Integrand size = 33, antiderivative size = 22 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=2 x-\frac {1}{5} \left (1-e^{28 x/5}\right ) x \log (x) \]
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Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 2225, 2207, 2634} \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=2 x-\frac {1}{5} x \log (x)-\frac {1}{28} e^{28 x/5} \log (x)+\frac {1}{140} e^{28 x/5} (28 x+5) \log (x) \]
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Rule 12
Rule 2207
Rule 2225
Rule 2634
Rubi steps \begin{align*} \text {integral}& = \frac {1}{25} \int \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx \\ & = \frac {9 x}{5}+\frac {1}{25} \int \left (-5+e^{28 x/5} (5+28 x)\right ) \log (x) \, dx+\frac {1}{5} \int e^{28 x/5} \, dx \\ & = \frac {1}{28} e^{28 x/5}+\frac {9 x}{5}-\frac {1}{28} e^{28 x/5} \log (x)-\frac {1}{5} x \log (x)+\frac {1}{140} e^{28 x/5} (5+28 x) \log (x)-\frac {1}{25} \int 5 \left (-1+e^{28 x/5}\right ) \, dx \\ & = \frac {1}{28} e^{28 x/5}+\frac {9 x}{5}-\frac {1}{28} e^{28 x/5} \log (x)-\frac {1}{5} x \log (x)+\frac {1}{140} e^{28 x/5} (5+28 x) \log (x)-\frac {1}{5} \int \left (-1+e^{28 x/5}\right ) \, dx \\ & = \frac {1}{28} e^{28 x/5}+2 x-\frac {1}{28} e^{28 x/5} \log (x)-\frac {1}{5} x \log (x)+\frac {1}{140} e^{28 x/5} (5+28 x) \log (x)-\frac {1}{5} \int e^{28 x/5} \, dx \\ & = 2 x-\frac {1}{28} e^{28 x/5} \log (x)-\frac {1}{5} x \log (x)+\frac {1}{140} e^{28 x/5} (5+28 x) \log (x) \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {1}{25} \left (50 x+5 \left (-1+e^{28 x/5}\right ) x \log (x)\right ) \]
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Time = 0.50 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86
method | result | size |
default | \(2 x +\frac {{\mathrm e}^{\frac {28 x}{5}} \ln \left (x \right ) x}{5}-\frac {x \ln \left (x \right )}{5}\) | \(19\) |
norman | \(2 x +\frac {{\mathrm e}^{\frac {28 x}{5}} \ln \left (x \right ) x}{5}-\frac {x \ln \left (x \right )}{5}\) | \(19\) |
parallelrisch | \(2 x +\frac {{\mathrm e}^{\frac {28 x}{5}} \ln \left (x \right ) x}{5}-\frac {x \ln \left (x \right )}{5}\) | \(19\) |
parts | \(2 x +\frac {{\mathrm e}^{\frac {28 x}{5}} \ln \left (x \right ) x}{5}-\frac {x \ln \left (x \right )}{5}\) | \(19\) |
risch | \(\frac {\left (5 \,{\mathrm e}^{\frac {28 x}{5}} x -5 x \right ) \ln \left (x \right )}{25}+2 x\) | \(20\) |
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Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {1}{5} \, {\left (x e^{\left (\frac {28}{5} \, x\right )} - x\right )} \log \left (x\right ) + 2 \, x \]
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Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {x e^{\frac {28 x}{5}} \log {\left (x \right )}}{5} - \frac {x \log {\left (x \right )}}{5} + 2 x \]
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Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {1}{5} \, {\left (x e^{\left (\frac {28}{5} \, x\right )} - x\right )} \log \left (x\right ) + 2 \, x \]
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {1}{5} \, {\left (x e^{\left (\frac {28}{5} \, x\right )} - x\right )} \log \left (x\right ) + 2 \, x \]
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Time = 9.54 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {x\,\left ({\mathrm {e}}^{\frac {28\,x}{5}}\,\ln \left (x\right )-\ln \left (x\right )+10\right )}{5} \]
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