Integrand size = 45, antiderivative size = 21 \[ \int \frac {x^2+e^{\frac {2 \left (x \log (5)+e \log (5) \log \left (x^2\right )\right )}{e x}} \left (-4 \log (5)+2 \log (5) \log \left (x^2\right )\right )}{x^2} \, dx=-5^{\frac {2 \left (\frac {x}{e}+\log \left (x^2\right )\right )}{x}}+x \]
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Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {14, 6838} \[ \int \frac {x^2+e^{\frac {2 \left (x \log (5)+e \log (5) \log \left (x^2\right )\right )}{e x}} \left (-4 \log (5)+2 \log (5) \log \left (x^2\right )\right )}{x^2} \, dx=x-5^{\frac {2 \left (e \log \left (x^2\right )+x\right )}{e x}} \]
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Rule 14
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {2\ 5^{\frac {2 \left (x+e \log \left (x^2\right )\right )}{e x}} \log (5) \left (-2+\log \left (x^2\right )\right )}{x^2}\right ) \, dx \\ & = x+(2 \log (5)) \int \frac {5^{\frac {2 \left (x+e \log \left (x^2\right )\right )}{e x}} \left (-2+\log \left (x^2\right )\right )}{x^2} \, dx \\ & = -5^{\frac {2 \left (x+e \log \left (x^2\right )\right )}{e x}}+x \\ \end{align*}
Time = 0.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {x^2+e^{\frac {2 \left (x \log (5)+e \log (5) \log \left (x^2\right )\right )}{e x}} \left (-4 \log (5)+2 \log (5) \log \left (x^2\right )\right )}{x^2} \, dx=-5^{\frac {2}{e}+\frac {2 \log \left (x^2\right )}{x}}+x \]
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05
method | result | size |
risch | \(x -5^{\frac {2 \ln \left (x^{2}\right )+2 \,{\mathrm e}^{-1} x}{x}}\) | \(22\) |
parallelrisch | \(-{\mathrm e}^{\frac {2 \ln \left (5\right ) \left ({\mathrm e} \ln \left (x^{2}\right )+x \right ) {\mathrm e}^{-1}}{x}}+x\) | \(27\) |
default | \(x -{\mathrm e}^{\frac {2 \left ({\mathrm e} \ln \left (5\right ) \ln \left (x^{2}\right )+x \ln \left (5\right )\right ) {\mathrm e}^{-1}}{x}}\) | \(30\) |
parts | \(x -{\mathrm e}^{\frac {2 \left ({\mathrm e} \ln \left (5\right ) \ln \left (x^{2}\right )+x \ln \left (5\right )\right ) {\mathrm e}^{-1}}{x}}\) | \(30\) |
norman | \(\frac {x^{2}-x \,{\mathrm e}^{\frac {2 \left ({\mathrm e} \ln \left (5\right ) \ln \left (x^{2}\right )+x \ln \left (5\right )\right ) {\mathrm e}^{-1}}{x}}}{x}\) | \(37\) |
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none
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {x^2+e^{\frac {2 \left (x \log (5)+e \log (5) \log \left (x^2\right )\right )}{e x}} \left (-4 \log (5)+2 \log (5) \log \left (x^2\right )\right )}{x^2} \, dx=x - e^{\left (\frac {2 \, {\left (e \log \left (5\right ) \log \left (x^{2}\right ) + x \log \left (5\right )\right )} e^{\left (-1\right )}}{x}\right )} \]
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Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {x^2+e^{\frac {2 \left (x \log (5)+e \log (5) \log \left (x^2\right )\right )}{e x}} \left (-4 \log (5)+2 \log (5) \log \left (x^2\right )\right )}{x^2} \, dx=x - e^{\frac {2 \left (x \log {\left (5 \right )} + e \log {\left (5 \right )} \log {\left (x^{2} \right )}\right )}{e x}} \]
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none
Time = 0.36 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {x^2+e^{\frac {2 \left (x \log (5)+e \log (5) \log \left (x^2\right )\right )}{e x}} \left (-4 \log (5)+2 \log (5) \log \left (x^2\right )\right )}{x^2} \, dx=-5^{2 \, e^{\left (-1\right )}} e^{\left (\frac {4 \, \log \left (5\right ) \log \left (x\right )}{x}\right )} + x \]
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Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {x^2+e^{\frac {2 \left (x \log (5)+e \log (5) \log \left (x^2\right )\right )}{e x}} \left (-4 \log (5)+2 \log (5) \log \left (x^2\right )\right )}{x^2} \, dx=x - e^{\left (\frac {2 \, {\left (e \log \left (5\right ) \log \left (x^{2}\right ) + x \log \left (5\right )\right )} e^{\left (-1\right )}}{x}\right )} \]
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Time = 8.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {x^2+e^{\frac {2 \left (x \log (5)+e \log (5) \log \left (x^2\right )\right )}{e x}} \left (-4 \log (5)+2 \log (5) \log \left (x^2\right )\right )}{x^2} \, dx=x-5^{2\,{\mathrm {e}}^{-1}}\,{\left (x^2\right )}^{\frac {2\,\ln \left (5\right )}{x}} \]
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