Integrand size = 43, antiderivative size = 30 \[ \int \frac {-16 e^{2 e^2}-80 x^4+40 e^5 x^4-5 e^{10} x^4+5 x^5}{5 x^5} \, dx=5+\frac {4 e^{2 e^2}}{5 x^4}+x-\left (4-e^5\right )^2 \log (x) \]
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Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {6, 12, 14} \[ \int \frac {-16 e^{2 e^2}-80 x^4+40 e^5 x^4-5 e^{10} x^4+5 x^5}{5 x^5} \, dx=\frac {4 e^{2 e^2}}{5 x^4}+x-\left (4-e^5\right )^2 \log (x) \]
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Rule 6
Rule 12
Rule 14
Rubi steps \begin{align*} \text {integral}& = \int \frac {-16 e^{2 e^2}-5 e^{10} x^4+\left (-80+40 e^5\right ) x^4+5 x^5}{5 x^5} \, dx \\ & = \int \frac {-16 e^{2 e^2}+\left (-80+40 e^5-5 e^{10}\right ) x^4+5 x^5}{5 x^5} \, dx \\ & = \frac {1}{5} \int \frac {-16 e^{2 e^2}+\left (-80+40 e^5-5 e^{10}\right ) x^4+5 x^5}{x^5} \, dx \\ & = \frac {1}{5} \int \left (5-\frac {16 e^{2 e^2}}{x^5}-\frac {5 \left (-4+e^5\right )^2}{x}\right ) \, dx \\ & = \frac {4 e^{2 e^2}}{5 x^4}+x-\left (4-e^5\right )^2 \log (x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {-16 e^{2 e^2}-80 x^4+40 e^5 x^4-5 e^{10} x^4+5 x^5}{5 x^5} \, dx=\frac {4 e^{2 e^2}}{5 x^4}+x-16 \log (x)+8 e^5 \log (x)-e^{10} \log (x) \]
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Time = 0.45 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90
method | result | size |
default | \(x +\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}}{5 x^{4}}+\frac {\left (40 \,{\mathrm e}^{5}-5 \,{\mathrm e}^{10}-80\right ) \ln \left (x \right )}{5}\) | \(27\) |
risch | \(x +\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}}{5 x^{4}}+8 \,{\mathrm e}^{5} \ln \left (x \right )-{\mathrm e}^{10} \ln \left (x \right )-16 \ln \left (x \right )\) | \(29\) |
norman | \(\frac {x^{5}+\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}}{5}}{x^{4}}+\left (8 \,{\mathrm e}^{5}-{\mathrm e}^{10}-16\right ) \ln \left (x \right )\) | \(32\) |
parallelrisch | \(-\frac {5 \ln \left (x \right ) x^{4} {\mathrm e}^{10}-40 \ln \left (x \right ) x^{4} {\mathrm e}^{5}+80 x^{4} \ln \left (x \right )-5 x^{5}-4 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}}{5 x^{4}}\) | \(46\) |
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Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {-16 e^{2 e^2}-80 x^4+40 e^5 x^4-5 e^{10} x^4+5 x^5}{5 x^5} \, dx=\frac {5 \, x^{5} - 5 \, {\left (x^{4} e^{10} - 8 \, x^{4} e^{5} + 16 \, x^{4}\right )} \log \left (x\right ) + 4 \, e^{\left (2 \, e^{2}\right )}}{5 \, x^{4}} \]
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Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-16 e^{2 e^2}-80 x^4+40 e^5 x^4-5 e^{10} x^4+5 x^5}{5 x^5} \, dx=x - \left (-4 + e^{5}\right )^{2} \log {\left (x \right )} + \frac {4 e^{2 e^{2}}}{5 x^{4}} \]
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Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-16 e^{2 e^2}-80 x^4+40 e^5 x^4-5 e^{10} x^4+5 x^5}{5 x^5} \, dx=-{\left (e^{10} - 8 \, e^{5} + 16\right )} \log \left (x\right ) + x + \frac {4 \, e^{\left (2 \, e^{2}\right )}}{5 \, x^{4}} \]
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Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {-16 e^{2 e^2}-80 x^4+40 e^5 x^4-5 e^{10} x^4+5 x^5}{5 x^5} \, dx=-{\left (e^{10} - 8 \, e^{5} + 16\right )} \log \left ({\left | x \right |}\right ) + x + \frac {4 \, e^{\left (2 \, e^{2}\right )}}{5 \, x^{4}} \]
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Time = 9.97 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-16 e^{2 e^2}-80 x^4+40 e^5 x^4-5 e^{10} x^4+5 x^5}{5 x^5} \, dx=x+\frac {4\,{\mathrm {e}}^{2\,{\mathrm {e}}^2}}{5\,x^4}-\ln \left (x\right )\,\left ({\mathrm {e}}^{10}-8\,{\mathrm {e}}^5+16\right ) \]
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