Integrand size = 20, antiderivative size = 19 \[ \int \frac {-4+4 x+5 e^{4+x} x}{5 x} \, dx=e^{4+x}+\frac {4}{5} (2+x-\log (4 x)) \]
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Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 14, 2225, 45} \[ \int \frac {-4+4 x+5 e^{4+x} x}{5 x} \, dx=\frac {4 x}{5}+e^{x+4}-\frac {4 \log (x)}{5} \]
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Rule 12
Rule 14
Rule 45
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {-4+4 x+5 e^{4+x} x}{x} \, dx \\ & = \frac {1}{5} \int \left (5 e^{4+x}+\frac {4 (-1+x)}{x}\right ) \, dx \\ & = \frac {4}{5} \int \frac {-1+x}{x} \, dx+\int e^{4+x} \, dx \\ & = e^{4+x}+\frac {4}{5} \int \left (1-\frac {1}{x}\right ) \, dx \\ & = e^{4+x}+\frac {4 x}{5}-\frac {4 \log (x)}{5} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-4+4 x+5 e^{4+x} x}{5 x} \, dx=e^{4+x}+\frac {4 x}{5}-\frac {4 \log (x)}{5} \]
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Time = 0.11 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68
method | result | size |
norman | \(\frac {4 x}{5}+{\mathrm e}^{4+x}-\frac {4 \ln \left (x \right )}{5}\) | \(13\) |
risch | \(\frac {4 x}{5}+{\mathrm e}^{4+x}-\frac {4 \ln \left (x \right )}{5}\) | \(13\) |
parallelrisch | \(\frac {4 x}{5}+{\mathrm e}^{4+x}-\frac {4 \ln \left (x \right )}{5}\) | \(13\) |
parts | \(\frac {4 x}{5}+{\mathrm e}^{4+x}-\frac {4 \ln \left (x \right )}{5}\) | \(13\) |
derivativedivides | \(-\frac {4 \ln \left (x \right )}{5}+\frac {16}{5}+\frac {4 x}{5}+{\mathrm e}^{4+x}\) | \(14\) |
default | \(-\frac {4 \ln \left (x \right )}{5}+\frac {16}{5}+\frac {4 x}{5}+{\mathrm e}^{4+x}\) | \(14\) |
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Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-4+4 x+5 e^{4+x} x}{5 x} \, dx=\frac {4}{5} \, x + e^{\left (x + 4\right )} - \frac {4}{5} \, \log \left (x\right ) \]
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Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-4+4 x+5 e^{4+x} x}{5 x} \, dx=\frac {4 x}{5} + e^{x + 4} - \frac {4 \log {\left (x \right )}}{5} \]
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Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-4+4 x+5 e^{4+x} x}{5 x} \, dx=\frac {4}{5} \, x + e^{\left (x + 4\right )} - \frac {4}{5} \, \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-4+4 x+5 e^{4+x} x}{5 x} \, dx=\frac {4}{5} \, x + e^{\left (x + 4\right )} - \frac {4}{5} \, \log \left (x\right ) \]
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Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-4+4 x+5 e^{4+x} x}{5 x} \, dx=\frac {4\,x}{5}+{\mathrm {e}}^{x+4}-\frac {4\,\ln \left (x\right )}{5} \]
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