Integrand size = 33, antiderivative size = 25 \[ \int \frac {-30+384 e^4-165 x-80 x^2}{-25+384 e^4-5 x-80 x^2} \, dx=x+\log \left (1+\frac {1+x}{4}+4 \left (-\frac {24 e^4}{5}+x^2\right )\right ) \]
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Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {1671, 642} \[ \int \frac {-30+384 e^4-165 x-80 x^2}{-25+384 e^4-5 x-80 x^2} \, dx=\log \left (80 x^2+5 x-384 e^4+25\right )+x \]
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Rule 642
Rule 1671
Rubi steps \begin{align*} \text {integral}& = \int \left (1-\frac {5 (1+32 x)}{-25+384 e^4-5 x-80 x^2}\right ) \, dx \\ & = x-5 \int \frac {1+32 x}{-25+384 e^4-5 x-80 x^2} \, dx \\ & = x+\log \left (25-384 e^4+5 x+80 x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {-30+384 e^4-165 x-80 x^2}{-25+384 e^4-5 x-80 x^2} \, dx=x+\log \left (25-384 e^4+5 x+80 x^2\right ) \]
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Time = 0.13 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64
method | result | size |
parallelrisch | \(x +\ln \left (-\frac {24 \,{\mathrm e}^{4}}{5}+x^{2}+\frac {x}{16}+\frac {5}{16}\right )\) | \(16\) |
default | \(x +\ln \left (-384 \,{\mathrm e}^{4}+80 x^{2}+5 x +25\right )\) | \(18\) |
norman | \(x +\ln \left (384 \,{\mathrm e}^{4}-80 x^{2}-5 x -25\right )\) | \(18\) |
risch | \(x +\ln \left (-384 \,{\mathrm e}^{4}+80 x^{2}+5 x +25\right )\) | \(18\) |
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Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {-30+384 e^4-165 x-80 x^2}{-25+384 e^4-5 x-80 x^2} \, dx=x + \log \left (80 \, x^{2} + 5 \, x - 384 \, e^{4} + 25\right ) \]
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Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {-30+384 e^4-165 x-80 x^2}{-25+384 e^4-5 x-80 x^2} \, dx=x + \log {\left (80 x^{2} + 5 x - 384 e^{4} + 25 \right )} \]
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Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {-30+384 e^4-165 x-80 x^2}{-25+384 e^4-5 x-80 x^2} \, dx=x + \log \left (80 \, x^{2} + 5 \, x - 384 \, e^{4} + 25\right ) \]
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Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {-30+384 e^4-165 x-80 x^2}{-25+384 e^4-5 x-80 x^2} \, dx=x + \log \left ({\left | 80 \, x^{2} + 5 \, x - 384 \, e^{4} + 25 \right |}\right ) \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {-30+384 e^4-165 x-80 x^2}{-25+384 e^4-5 x-80 x^2} \, dx=x+\ln \left (80\,x^2+5\,x-384\,{\mathrm {e}}^4+25\right ) \]
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