Integrand size = 29, antiderivative size = 17 \[ \int \frac {e^{25} (1+4 x)}{4 \left (-4 x+4 x^2+x \log (6 x)\right )} \, dx=\frac {1}{4} e^{25} \log (-4+4 x+\log (6 x)) \]
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\[ \int \frac {e^{25} (1+4 x)}{4 \left (-4 x+4 x^2+x \log (6 x)\right )} \, dx=\int \frac {e^{25} (1+4 x)}{4 \left (-4 x+4 x^2+x \log (6 x)\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} e^{25} \int \frac {1+4 x}{-4 x+4 x^2+x \log (6 x)} \, dx \\ & = \frac {1}{4} e^{25} \int \left (\frac {4}{-4+4 x+\log (6 x)}+\frac {1}{x (-4+4 x+\log (6 x))}\right ) \, dx \\ & = \frac {1}{4} e^{25} \int \frac {1}{x (-4+4 x+\log (6 x))} \, dx+e^{25} \int \frac {1}{-4+4 x+\log (6 x)} \, dx \\ & = \frac {1}{4} e^{25} \int \frac {1}{x (-4+4 x+\log (6 x))} \, dx+\frac {1}{2} e^{25} \text {Subst}\left (\int \frac {1}{-4+2 x+\log (3 x)} \, dx,x,2 x\right ) \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {e^{25} (1+4 x)}{4 \left (-4 x+4 x^2+x \log (6 x)\right )} \, dx=\frac {1}{4} e^{25} \log (12-12 x-3 \log (6 x)) \]
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Time = 1.66 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {\ln \left (\ln \left (6 x \right )+4 x -4\right ) {\mathrm e}^{25}}{4}\) | \(15\) |
risch | \(\frac {\ln \left (\ln \left (6 x \right )+4 x -4\right ) {\mathrm e}^{25}}{4}\) | \(15\) |
default | \(\frac {{\mathrm e}^{25} \ln \left (\ln \left (2\right )+\ln \left (3\right )+\ln \left (x \right )+4 x -4\right )}{4}\) | \(17\) |
parallelrisch | \({\mathrm e}^{\ln \left (\ln \left (\ln \left (6 x \right )+4 x -4\right )\right )-2 \ln \left (2\right )+25}\) | \(19\) |
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Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {e^{25} (1+4 x)}{4 \left (-4 x+4 x^2+x \log (6 x)\right )} \, dx=e^{\left (-2 \, \log \left (2\right ) + 25\right )} \log \left (4 \, x + \log \left (6 \, x\right ) - 4\right ) \]
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Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{25} (1+4 x)}{4 \left (-4 x+4 x^2+x \log (6 x)\right )} \, dx=\frac {e^{25} \log {\left (4 x + \log {\left (6 x \right )} - 4 \right )}}{4} \]
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Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {e^{25} (1+4 x)}{4 \left (-4 x+4 x^2+x \log (6 x)\right )} \, dx=\frac {1}{4} \, e^{25} \log \left (4 \, x + \log \left (3\right ) + \log \left (2\right ) + \log \left (x\right ) - 4\right ) \]
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Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{25} (1+4 x)}{4 \left (-4 x+4 x^2+x \log (6 x)\right )} \, dx=\frac {1}{4} \, e^{25} \log \left (4 \, x + \log \left (6\right ) + \log \left (x\right ) - 4\right ) \]
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Time = 13.47 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{25} (1+4 x)}{4 \left (-4 x+4 x^2+x \log (6 x)\right )} \, dx=\frac {{\mathrm {e}}^{25}\,\ln \left (4\,x+\ln \left (6\,x\right )-4\right )}{4} \]
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