Integrand size = 26, antiderivative size = 26 \[ \int \frac {1}{3} \left (1+2 x+e^{5+x} (1+x)-\log \left (\frac {x}{4}\right )\right ) \, dx=2+\frac {1}{3} \left (-1+x \left (2+e^{5+x}+x-\log \left (\frac {x}{4}\right )\right )\right ) \]
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Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 2207, 2225, 2332} \[ \int \frac {1}{3} \left (1+2 x+e^{5+x} (1+x)-\log \left (\frac {x}{4}\right )\right ) \, dx=\frac {x^2}{3}+\frac {2 x}{3}-\frac {e^{x+5}}{3}+\frac {1}{3} e^{x+5} (x+1)-\frac {1}{3} x \log \left (\frac {x}{4}\right ) \]
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Rule 12
Rule 2207
Rule 2225
Rule 2332
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \left (1+2 x+e^{5+x} (1+x)-\log \left (\frac {x}{4}\right )\right ) \, dx \\ & = \frac {x}{3}+\frac {x^2}{3}+\frac {1}{3} \int e^{5+x} (1+x) \, dx-\frac {1}{3} \int \log \left (\frac {x}{4}\right ) \, dx \\ & = \frac {2 x}{3}+\frac {x^2}{3}+\frac {1}{3} e^{5+x} (1+x)-\frac {1}{3} x \log \left (\frac {x}{4}\right )-\frac {1}{3} \int e^{5+x} \, dx \\ & = -\frac {e^{5+x}}{3}+\frac {2 x}{3}+\frac {x^2}{3}+\frac {1}{3} e^{5+x} (1+x)-\frac {1}{3} x \log \left (\frac {x}{4}\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {1}{3} \left (1+2 x+e^{5+x} (1+x)-\log \left (\frac {x}{4}\right )\right ) \, dx=\frac {1}{3} \left (2 x+e^{5+x} x+x^2+x \log (4)-x \log (x)\right ) \]
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Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
norman | \(\frac {2 x}{3}+\frac {x^{2}}{3}+\frac {x \,{\mathrm e}^{5+x}}{3}-\frac {x \ln \left (\frac {x}{4}\right )}{3}\) | \(24\) |
risch | \(\frac {2 x}{3}+\frac {x^{2}}{3}+\frac {x \,{\mathrm e}^{5+x}}{3}-\frac {x \ln \left (\frac {x}{4}\right )}{3}\) | \(24\) |
parallelrisch | \(\frac {2 x}{3}+\frac {x^{2}}{3}+\frac {x \,{\mathrm e}^{5+x}}{3}-\frac {x \ln \left (\frac {x}{4}\right )}{3}\) | \(24\) |
default | \(\frac {2 x}{3}+\frac {{\mathrm e}^{5+x} \left (5+x \right )}{3}-\frac {5 \,{\mathrm e}^{5+x}}{3}+\frac {x^{2}}{3}-\frac {x \ln \left (\frac {x}{4}\right )}{3}\) | \(32\) |
parts | \(\frac {2 x}{3}+\frac {{\mathrm e}^{5+x} \left (5+x \right )}{3}-\frac {5 \,{\mathrm e}^{5+x}}{3}+\frac {x^{2}}{3}-\frac {x \ln \left (\frac {x}{4}\right )}{3}\) | \(32\) |
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Time = 0.41 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {1}{3} \left (1+2 x+e^{5+x} (1+x)-\log \left (\frac {x}{4}\right )\right ) \, dx=\frac {1}{3} \, x^{2} + \frac {1}{3} \, x e^{\left (x + 5\right )} - \frac {1}{3} \, x \log \left (\frac {1}{4} \, x\right ) + \frac {2}{3} \, x \]
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Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {1}{3} \left (1+2 x+e^{5+x} (1+x)-\log \left (\frac {x}{4}\right )\right ) \, dx=\frac {x^{2}}{3} + \frac {x e^{x + 5}}{3} - \frac {x \log {\left (\frac {x}{4} \right )}}{3} + \frac {2 x}{3} \]
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Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {1}{3} \left (1+2 x+e^{5+x} (1+x)-\log \left (\frac {x}{4}\right )\right ) \, dx=\frac {1}{3} \, x^{2} + \frac {1}{3} \, x e^{\left (x + 5\right )} - \frac {1}{3} \, x \log \left (\frac {1}{4} \, x\right ) + \frac {2}{3} \, x \]
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Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {1}{3} \left (1+2 x+e^{5+x} (1+x)-\log \left (\frac {x}{4}\right )\right ) \, dx=\frac {1}{3} \, x^{2} + \frac {1}{3} \, x e^{\left (x + 5\right )} - \frac {1}{3} \, x \log \left (\frac {1}{4} \, x\right ) + \frac {2}{3} \, x \]
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Time = 13.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {1}{3} \left (1+2 x+e^{5+x} (1+x)-\log \left (\frac {x}{4}\right )\right ) \, dx=\frac {x\,\left (x+{\mathrm {e}}^{x+5}-\ln \left (\frac {x}{4}\right )+2\right )}{3} \]
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