\left (5\,{\mathrm {e}}^{8\,x^3}-18\,{\mathrm {e}}^{4\,x^3}+45\right )}{135} \]
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