Integrand size = 33, antiderivative size = 25 \[ \int \frac {54-18 x+3 x^2+e^{x/3} x^3+18 x \log (2)}{3 x^3} \, dx=e^{x/3}-\frac {(-3+x-x \log (2))^2}{x^2}+\log (x) \]
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Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6, 12, 14, 2225} \[ \int \frac {54-18 x+3 x^2+e^{x/3} x^3+18 x \log (2)}{3 x^3} \, dx=-\frac {9}{x^2}+e^{x/3}+\log (x)+\frac {6 (1-\log (2))}{x} \]
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Rule 6
Rule 12
Rule 14
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \int \frac {54+3 x^2+e^{x/3} x^3+x (-18+18 \log (2))}{3 x^3} \, dx \\ & = \frac {1}{3} \int \frac {54+3 x^2+e^{x/3} x^3+x (-18+18 \log (2))}{x^3} \, dx \\ & = \frac {1}{3} \int \left (e^{x/3}+\frac {3 \left (18+x^2-6 x (1-\log (2))\right )}{x^3}\right ) \, dx \\ & = \frac {1}{3} \int e^{x/3} \, dx+\int \frac {18+x^2-6 x (1-\log (2))}{x^3} \, dx \\ & = e^{x/3}+\int \left (\frac {18}{x^3}+\frac {1}{x}+\frac {6 (-1+\log (2))}{x^2}\right ) \, dx \\ & = e^{x/3}-\frac {9}{x^2}+\frac {6 (1-\log (2))}{x}+\log (x) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {54-18 x+3 x^2+e^{x/3} x^3+18 x \log (2)}{3 x^3} \, dx=e^{x/3}-\frac {9}{x^2}-\frac {6 (-1+\log (2))}{x}+\log (x) \]
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Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\frac {\left (-18 \ln \left (2\right )+18\right ) x -27}{3 x^{2}}+\ln \left (x \right )+{\mathrm e}^{\frac {x}{3}}\) | \(23\) |
parts | \(\ln \left (x \right )-\frac {6 \ln \left (2\right )-6}{x}-\frac {9}{x^{2}}+{\mathrm e}^{\frac {x}{3}}\) | \(24\) |
norman | \(\frac {-9+x^{2} {\mathrm e}^{\frac {x}{3}}+\left (-6 \ln \left (2\right )+6\right ) x}{x^{2}}+\ln \left (x \right )\) | \(26\) |
derivativedivides | \(\ln \left (\frac {x}{3}\right )-\frac {9}{x^{2}}+\frac {6}{x}-\frac {6 \ln \left (2\right )}{x}+{\mathrm e}^{\frac {x}{3}}\) | \(27\) |
default | \(\ln \left (\frac {x}{3}\right )-\frac {9}{x^{2}}+\frac {6}{x}-\frac {6 \ln \left (2\right )}{x}+{\mathrm e}^{\frac {x}{3}}\) | \(27\) |
parallelrisch | \(-\frac {-3 x^{2} \ln \left (x \right )-3 x^{2} {\mathrm e}^{\frac {x}{3}}+27+18 x \ln \left (2\right )-18 x}{3 x^{2}}\) | \(32\) |
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Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {54-18 x+3 x^2+e^{x/3} x^3+18 x \log (2)}{3 x^3} \, dx=\frac {x^{2} e^{\left (\frac {1}{3} \, x\right )} + x^{2} \log \left (x\right ) - 6 \, x \log \left (2\right ) + 6 \, x - 9}{x^{2}} \]
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Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {54-18 x+3 x^2+e^{x/3} x^3+18 x \log (2)}{3 x^3} \, dx=e^{\frac {x}{3}} + \log {\left (x \right )} + \frac {x \left (6 - 6 \log {\left (2 \right )}\right ) - 9}{x^{2}} \]
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Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {54-18 x+3 x^2+e^{x/3} x^3+18 x \log (2)}{3 x^3} \, dx=-\frac {6 \, \log \left (2\right )}{x} + \frac {6}{x} - \frac {9}{x^{2}} + e^{\left (\frac {1}{3} \, x\right )} + \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {54-18 x+3 x^2+e^{x/3} x^3+18 x \log (2)}{3 x^3} \, dx=\frac {x^{2} e^{\left (\frac {1}{3} \, x\right )} + x^{2} \log \left (\frac {1}{3} \, x\right ) - 6 \, x \log \left (2\right ) + 6 \, x - 9}{x^{2}} \]
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Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {54-18 x+3 x^2+e^{x/3} x^3+18 x \log (2)}{3 x^3} \, dx={\mathrm {e}}^{x/3}+\ln \left (x\right )-\frac {x\,\left (\ln \left (64\right )-6\right )+9}{x^2} \]
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