Integrand size = 77, antiderivative size = 15 \[ \int \frac {2 e^{5/2} x^2+e^{5/4} \left (2 x+2 x^2\right )+\left (2+4 x+2 x^2+e^{5/4} \left (2 x+4 x^2\right )\right ) \log (x)+\left (2 x+2 x^2\right ) \log ^2(x)}{e^{5/2} x} \, dx=\left (x+\frac {(1+x) \log (x)}{e^{5/4}}\right )^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(179\) vs. \(2(15)=30\).
Time = 0.11 (sec) , antiderivative size = 179, normalized size of antiderivative = 11.93, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.117, Rules used = {12, 14, 2404, 2332, 2338, 2341, 2367, 2333, 2342} \[ \int \frac {2 e^{5/2} x^2+e^{5/4} \left (2 x+2 x^2\right )+\left (2+4 x+2 x^2+e^{5/4} \left (2 x+4 x^2\right )\right ) \log (x)+\left (2 x+2 x^2\right ) \log ^2(x)}{e^{5/2} x} \, dx=-\frac {\left (1+2 e^{5/4}\right ) x^2}{2 e^{5/2}}+\frac {x^2}{2 e^{5/2}}+\frac {x^2 \log ^2(x)}{e^{5/2}}+\frac {\left (1+2 e^{5/4}\right ) x^2 \log (x)}{e^{5/2}}-\frac {x^2 \log (x)}{e^{5/2}}-\frac {2 \left (2+e^{5/4}\right ) x}{e^{5/2}}+\frac {4 x}{e^{5/2}}+\frac {\left (\left (1+e^{5/4}\right ) x+1\right )^2}{e^{5/4} \left (1+e^{5/4}\right )}+\frac {2 x \log ^2(x)}{e^{5/2}}+\frac {\log ^2(x)}{e^{5/2}}+\frac {2 \left (2+e^{5/4}\right ) x \log (x)}{e^{5/2}}-\frac {4 x \log (x)}{e^{5/2}} \]
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Rule 12
Rule 14
Rule 2332
Rule 2333
Rule 2338
Rule 2341
Rule 2342
Rule 2367
Rule 2404
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {2 e^{5/2} x^2+e^{5/4} \left (2 x+2 x^2\right )+\left (2+4 x+2 x^2+e^{5/4} \left (2 x+4 x^2\right )\right ) \log (x)+\left (2 x+2 x^2\right ) \log ^2(x)}{x} \, dx}{e^{5/2}} \\ & = \frac {\int \left (2 e^{5/4} \left (1+\left (1+e^{5/4}\right ) x\right )+\frac {2 \left (1+\left (2+e^{5/4}\right ) x+\left (1+2 e^{5/4}\right ) x^2\right ) \log (x)}{x}+2 (1+x) \log ^2(x)\right ) \, dx}{e^{5/2}} \\ & = \frac {\left (1+\left (1+e^{5/4}\right ) x\right )^2}{e^{5/4} \left (1+e^{5/4}\right )}+\frac {2 \int \frac {\left (1+\left (2+e^{5/4}\right ) x+\left (1+2 e^{5/4}\right ) x^2\right ) \log (x)}{x} \, dx}{e^{5/2}}+\frac {2 \int (1+x) \log ^2(x) \, dx}{e^{5/2}} \\ & = \frac {\left (1+\left (1+e^{5/4}\right ) x\right )^2}{e^{5/4} \left (1+e^{5/4}\right )}+\frac {2 \int \left (\left (2+e^{5/4}\right ) \log (x)+\frac {\log (x)}{x}+\left (1+2 e^{5/4}\right ) x \log (x)\right ) \, dx}{e^{5/2}}+\frac {2 \int \left (\log ^2(x)+x \log ^2(x)\right ) \, dx}{e^{5/2}} \\ & = \frac {\left (1+\left (1+e^{5/4}\right ) x\right )^2}{e^{5/4} \left (1+e^{5/4}\right )}+\frac {2 \int \frac {\log (x)}{x} \, dx}{e^{5/2}}+\frac {2 \int \log ^2(x) \, dx}{e^{5/2}}+\frac {2 \int x \log ^2(x) \, dx}{e^{5/2}}+\frac {\left (2 \left (2+e^{5/4}\right )\right ) \int \log (x) \, dx}{e^{5/2}}+\frac {\left (2 \left (1+2 e^{5/4}\right )\right ) \int x \log (x) \, dx}{e^{5/2}} \\ & = -\frac {2 \left (2+e^{5/4}\right ) x}{e^{5/2}}-\frac {\left (1+2 e^{5/4}\right ) x^2}{2 e^{5/2}}+\frac {\left (1+\left (1+e^{5/4}\right ) x\right )^2}{e^{5/4} \left (1+e^{5/4}\right )}+\frac {2 \left (2+e^{5/4}\right ) x \log (x)}{e^{5/2}}+\frac {\left (1+2 e^{5/4}\right ) x^2 \log (x)}{e^{5/2}}+\frac {\log ^2(x)}{e^{5/2}}+\frac {2 x \log ^2(x)}{e^{5/2}}+\frac {x^2 \log ^2(x)}{e^{5/2}}-\frac {2 \int x \log (x) \, dx}{e^{5/2}}-\frac {4 \int \log (x) \, dx}{e^{5/2}} \\ & = \frac {4 x}{e^{5/2}}-\frac {2 \left (2+e^{5/4}\right ) x}{e^{5/2}}+\frac {x^2}{2 e^{5/2}}-\frac {\left (1+2 e^{5/4}\right ) x^2}{2 e^{5/2}}+\frac {\left (1+\left (1+e^{5/4}\right ) x\right )^2}{e^{5/4} \left (1+e^{5/4}\right )}-\frac {4 x \log (x)}{e^{5/2}}+\frac {2 \left (2+e^{5/4}\right ) x \log (x)}{e^{5/2}}-\frac {x^2 \log (x)}{e^{5/2}}+\frac {\left (1+2 e^{5/4}\right ) x^2 \log (x)}{e^{5/2}}+\frac {\log ^2(x)}{e^{5/2}}+\frac {2 x \log ^2(x)}{e^{5/2}}+\frac {x^2 \log ^2(x)}{e^{5/2}} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47 \[ \int \frac {2 e^{5/2} x^2+e^{5/4} \left (2 x+2 x^2\right )+\left (2+4 x+2 x^2+e^{5/4} \left (2 x+4 x^2\right )\right ) \log (x)+\left (2 x+2 x^2\right ) \log ^2(x)}{e^{5/2} x} \, dx=\frac {\left (e^{5/4} x+(1+x) \log (x)\right )^2}{e^{5/2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(29\) vs. \(2(14)=28\).
Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 2.00
method | result | size |
risch | \({\mathrm e}^{-\frac {5}{2}} \left (x^{2}+2 x +1\right ) \ln \left (x \right )^{2}+2 \,{\mathrm e}^{-\frac {5}{4}} \left (1+x \right ) x \ln \left (x \right )+x^{2}\) | \(30\) |
parallelrisch | \({\mathrm e}^{-\frac {5}{2}} \left (x^{2} {\mathrm e}^{\frac {5}{2}}+2 \ln \left (x \right ) {\mathrm e}^{\frac {5}{4}} x^{2}+x^{2} \ln \left (x \right )^{2}+2 \ln \left (x \right ) {\mathrm e}^{\frac {5}{4}} x +2 x \ln \left (x \right )^{2}+\ln \left (x \right )^{2}\right )\) | \(50\) |
norman | \(\left ({\mathrm e}^{-\frac {5}{4}} \ln \left (x \right )^{2}+{\mathrm e}^{\frac {5}{4}} x^{2}+{\mathrm e}^{-\frac {5}{4}} x^{2} \ln \left (x \right )^{2}+2 x \ln \left (x \right )+2 x^{2} \ln \left (x \right )+2 \,{\mathrm e}^{-\frac {5}{4}} x \ln \left (x \right )^{2}\right ) {\mathrm e}^{-\frac {5}{4}}\) | \(57\) |
default | \({\mathrm e}^{-\frac {5}{2}} \left (x^{2} \ln \left (x \right )^{2}+4 \,{\mathrm e}^{\frac {5}{4}} \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+x^{2} {\mathrm e}^{\frac {5}{2}}+2 x \ln \left (x \right )^{2}+2 \,{\mathrm e}^{\frac {5}{4}} \left (x \ln \left (x \right )-x \right )+{\mathrm e}^{\frac {5}{4}} x^{2}+2 x \,{\mathrm e}^{\frac {5}{4}}+\ln \left (x \right )^{2}\right )\) | \(74\) |
parts | \(2 \,{\mathrm e}^{-\frac {5}{4}} \left (\frac {{\mathrm e}^{\frac {5}{4}} x^{2}}{2}+\frac {x^{2}}{2}+x \right )+2 \,{\mathrm e}^{-\frac {5}{2}} \left (\frac {x^{2} \ln \left (x \right )^{2}}{2}-\frac {x^{2} \ln \left (x \right )}{2}+\frac {x^{2}}{4}+x \ln \left (x \right )^{2}-2 x \ln \left (x \right )+2 x \right )+2 \,{\mathrm e}^{-\frac {5}{2}} \left (2 \,{\mathrm e}^{\frac {5}{4}} \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+{\mathrm e}^{\frac {5}{4}} \left (x \ln \left (x \right )-x \right )+\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}+2 x \ln \left (x \right )-2 x +\frac {\ln \left (x \right )^{2}}{2}\right )\) | \(125\) |
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Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (12) = 24\).
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.27 \[ \int \frac {2 e^{5/2} x^2+e^{5/4} \left (2 x+2 x^2\right )+\left (2+4 x+2 x^2+e^{5/4} \left (2 x+4 x^2\right )\right ) \log (x)+\left (2 x+2 x^2\right ) \log ^2(x)}{e^{5/2} x} \, dx={\left (x^{2} e^{\frac {5}{2}} + 2 \, {\left (x^{2} + x\right )} e^{\frac {5}{4}} \log \left (x\right ) + {\left (x^{2} + 2 \, x + 1\right )} \log \left (x\right )^{2}\right )} e^{\left (-\frac {5}{2}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (14) = 28\).
Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.47 \[ \int \frac {2 e^{5/2} x^2+e^{5/4} \left (2 x+2 x^2\right )+\left (2+4 x+2 x^2+e^{5/4} \left (2 x+4 x^2\right )\right ) \log (x)+\left (2 x+2 x^2\right ) \log ^2(x)}{e^{5/2} x} \, dx=x^{2} + \frac {\left (2 x^{2} + 2 x\right ) \log {\left (x \right )}}{e^{\frac {5}{4}}} + \frac {\left (x^{2} + 2 x + 1\right ) \log {\left (x \right )}^{2}}{e^{\frac {5}{2}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (12) = 24\).
Time = 0.19 (sec) , antiderivative size = 108, normalized size of antiderivative = 7.20 \[ \int \frac {2 e^{5/2} x^2+e^{5/4} \left (2 x+2 x^2\right )+\left (2+4 x+2 x^2+e^{5/4} \left (2 x+4 x^2\right )\right ) \log (x)+\left (2 x+2 x^2\right ) \log ^2(x)}{e^{5/2} x} \, dx=\frac {1}{2} \, {\left ({\left (2 \, \log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 1\right )} x^{2} + 2 \, x^{2} e^{\frac {5}{2}} + 2 \, x^{2} e^{\frac {5}{4}} + 2 \, x^{2} \log \left (x\right ) + 4 \, {\left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 2\right )} x - x^{2} + 2 \, {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} e^{\frac {5}{4}} + 4 \, {\left (x \log \left (x\right ) - x\right )} e^{\frac {5}{4}} + 4 \, x e^{\frac {5}{4}} + 8 \, x \log \left (x\right ) + 2 \, \log \left (x\right )^{2} - 8 \, x\right )} e^{\left (-\frac {5}{2}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (12) = 24\).
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 3.00 \[ \int \frac {2 e^{5/2} x^2+e^{5/4} \left (2 x+2 x^2\right )+\left (2+4 x+2 x^2+e^{5/4} \left (2 x+4 x^2\right )\right ) \log (x)+\left (2 x+2 x^2\right ) \log ^2(x)}{e^{5/2} x} \, dx={\left (2 \, x^{2} e^{\frac {5}{4}} \log \left (x\right ) + x^{2} \log \left (x\right )^{2} + x^{2} e^{\frac {5}{2}} + 2 \, x e^{\frac {5}{4}} \log \left (x\right ) + 2 \, x \log \left (x\right )^{2} + \log \left (x\right )^{2}\right )} e^{\left (-\frac {5}{2}\right )} \]
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Time = 14.64 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {2 e^{5/2} x^2+e^{5/4} \left (2 x+2 x^2\right )+\left (2+4 x+2 x^2+e^{5/4} \left (2 x+4 x^2\right )\right ) \log (x)+\left (2 x+2 x^2\right ) \log ^2(x)}{e^{5/2} x} \, dx={\mathrm {e}}^{-\frac {5}{2}}\,{\left (\ln \left (x\right )+x\,{\mathrm {e}}^{5/4}+x\,\ln \left (x\right )\right )}^2 \]
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