Integrand size = 57, antiderivative size = 22 \[ \int \frac {-8+2 x^2+\frac {e^x \left (4 x^3+2 x^4\right )}{x}+\frac {e^{2 x} \left (2 x^4+2 x^5\right )}{x^2}-2 \log \left (\frac {x}{3}\right )}{x} \, dx=\left (x+e^x x\right )^2-\left (4+\log \left (\frac {x}{3}\right )\right )^2 \]
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Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68, number of steps used = 21, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {14, 2227, 2207, 2225, 2338} \[ \int \frac {-8+2 x^2+\frac {e^x \left (4 x^3+2 x^4\right )}{x}+\frac {e^{2 x} \left (2 x^4+2 x^5\right )}{x^2}-2 \log \left (\frac {x}{3}\right )}{x} \, dx=2 e^x x^2+e^{2 x} x^2+x^2-\log ^2(x)-2 (4-\log (3)) \log (x) \]
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Rule 14
Rule 2207
Rule 2225
Rule 2227
Rule 2338
Rubi steps \begin{align*} \text {integral}& = \int \left (2 e^{2 x} x (1+x)+2 e^x x (2+x)+\frac {2 \left (x^2-4 \left (1-\frac {\log (3)}{4}\right )-\log (x)\right )}{x}\right ) \, dx \\ & = 2 \int e^{2 x} x (1+x) \, dx+2 \int e^x x (2+x) \, dx+2 \int \frac {x^2-4 \left (1-\frac {\log (3)}{4}\right )-\log (x)}{x} \, dx \\ & = 2 \int \left (2 e^x x+e^x x^2\right ) \, dx+2 \int \left (e^{2 x} x+e^{2 x} x^2\right ) \, dx+2 \int \left (\frac {-4+x^2+\log (3)}{x}-\frac {\log (x)}{x}\right ) \, dx \\ & = 2 \int e^{2 x} x \, dx+2 \int e^x x^2 \, dx+2 \int e^{2 x} x^2 \, dx+2 \int \frac {-4+x^2+\log (3)}{x} \, dx-2 \int \frac {\log (x)}{x} \, dx+4 \int e^x x \, dx \\ & = 4 e^x x+e^{2 x} x+2 e^x x^2+e^{2 x} x^2-\log ^2(x)-2 \int e^{2 x} x \, dx+2 \int \left (x+\frac {-4+\log (3)}{x}\right ) \, dx-4 \int e^x \, dx-4 \int e^x x \, dx-\int e^{2 x} \, dx \\ & = -4 e^x-\frac {e^{2 x}}{2}+x^2+2 e^x x^2+e^{2 x} x^2-2 (4-\log (3)) \log (x)-\log ^2(x)+4 \int e^x \, dx+\int e^{2 x} \, dx \\ & = x^2+2 e^x x^2+e^{2 x} x^2-2 (4-\log (3)) \log (x)-\log ^2(x) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {-8+2 x^2+\frac {e^x \left (4 x^3+2 x^4\right )}{x}+\frac {e^{2 x} \left (2 x^4+2 x^5\right )}{x^2}-2 \log \left (\frac {x}{3}\right )}{x} \, dx=-16+\left (1+e^x\right )^2 x^2-8 \log \left (\frac {x}{3}\right )-\log ^2\left (\frac {x}{3}\right ) \]
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Time = 0.36 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64
method | result | size |
risch | \(-\ln \left (x \right )^{2}+x^{2}+2 \ln \left (3\right ) \ln \left (x \right )-8 \ln \left (x \right )+{\mathrm e}^{2 x} x^{2}+2 \,{\mathrm e}^{x} x^{2}\) | \(36\) |
default | \(x^{2}-8 \ln \left (x \right )-\ln \left (\frac {x}{3}\right )^{2}+2 \,{\mathrm e}^{x} x^{2}+{\mathrm e}^{2 x} x^{2}\) | \(37\) |
parts | \(x^{2}-8 \ln \left (x \right )-\ln \left (\frac {x}{3}\right )^{2}+2 \,{\mathrm e}^{x} x^{2}+{\mathrm e}^{2 x} x^{2}\) | \(37\) |
parallelrisch | \({\mathrm e}^{2 x} x^{2}+2 x^{3} {\mathrm e}^{x -\ln \left (x \right )}-\ln \left (\frac {x}{3}\right )^{2}-8 \ln \left (x \right )+x^{2}\) | \(42\) |
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Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (26) = 52\).
Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.50 \[ \int \frac {-8+2 x^2+\frac {e^x \left (4 x^3+2 x^4\right )}{x}+\frac {e^{2 x} \left (2 x^4+2 x^5\right )}{x^2}-2 \log \left (\frac {x}{3}\right )}{x} \, dx=x^{4} e^{\left (2 \, x - 2 \, \log \left (3\right ) - 2 \, \log \left (\frac {1}{3} \, x\right )\right )} + 2 \, x^{3} e^{\left (x - \log \left (3\right ) - \log \left (\frac {1}{3} \, x\right )\right )} + x^{2} - \log \left (\frac {1}{3} \, x\right )^{2} - 8 \, \log \left (\frac {1}{3} \, x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (15) = 30\).
Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {-8+2 x^2+\frac {e^x \left (4 x^3+2 x^4\right )}{x}+\frac {e^{2 x} \left (2 x^4+2 x^5\right )}{x^2}-2 \log \left (\frac {x}{3}\right )}{x} \, dx=x^{2} e^{2 x} + 2 x^{2} e^{x} + x^{2} - \log {\left (\frac {x}{3} \right )}^{2} - 8 \log {\left (x \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (26) = 52\).
Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.82 \[ \int \frac {-8+2 x^2+\frac {e^x \left (4 x^3+2 x^4\right )}{x}+\frac {e^{2 x} \left (2 x^4+2 x^5\right )}{x^2}-2 \log \left (\frac {x}{3}\right )}{x} \, dx=x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + \frac {1}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + 2 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 4 \, {\left (x - 1\right )} e^{x} - \log \left (\frac {1}{3} \, x\right )^{2} - 8 \, \log \left (x\right ) \]
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none
Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {-8+2 x^2+\frac {e^x \left (4 x^3+2 x^4\right )}{x}+\frac {e^{2 x} \left (2 x^4+2 x^5\right )}{x^2}-2 \log \left (\frac {x}{3}\right )}{x} \, dx=x^{2} e^{\left (2 \, x\right )} + 2 \, x^{2} e^{x} + x^{2} + 2 \, \log \left (3\right ) \log \left (x\right ) - \log \left (x\right )^{2} - 8 \, \log \left (x\right ) \]
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Time = 11.85 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {-8+2 x^2+\frac {e^x \left (4 x^3+2 x^4\right )}{x}+\frac {e^{2 x} \left (2 x^4+2 x^5\right )}{x^2}-2 \log \left (\frac {x}{3}\right )}{x} \, dx=2\,x^2\,{\mathrm {e}}^x-8\,\ln \left (x\right )-{\left (\ln \left (3\right )-\ln \left (x\right )\right )}^2+x^2\,{\mathrm {e}}^{2\,x}+x^2 \]
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