Integrand size = 26, antiderivative size = 20 \[ \int \frac {e \left (-9+18 x+3 x^2\right )+e x^2 \log \left (x^2\right )}{x^2} \, dx=e \left (5-x+(9+x) \left (2+\frac {1}{x}+\log \left (x^2\right )\right )\right ) \]
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Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {14, 2332} \[ \int \frac {e \left (-9+18 x+3 x^2\right )+e x^2 \log \left (x^2\right )}{x^2} \, dx=e x \log \left (x^2\right )+e x+\frac {9 e}{x}+18 e \log (x) \]
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Rule 14
Rule 2332
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3 e \left (-3+6 x+x^2\right )}{x^2}+e \log \left (x^2\right )\right ) \, dx \\ & = e \int \log \left (x^2\right ) \, dx+(3 e) \int \frac {-3+6 x+x^2}{x^2} \, dx \\ & = -2 e x+e x \log \left (x^2\right )+(3 e) \int \left (1-\frac {3}{x^2}+\frac {6}{x}\right ) \, dx \\ & = \frac {9 e}{x}+e x+18 e \log (x)+e x \log \left (x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {e \left (-9+18 x+3 x^2\right )+e x^2 \log \left (x^2\right )}{x^2} \, dx=\frac {9 e}{x}+e x+18 e \log (x)+e x \log \left (x^2\right ) \]
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Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30
method | result | size |
risch | \(x \,{\mathrm e} \ln \left (x^{2}\right )+\frac {{\mathrm e} \left (18 x \ln \left (x \right )+x^{2}+9\right )}{x}\) | \(26\) |
default | \({\mathrm e} \left (x \ln \left (x^{2}\right )-2 x \right )+3 \,{\mathrm e} \left (x +6 \ln \left (x \right )+\frac {3}{x}\right )\) | \(30\) |
parts | \({\mathrm e} \left (x \ln \left (x^{2}\right )-2 x \right )+3 \,{\mathrm e} \left (x +6 \ln \left (x \right )+\frac {3}{x}\right )\) | \(30\) |
norman | \(\frac {x^{2} {\mathrm e}+x^{2} {\mathrm e} \ln \left (x^{2}\right )+9 x \,{\mathrm e} \ln \left (x^{2}\right )+9 \,{\mathrm e}}{x}\) | \(35\) |
parallelrisch | \(\frac {2 x^{2} {\mathrm e} \ln \left (x^{2}\right )+2 x^{2} {\mathrm e}+18 x \,{\mathrm e} \ln \left (x^{2}\right )+18 \,{\mathrm e}}{2 x}\) | \(38\) |
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Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e \left (-9+18 x+3 x^2\right )+e x^2 \log \left (x^2\right )}{x^2} \, dx=\frac {{\left (x^{2} + 9 \, x\right )} e \log \left (x^{2}\right ) + {\left (x^{2} + 9\right )} e}{x} \]
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Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {e \left (-9+18 x+3 x^2\right )+e x^2 \log \left (x^2\right )}{x^2} \, dx=e x \log {\left (x^{2} \right )} + e x + 18 e \log {\left (x \right )} + \frac {9 e}{x} \]
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Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {e \left (-9+18 x+3 x^2\right )+e x^2 \log \left (x^2\right )}{x^2} \, dx={\left (x \log \left (x^{2}\right ) - 2 \, x\right )} e + 3 \, x e + 18 \, e \log \left (x\right ) + \frac {9 \, e}{x} \]
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Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {e \left (-9+18 x+3 x^2\right )+e x^2 \log \left (x^2\right )}{x^2} \, dx=\frac {x^{2} e \log \left (x^{2}\right ) + x^{2} e + 18 \, x e \log \left (x\right ) + 9 \, e}{x} \]
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Time = 11.91 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {e \left (-9+18 x+3 x^2\right )+e x^2 \log \left (x^2\right )}{x^2} \, dx=9\,\ln \left (x^2\right )\,\mathrm {e}+\frac {9\,\mathrm {e}}{x}+x\,\mathrm {e}\,\left (\ln \left (x^2\right )+1\right ) \]
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