Integrand size = 43, antiderivative size = 22 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=x-\frac {1}{5} \left (2-\frac {5}{4} e^{\frac {1}{x^2}-2 x}\right ) x \]
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Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 14, 2326} \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {e^{\frac {1}{x^2}-2 x} \left (x^3+1\right )}{4 \left (\frac {1}{x^3}+1\right ) x^2}+\frac {3 x}{5} \]
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Rule 12
Rule 14
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {1}{20} \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{x^2} \, dx \\ & = \frac {1}{20} \int \left (12-\frac {5 e^{\frac {1}{x^2}-2 x} \left (2-x^2+2 x^3\right )}{x^2}\right ) \, dx \\ & = \frac {3 x}{5}-\frac {1}{4} \int \frac {e^{\frac {1}{x^2}-2 x} \left (2-x^2+2 x^3\right )}{x^2} \, dx \\ & = \frac {3 x}{5}+\frac {e^{\frac {1}{x^2}-2 x} \left (1+x^3\right )}{4 \left (1+\frac {1}{x^3}\right ) x^2} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\left (\frac {3}{5}+\frac {1}{4} e^{\frac {1}{x^2}-2 x}\right ) x \]
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Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {x \,{\mathrm e}^{-\frac {2 x^{3}-1}{x^{2}}}}{4}+\frac {3 x}{5}\) | \(21\) |
parallelrisch | \(\frac {x \,{\mathrm e}^{\frac {x^{2} \ln \left (5\right )-2 x^{3}+1}{x^{2}}}}{20}+\frac {3 x}{5}\) | \(26\) |
parts | \(\frac {x \,{\mathrm e}^{\frac {x^{2} \ln \left (5\right )-2 x^{3}+1}{x^{2}}}}{20}+\frac {3 x}{5}\) | \(26\) |
norman | \(\frac {\frac {3 x^{2}}{5}+\frac {{\mathrm e}^{\frac {x^{2} \ln \left (5\right )-2 x^{3}+1}{x^{2}}} x^{2}}{20}}{x}\) | \(34\) |
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Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {1}{20} \, x e^{\left (-\frac {2 \, x^{3} - x^{2} \log \left (5\right ) - 1}{x^{2}}\right )} + \frac {3}{5} \, x \]
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Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {x e^{\frac {- 2 x^{3} + x^{2} \log {\left (5 \right )} + 1}{x^{2}}}}{20} + \frac {3 x}{5} \]
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Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {1}{4} \, x e^{\left (-2 \, x + \frac {1}{x^{2}}\right )} + \frac {3}{5} \, x \]
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Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {1}{4} \, x e^{\left (-\frac {2 \, x^{3} - 1}{x^{2}}\right )} + \frac {3}{5} \, x \]
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Time = 12.84 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {x\,\left (5\,{\mathrm {e}}^{\frac {1}{x^2}-2\,x}+12\right )}{20} \]
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