Integrand size = 43, antiderivative size = 29 \[ \int \frac {100-40 e^2+4 e^4-x^3+\left (10 x^4+2 x^5\right ) \log (x)+x^5 \log ^2(x)}{x^5} \, dx=\frac {-\frac {\left (5-e^2\right )^2}{x^2}+x}{x^2}+(5+x) \log ^2(x) \]
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Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {14, 2388, 2338, 2332, 2333} \[ \int \frac {100-40 e^2+4 e^4-x^3+\left (10 x^4+2 x^5\right ) \log (x)+x^5 \log ^2(x)}{x^5} \, dx=-\frac {\left (5-e^2\right )^2}{x^4}+\frac {1}{x}+x \log ^2(x)+5 \log ^2(x) \]
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Rule 14
Rule 2332
Rule 2333
Rule 2338
Rule 2388
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {100-40 e^2+4 e^4-x^3}{x^5}+\frac {2 (5+x) \log (x)}{x}+\log ^2(x)\right ) \, dx \\ & = 2 \int \frac {(5+x) \log (x)}{x} \, dx+\int \frac {100-40 e^2+4 e^4-x^3}{x^5} \, dx+\int \log ^2(x) \, dx \\ & = x \log ^2(x)+10 \int \frac {\log (x)}{x} \, dx+\int \left (\frac {4 \left (-5+e^2\right )^2}{x^5}-\frac {1}{x^2}\right ) \, dx \\ & = -\frac {\left (5-e^2\right )^2}{x^4}+\frac {1}{x}+5 \log ^2(x)+x \log ^2(x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {100-40 e^2+4 e^4-x^3+\left (10 x^4+2 x^5\right ) \log (x)+x^5 \log ^2(x)}{x^5} \, dx=-\frac {25}{x^4}+\frac {10 e^2}{x^4}-\frac {e^4}{x^4}+\frac {1}{x}+5 \log ^2(x)+x \log ^2(x) \]
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Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\left (5+x \right ) \ln \left (x \right )^{2}-\frac {-x^{3}+{\mathrm e}^{4}-10 \,{\mathrm e}^{2}+25}{x^{4}}\) | \(28\) |
parts | \(x \ln \left (x \right )^{2}+\frac {1}{x}-\frac {-40 \,{\mathrm e}^{2}+4 \,{\mathrm e}^{4}+100}{4 x^{4}}+5 \ln \left (x \right )^{2}\) | \(32\) |
norman | \(\frac {x^{5} \ln \left (x \right )^{2}+5 x^{4} \ln \left (x \right )^{2}+x^{3}+10 \,{\mathrm e}^{2}-{\mathrm e}^{4}-25}{x^{4}}\) | \(37\) |
default | \(x \ln \left (x \right )^{2}+5 \ln \left (x \right )^{2}+\frac {1}{x}-\frac {{\mathrm e}^{4}}{x^{4}}+\frac {10 \,{\mathrm e}^{2}}{x^{4}}-\frac {25}{x^{4}}\) | \(38\) |
parallelrisch | \(-\frac {-x^{5} \ln \left (x \right )^{2}-5 x^{4} \ln \left (x \right )^{2}+25-x^{3}+{\mathrm e}^{4}-10 \,{\mathrm e}^{2}}{x^{4}}\) | \(39\) |
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Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {100-40 e^2+4 e^4-x^3+\left (10 x^4+2 x^5\right ) \log (x)+x^5 \log ^2(x)}{x^5} \, dx=\frac {x^{3} + {\left (x^{5} + 5 \, x^{4}\right )} \log \left (x\right )^{2} - e^{4} + 10 \, e^{2} - 25}{x^{4}} \]
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Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {100-40 e^2+4 e^4-x^3+\left (10 x^4+2 x^5\right ) \log (x)+x^5 \log ^2(x)}{x^5} \, dx=\left (x + 5\right ) \log {\left (x \right )}^{2} - \frac {- x^{3} - 10 e^{2} + 25 + e^{4}}{x^{4}} \]
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Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {100-40 e^2+4 e^4-x^3+\left (10 x^4+2 x^5\right ) \log (x)+x^5 \log ^2(x)}{x^5} \, dx={\left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 2\right )} x + 2 \, x \log \left (x\right ) + 5 \, \log \left (x\right )^{2} - 2 \, x + \frac {1}{x} - \frac {e^{4}}{x^{4}} + \frac {10 \, e^{2}}{x^{4}} - \frac {25}{x^{4}} \]
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Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {100-40 e^2+4 e^4-x^3+\left (10 x^4+2 x^5\right ) \log (x)+x^5 \log ^2(x)}{x^5} \, dx=\frac {x^{5} \log \left (x\right )^{2} + 5 \, x^{4} \log \left (x\right )^{2} + x^{3} - e^{4} + 10 \, e^{2} - 25}{x^{4}} \]
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Time = 8.40 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {100-40 e^2+4 e^4-x^3+\left (10 x^4+2 x^5\right ) \log (x)+x^5 \log ^2(x)}{x^5} \, dx=x\,{\ln \left (x\right )}^2+5\,{\ln \left (x\right )}^2+\frac {x^4-x\,{\left ({\mathrm {e}}^2-5\right )}^2}{x^5} \]
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