Integrand size = 48, antiderivative size = 23 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{3 x \left (4-\log \left (5+x-\frac {2+x^2}{x}\right )\right )} \]
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Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6838} \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{12 x} \left (-\frac {2-5 x}{x}\right )^{-3 x} \]
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Rule 6838
Rubi steps \begin{align*} \text {integral}& = e^{12 x} \left (-\frac {2-5 x}{x}\right )^{-3 x} \\ \end{align*}
Time = 0.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{12 x} \left (5-\frac {2}{x}\right )^{-3 x} \]
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Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83
method | result | size |
norman | \({\mathrm e}^{-3 x \ln \left (\frac {5 x -2}{x}\right )+12 x}\) | \(19\) |
risch | \(\left (\frac {5 x -2}{x}\right )^{-3 x} {\mathrm e}^{12 x}\) | \(19\) |
parallelrisch | \({\mathrm e}^{-3 x \ln \left (\frac {5 x -2}{x}\right )+12 x}\) | \(19\) |
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none
Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{\left (-3 \, x \log \left (\frac {5 \, x - 2}{x}\right ) + 12 \, x\right )} \]
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Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{- 3 x \log {\left (\frac {5 x - 2}{x} \right )} + 12 x} \]
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none
Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{\left (-3 \, x \log \left (5 \, x - 2\right ) + 3 \, x \log \left (x\right ) + 12 \, x\right )} \]
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none
Time = 0.33 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{\left (-3 \, x \log \left (-\frac {2}{x} + 5\right ) + 12 \, x\right )} \]
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Time = 8.35 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=\frac {{\mathrm {e}}^{12\,x}}{{\left (5-\frac {2}{x}\right )}^{3\,x}} \]
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