Integrand size = 56, antiderivative size = 24 \[ \int \frac {e^{2+x+x^2}+5 x+5 x^2+\log (3)+\left (5 x^2+e^{2+x+x^2} \left (-1+x+2 x^2\right )-\log (3)\right ) \log (x)}{5 x^2} \, dx=\left (1+x+\frac {e^{2+x+x^2}+\log (3)}{5 x}\right ) \log (x) \]
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Leaf count is larger than twice the leaf count of optimal. \(52\) vs. \(2(24)=48\).
Time = 0.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 14, 2326, 2372} \[ \int \frac {e^{2+x+x^2}+5 x+5 x^2+\log (3)+\left (5 x^2+e^{2+x+x^2} \left (-1+x+2 x^2\right )-\log (3)\right ) \log (x)}{5 x^2} \, dx=\frac {e^{x^2+x+2} \left (2 x^2 \log (x)+x \log (x)\right )}{5 x^2 (2 x+1)}+x \log (x)+\frac {\log (3) \log (x)}{5 x}+\log (x) \]
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Rule 12
Rule 14
Rule 2326
Rule 2372
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{2+x+x^2}+5 x+5 x^2+\log (3)+\left (5 x^2+e^{2+x+x^2} \left (-1+x+2 x^2\right )-\log (3)\right ) \log (x)}{x^2} \, dx \\ & = \frac {1}{5} \int \left (\frac {e^{2+x+x^2} \left (1-\log (x)+x \log (x)+2 x^2 \log (x)\right )}{x^2}+\frac {5 x+5 x^2+\log (3)+5 x^2 \log (x)-\log (3) \log (x)}{x^2}\right ) \, dx \\ & = \frac {1}{5} \int \frac {e^{2+x+x^2} \left (1-\log (x)+x \log (x)+2 x^2 \log (x)\right )}{x^2} \, dx+\frac {1}{5} \int \frac {5 x+5 x^2+\log (3)+5 x^2 \log (x)-\log (3) \log (x)}{x^2} \, dx \\ & = \frac {e^{2+x+x^2} \left (x \log (x)+2 x^2 \log (x)\right )}{5 x^2 (1+2 x)}+\frac {1}{5} \int \left (\frac {5 x+5 x^2+\log (3)}{x^2}+\frac {\left (5 x^2-\log (3)\right ) \log (x)}{x^2}\right ) \, dx \\ & = \frac {e^{2+x+x^2} \left (x \log (x)+2 x^2 \log (x)\right )}{5 x^2 (1+2 x)}+\frac {1}{5} \int \frac {5 x+5 x^2+\log (3)}{x^2} \, dx+\frac {1}{5} \int \frac {\left (5 x^2-\log (3)\right ) \log (x)}{x^2} \, dx \\ & = x \log (x)+\frac {\log (3) \log (x)}{5 x}+\frac {e^{2+x+x^2} \left (x \log (x)+2 x^2 \log (x)\right )}{5 x^2 (1+2 x)}-\frac {1}{5} \int \left (5+\frac {\log (3)}{x^2}\right ) \, dx+\frac {1}{5} \int \left (5+\frac {5}{x}+\frac {\log (3)}{x^2}\right ) \, dx \\ & = \log (x)+x \log (x)+\frac {\log (3) \log (x)}{5 x}+\frac {e^{2+x+x^2} \left (x \log (x)+2 x^2 \log (x)\right )}{5 x^2 (1+2 x)} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2+x+x^2}+5 x+5 x^2+\log (3)+\left (5 x^2+e^{2+x+x^2} \left (-1+x+2 x^2\right )-\log (3)\right ) \log (x)}{5 x^2} \, dx=\frac {\left (e^{2+x+x^2}+5 x+5 x^2+\log (3)\right ) \log (x)}{5 x} \]
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Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\frac {\left (5 x^{2}+\ln \left (3\right )+{\mathrm e}^{x^{2}+x +2}\right ) \ln \left (x \right )}{5 x}+\ln \left (x \right )\) | \(26\) |
norman | \(\frac {x^{2} \ln \left (x \right )+x \ln \left (x \right )+\frac {\ln \left (3\right ) \ln \left (x \right )}{5}+\frac {\ln \left (x \right ) {\mathrm e}^{x^{2}+x +2}}{5}}{x}\) | \(33\) |
parallelrisch | \(\frac {5 x^{2} \ln \left (x \right )+\ln \left (3\right ) \ln \left (x \right )+5 x \ln \left (x \right )+\ln \left (x \right ) {\mathrm e}^{x^{2}+x +2}}{5 x}\) | \(34\) |
default | \(\frac {{\mathrm e}^{x^{2}+x +2} \ln \left (x \right )}{5 x}+\ln \left (x \right )-\frac {\ln \left (3\right )}{5 x}+x \ln \left (x \right )-\frac {\ln \left (3\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )}{5}\) | \(46\) |
parts | \(\frac {{\mathrm e}^{x^{2}+x +2} \ln \left (x \right )}{5 x}+\ln \left (x \right )-\frac {\ln \left (3\right )}{5 x}+x \ln \left (x \right )-\frac {\ln \left (3\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )}{5}\) | \(46\) |
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Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{2+x+x^2}+5 x+5 x^2+\log (3)+\left (5 x^2+e^{2+x+x^2} \left (-1+x+2 x^2\right )-\log (3)\right ) \log (x)}{5 x^2} \, dx=\frac {{\left (5 \, x^{2} + 5 \, x + e^{\left (x^{2} + x + 2\right )} + \log \left (3\right )\right )} \log \left (x\right )}{5 \, x} \]
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Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{2+x+x^2}+5 x+5 x^2+\log (3)+\left (5 x^2+e^{2+x+x^2} \left (-1+x+2 x^2\right )-\log (3)\right ) \log (x)}{5 x^2} \, dx=\log {\left (x \right )} + \frac {\left (5 x^{2} + \log {\left (3 \right )}\right ) \log {\left (x \right )}}{5 x} + \frac {e^{x^{2} + x + 2} \log {\left (x \right )}}{5 x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (23) = 46\).
Time = 0.33 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {e^{2+x+x^2}+5 x+5 x^2+\log (3)+\left (5 x^2+e^{2+x+x^2} \left (-1+x+2 x^2\right )-\log (3)\right ) \log (x)}{5 x^2} \, dx=x - \frac {5 \, x^{2} - {\left (5 \, x^{2} + \log \left (3\right )\right )} \log \left (x\right ) - e^{\left (x^{2} + x + 2\right )} \log \left (x\right ) - \log \left (3\right )}{5 \, x} - \frac {\log \left (3\right )}{5 \, x} + \log \left (x\right ) \]
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Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{2+x+x^2}+5 x+5 x^2+\log (3)+\left (5 x^2+e^{2+x+x^2} \left (-1+x+2 x^2\right )-\log (3)\right ) \log (x)}{5 x^2} \, dx=\frac {5 \, x^{2} \log \left (x\right ) + 5 \, x \log \left (x\right ) + e^{\left (x^{2} + x + 2\right )} \log \left (x\right ) + \log \left (3\right ) \log \left (x\right )}{5 \, x} \]
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Time = 12.72 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{2+x+x^2}+5 x+5 x^2+\log (3)+\left (5 x^2+e^{2+x+x^2} \left (-1+x+2 x^2\right )-\log (3)\right ) \log (x)}{5 x^2} \, dx=\frac {\ln \left (x\right )\,\left (5\,x+{\mathrm {e}}^{x^2+x+2}+\ln \left (3\right )+5\,x^2\right )}{5\,x} \]
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