Integrand size = 38, antiderivative size = 28 \[ \int \frac {16-8 x+x^2+e^{3 x} \left (-13 x+3 x^2\right )}{16 x-8 x^2+x^3} \, dx=4-e^2+\frac {e^{3 x}}{-4+x}+\log \left (-x+\frac {x}{e^4}\right ) \]
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Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1608, 27, 6874, 2228} \[ \int \frac {16-8 x+x^2+e^{3 x} \left (-13 x+3 x^2\right )}{16 x-8 x^2+x^3} \, dx=\log (x)-\frac {e^{3 x}}{4-x} \]
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Rule 27
Rule 1608
Rule 2228
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {16-8 x+x^2+e^{3 x} \left (-13 x+3 x^2\right )}{x \left (16-8 x+x^2\right )} \, dx \\ & = \int \frac {16-8 x+x^2+e^{3 x} \left (-13 x+3 x^2\right )}{(-4+x)^2 x} \, dx \\ & = \int \left (\frac {1}{x}+\frac {e^{3 x} (-13+3 x)}{(-4+x)^2}\right ) \, dx \\ & = \log (x)+\int \frac {e^{3 x} (-13+3 x)}{(-4+x)^2} \, dx \\ & = -\frac {e^{3 x}}{4-x}+\log (x) \\ \end{align*}
Time = 0.41 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.50 \[ \int \frac {16-8 x+x^2+e^{3 x} \left (-13 x+3 x^2\right )}{16 x-8 x^2+x^3} \, dx=\frac {e^{3 x}}{-4+x}+\log (x) \]
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Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.50
method | result | size |
norman | \(\frac {{\mathrm e}^{3 x}}{x -4}+\ln \left (x \right )\) | \(14\) |
risch | \(\frac {{\mathrm e}^{3 x}}{x -4}+\ln \left (x \right )\) | \(14\) |
parts | \(\ln \left (x \right )+\frac {3 \,{\mathrm e}^{3 x}}{3 x -12}\) | \(17\) |
derivativedivides | \(\ln \left (3 x \right )+\frac {3 \,{\mathrm e}^{3 x}}{3 x -12}\) | \(19\) |
default | \(\ln \left (3 x \right )+\frac {3 \,{\mathrm e}^{3 x}}{3 x -12}\) | \(19\) |
parallelrisch | \(\frac {x \ln \left (x \right )-4 \ln \left (x \right )+{\mathrm e}^{3 x}}{x -4}\) | \(20\) |
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Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {16-8 x+x^2+e^{3 x} \left (-13 x+3 x^2\right )}{16 x-8 x^2+x^3} \, dx=\frac {{\left (x - 4\right )} \log \left (x\right ) + e^{\left (3 \, x\right )}}{x - 4} \]
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Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.36 \[ \int \frac {16-8 x+x^2+e^{3 x} \left (-13 x+3 x^2\right )}{16 x-8 x^2+x^3} \, dx=\log {\left (x \right )} + \frac {e^{3 x}}{x - 4} \]
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\[ \int \frac {16-8 x+x^2+e^{3 x} \left (-13 x+3 x^2\right )}{16 x-8 x^2+x^3} \, dx=\int { \frac {x^{2} + {\left (3 \, x^{2} - 13 \, x\right )} e^{\left (3 \, x\right )} - 8 \, x + 16}{x^{3} - 8 \, x^{2} + 16 \, x} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {16-8 x+x^2+e^{3 x} \left (-13 x+3 x^2\right )}{16 x-8 x^2+x^3} \, dx=\frac {x \log \left (x\right ) + e^{\left (3 \, x\right )} - 4 \, \log \left (x\right )}{x - 4} \]
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Time = 0.09 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.46 \[ \int \frac {16-8 x+x^2+e^{3 x} \left (-13 x+3 x^2\right )}{16 x-8 x^2+x^3} \, dx=\ln \left (x\right )+\frac {{\mathrm {e}}^{3\,x}}{x-4} \]
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