Integrand size = 172, antiderivative size = 33 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{5+\log (5)+\log \left (-4+3 \left (e^2+e^{\frac {x-4 x^2}{x}}-x\right )\right )} \]
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Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6820, 12, 6818} \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{\log \left (-3 x+3 e^{1-4 x}+3 e^2-4\right )+5+\log (5)} \]
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Rule 12
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {15 \left (4 e+e^{4 x}\right )}{\left (3 e+3 e^{2+4 x}-e^{4 x} (4+3 x)\right ) \left (5 \left (1+\frac {\log (5)}{5}\right )+\log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )\right )^2} \, dx \\ & = 15 \int \frac {4 e+e^{4 x}}{\left (3 e+3 e^{2+4 x}-e^{4 x} (4+3 x)\right ) \left (5 \left (1+\frac {\log (5)}{5}\right )+\log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )\right )^2} \, dx \\ & = \frac {5}{5+\log (5)+\log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{5+\log (5)+\log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \]
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Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82
method | result | size |
norman | \(\frac {5}{\ln \left (5\right )+\ln \left (3 \,{\mathrm e}^{-4 x +1}+3 \,{\mathrm e}^{2}-3 x -4\right )+5}\) | \(27\) |
risch | \(\frac {5}{\ln \left (5\right )+\ln \left (3 \,{\mathrm e}^{-4 x +1}+3 \,{\mathrm e}^{2}-3 x -4\right )+5}\) | \(27\) |
parallelrisch | \(\frac {5}{\ln \left (5\right )+\ln \left (3 \,{\mathrm e}^{-4 x +1}+3 \,{\mathrm e}^{2}-3 x -4\right )+5}\) | \(27\) |
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Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{\log \left (5\right ) + \log \left (-3 \, x + 3 \, e^{2} + 3 \, e^{\left (-4 \, x + 1\right )} - 4\right ) + 5} \]
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Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{\log {\left (- 3 x + 3 e^{1 - 4 x} - 4 + 3 e^{2} \right )} + \log {\left (5 \right )} + 5} \]
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Time = 0.49 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=-\frac {5}{4 \, x - \log \left (5\right ) - \log \left (-{\left (3 \, x - 3 \, e^{2} + 4\right )} e^{\left (4 \, x\right )} + 3 \, e\right ) - 5} \]
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Time = 0.84 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{\log \left (5\right ) + \log \left (-3 \, x + 3 \, e^{2} + 3 \, e^{\left (-4 \, x + 1\right )} - 4\right ) + 5} \]
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Time = 21.55 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{\ln \left (15\,{\mathrm {e}}^2-15\,x+15\,{\mathrm {e}}^{-4\,x}\,\mathrm {e}-20\right )+5} \]
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