Integrand size = 34, antiderivative size = 27 \[ \int \frac {e^9+5 x^2+e^x x^3+x^2 \log \left (\frac {e^{e^x}}{5}\right )}{x^2} \, dx=x \left (5-\frac {e^9 (1-x)}{x^2}+\log \left (\frac {e^{e^x}}{5}\right )\right ) \]
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Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {14, 2207, 2225, 2320, 2189, 29} \[ \int \frac {e^9+5 x^2+e^x x^3+x^2 \log \left (\frac {e^{e^x}}{5}\right )}{x^2} \, dx=e^x x+5 x-\frac {e^9}{x}-\left (e^x-\log \left (\frac {e^{e^x}}{5}\right )\right ) \log \left (e^x\right ) \]
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Rule 14
Rule 29
Rule 2189
Rule 2207
Rule 2225
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \int \left (e^x x+\frac {e^9+5 x^2+x^2 \log \left (\frac {e^{e^x}}{5}\right )}{x^2}\right ) \, dx \\ & = \int e^x x \, dx+\int \frac {e^9+5 x^2+x^2 \log \left (\frac {e^{e^x}}{5}\right )}{x^2} \, dx \\ & = e^x x-\int e^x \, dx+\int \left (\frac {e^9+5 x^2}{x^2}+\log \left (\frac {e^{e^x}}{5}\right )\right ) \, dx \\ & = -e^x+e^x x+\int \frac {e^9+5 x^2}{x^2} \, dx+\int \log \left (\frac {e^{e^x}}{5}\right ) \, dx \\ & = -e^x+e^x x+\int \left (5+\frac {e^9}{x^2}\right ) \, dx+\text {Subst}\left (\int \frac {\log \left (\frac {e^x}{5}\right )}{x} \, dx,x,e^x\right ) \\ & = -\frac {e^9}{x}+5 x+e^x x-\left (e^x-\log \left (\frac {e^{e^x}}{5}\right )\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right ) \\ & = -\frac {e^9}{x}+5 x+e^x x-x \left (e^x-\log \left (\frac {e^{e^x}}{5}\right )\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^9+5 x^2+e^x x^3+x^2 \log \left (\frac {e^{e^x}}{5}\right )}{x^2} \, dx=-\frac {e^9}{x}+5 x+x \log \left (\frac {e^{e^x}}{5}\right ) \]
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Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
parallelrisch | \(-\frac {-x^{2} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{x}}}{5}\right )-5 x^{2}+{\mathrm e}^{9}}{x}\) | \(25\) |
risch | \(x \ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )-\frac {2 x^{2} \ln \left (5\right )-10 x^{2}+2 \,{\mathrm e}^{9}}{2 x}\) | \(30\) |
default | \(5 x -\frac {{\mathrm e}^{9}}{x}+{\mathrm e}^{x} x +\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{x}}}{5}\right ) \ln \left ({\mathrm e}^{x}\right )-\ln \left ({\mathrm e}^{x}\right ) {\mathrm e}^{x}\) | \(33\) |
parts | \(5 x -\frac {{\mathrm e}^{9}}{x}+{\mathrm e}^{x} x +\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{x}}}{5}\right ) \ln \left ({\mathrm e}^{x}\right )-\ln \left ({\mathrm e}^{x}\right ) {\mathrm e}^{x}\) | \(33\) |
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Time = 0.38 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^9+5 x^2+e^x x^3+x^2 \log \left (\frac {e^{e^x}}{5}\right )}{x^2} \, dx=\frac {x^{2} e^{x} - x^{2} \log \left (5\right ) + 5 \, x^{2} - e^{9}}{x} \]
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Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {e^9+5 x^2+e^x x^3+x^2 \log \left (\frac {e^{e^x}}{5}\right )}{x^2} \, dx=x e^{x} + x \left (5 - \log {\left (5 \right )}\right ) - \frac {e^{9}}{x} \]
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Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^9+5 x^2+e^x x^3+x^2 \log \left (\frac {e^{e^x}}{5}\right )}{x^2} \, dx={\left (x - 1\right )} e^{x} - x e^{x} + x \log \left (\frac {1}{5} \, e^{\left (e^{x}\right )}\right ) + 5 \, x - \frac {e^{9}}{x} + e^{x} \]
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Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^9+5 x^2+e^x x^3+x^2 \log \left (\frac {e^{e^x}}{5}\right )}{x^2} \, dx=\frac {x^{2} e^{x} - x^{2} \log \left (5\right ) + 5 \, x^{2} - e^{9}}{x} \]
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Time = 0.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {e^9+5 x^2+e^x x^3+x^2 \log \left (\frac {e^{e^x}}{5}\right )}{x^2} \, dx=x\,\left ({\mathrm {e}}^x-\ln \left (5\right )+5\right )-\frac {{\mathrm {e}}^9}{x} \]
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