Integrand size = 21, antiderivative size = 22 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=-3+\frac {2 e^{x^2}}{5}+2 x+x^2 \log ^2(3) \]
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Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 2240} \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=\frac {2 e^{x^2}}{5}+x^2 \log ^2(3)+2 x \]
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Rule 12
Rule 2240
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx \\ & = 2 x+x^2 \log ^2(3)+\frac {4}{5} \int e^{x^2} x \, dx \\ & = \frac {2 e^{x^2}}{5}+2 x+x^2 \log ^2(3) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=\frac {2 e^{x^2}}{5}+2 x+x^2 \log ^2(3) \]
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Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86
method | result | size |
default | \(2 x +x^{2} \ln \left (3\right )^{2}+\frac {2 \,{\mathrm e}^{x^{2}}}{5}\) | \(19\) |
norman | \(2 x +x^{2} \ln \left (3\right )^{2}+\frac {2 \,{\mathrm e}^{x^{2}}}{5}\) | \(19\) |
risch | \(2 x +x^{2} \ln \left (3\right )^{2}+\frac {2 \,{\mathrm e}^{x^{2}}}{5}\) | \(19\) |
parallelrisch | \(2 x +x^{2} \ln \left (3\right )^{2}+\frac {2 \,{\mathrm e}^{x^{2}}}{5}\) | \(19\) |
parts | \(2 x +x^{2} \ln \left (3\right )^{2}+\frac {2 \,{\mathrm e}^{x^{2}}}{5}\) | \(19\) |
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Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=x^{2} \log \left (3\right )^{2} + 2 \, x + \frac {2}{5} \, e^{\left (x^{2}\right )} \]
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Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=x^{2} \log {\left (3 \right )}^{2} + 2 x + \frac {2 e^{x^{2}}}{5} \]
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Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=x^{2} \log \left (3\right )^{2} + 2 \, x + \frac {2}{5} \, e^{\left (x^{2}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=x^{2} \log \left (3\right )^{2} + 2 \, x + \frac {2}{5} \, e^{\left (x^{2}\right )} \]
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Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=2\,x+\frac {2\,{\mathrm {e}}^{x^2}}{5}+x^2\,{\ln \left (3\right )}^2 \]
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