Integrand size = 49, antiderivative size = 30 \[ \int \frac {e^{-2 x} \left (e^5 \left (3 x^3-2 x^4\right )+e^{2 x} \left (2+e^5 \left (-x-2 x^2\right )\right )\right )}{2 x} \, dx=-\frac {1}{2} e^5 \left (x+e^{-2 x} \left (e^{2 x}-x\right ) x^2\right )+\log (x) \]
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Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {12, 6820, 2227, 2207, 2225} \[ \int \frac {e^{-2 x} \left (e^5 \left (3 x^3-2 x^4\right )+e^{2 x} \left (2+e^5 \left (-x-2 x^2\right )\right )\right )}{2 x} \, dx=\frac {1}{2} e^{5-2 x} x^3-\frac {1}{8} e^5 (2 x+1)^2+\log (x) \]
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Rule 12
Rule 2207
Rule 2225
Rule 2227
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {e^{-2 x} \left (e^5 \left (3 x^3-2 x^4\right )+e^{2 x} \left (2+e^5 \left (-x-2 x^2\right )\right )\right )}{x} \, dx \\ & = \frac {1}{2} \int \left (\frac {2}{x}+e^{5-2 x} (3-2 x) x^2-e^5 (1+2 x)\right ) \, dx \\ & = -\frac {1}{8} e^5 (1+2 x)^2+\log (x)+\frac {1}{2} \int e^{5-2 x} (3-2 x) x^2 \, dx \\ & = -\frac {1}{8} e^5 (1+2 x)^2+\log (x)+\frac {1}{2} \int \left (3 e^{5-2 x} x^2-2 e^{5-2 x} x^3\right ) \, dx \\ & = -\frac {1}{8} e^5 (1+2 x)^2+\log (x)+\frac {3}{2} \int e^{5-2 x} x^2 \, dx-\int e^{5-2 x} x^3 \, dx \\ & = -\frac {3}{4} e^{5-2 x} x^2+\frac {1}{2} e^{5-2 x} x^3-\frac {1}{8} e^5 (1+2 x)^2+\log (x)+\frac {3}{2} \int e^{5-2 x} x \, dx-\frac {3}{2} \int e^{5-2 x} x^2 \, dx \\ & = -\frac {3}{4} e^{5-2 x} x+\frac {1}{2} e^{5-2 x} x^3-\frac {1}{8} e^5 (1+2 x)^2+\log (x)+\frac {3}{4} \int e^{5-2 x} \, dx-\frac {3}{2} \int e^{5-2 x} x \, dx \\ & = -\frac {3}{8} e^{5-2 x}+\frac {1}{2} e^{5-2 x} x^3-\frac {1}{8} e^5 (1+2 x)^2+\log (x)-\frac {3}{4} \int e^{5-2 x} \, dx \\ & = \frac {1}{2} e^{5-2 x} x^3-\frac {1}{8} e^5 (1+2 x)^2+\log (x) \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-2 x} \left (e^5 \left (3 x^3-2 x^4\right )+e^{2 x} \left (2+e^5 \left (-x-2 x^2\right )\right )\right )}{2 x} \, dx=-\frac {e^5 x}{2}-\frac {e^5 x^2}{2}+\frac {1}{2} e^{5-2 x} x^3+\log (x) \]
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Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90
method | result | size |
risch | \(-\frac {x^{2} {\mathrm e}^{5}}{2}-\frac {x \,{\mathrm e}^{5}}{2}+\ln \left (x \right )+\frac {x^{3} {\mathrm e}^{5-2 x}}{2}\) | \(27\) |
parts | \(-\frac {x^{2} {\mathrm e}^{5}}{2}-\frac {x \,{\mathrm e}^{5}}{2}+\ln \left (x \right )+\frac {x^{3} {\mathrm e}^{-2 x} {\mathrm e}^{5}}{2}\) | \(27\) |
norman | \(\left (\frac {x^{3} {\mathrm e}^{5}}{2}-\frac {x \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{5} {\mathrm e}^{2 x} x^{2}}{2}\right ) {\mathrm e}^{-2 x}+\ln \left (x \right )\) | \(37\) |
parallelrisch | \(\frac {\left (-{\mathrm e}^{5} {\mathrm e}^{2 x} x^{2}+x^{3} {\mathrm e}^{5}-x \,{\mathrm e}^{5} {\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x} \ln \left (x \right )\right ) {\mathrm e}^{-2 x}}{2}\) | \(42\) |
default | \(\ln \left (x \right )-\frac {x^{2} {\mathrm e}^{5}}{2}+\frac {3 \,{\mathrm e}^{5} \left (-\frac {x^{2} {\mathrm e}^{-2 x}}{2}-\frac {{\mathrm e}^{-2 x} x}{2}-\frac {{\mathrm e}^{-2 x}}{4}\right )}{2}-{\mathrm e}^{5} \left (-\frac {x^{3} {\mathrm e}^{-2 x}}{2}-\frac {3 x^{2} {\mathrm e}^{-2 x}}{4}-\frac {3 \,{\mathrm e}^{-2 x} x}{4}-\frac {3 \,{\mathrm e}^{-2 x}}{8}\right )-\frac {x \,{\mathrm e}^{5}}{2}\) | \(79\) |
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Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-2 x} \left (e^5 \left (3 x^3-2 x^4\right )+e^{2 x} \left (2+e^5 \left (-x-2 x^2\right )\right )\right )}{2 x} \, dx=\frac {1}{2} \, {\left (x^{3} e^{5} - {\left (x^{2} + x\right )} e^{\left (2 \, x + 5\right )} + 2 \, e^{\left (2 \, x\right )} \log \left (x\right )\right )} e^{\left (-2 \, x\right )} \]
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Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-2 x} \left (e^5 \left (3 x^3-2 x^4\right )+e^{2 x} \left (2+e^5 \left (-x-2 x^2\right )\right )\right )}{2 x} \, dx=\frac {x^{3} e^{5} e^{- 2 x}}{2} - \frac {x^{2} e^{5}}{2} - \frac {x e^{5}}{2} + \log {\left (x \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (27) = 54\).
Time = 0.18 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.20 \[ \int \frac {e^{-2 x} \left (e^5 \left (3 x^3-2 x^4\right )+e^{2 x} \left (2+e^5 \left (-x-2 x^2\right )\right )\right )}{2 x} \, dx=-\frac {1}{2} \, x^{2} e^{5} - \frac {1}{2} \, x e^{5} + \frac {1}{8} \, {\left (4 \, x^{3} e^{5} + 6 \, x^{2} e^{5} + 6 \, x e^{5} + 3 \, e^{5}\right )} e^{\left (-2 \, x\right )} - \frac {3}{8} \, {\left (2 \, x^{2} e^{5} + 2 \, x e^{5} + e^{5}\right )} e^{\left (-2 \, x\right )} + \log \left (x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-2 x} \left (e^5 \left (3 x^3-2 x^4\right )+e^{2 x} \left (2+e^5 \left (-x-2 x^2\right )\right )\right )}{2 x} \, dx=\frac {1}{2} \, x^{3} e^{\left (-2 \, x + 5\right )} - \frac {1}{2} \, x^{2} e^{5} - \frac {1}{2} \, x e^{5} + \log \left (2 \, x\right ) \]
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Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-2 x} \left (e^5 \left (3 x^3-2 x^4\right )+e^{2 x} \left (2+e^5 \left (-x-2 x^2\right )\right )\right )}{2 x} \, dx=\ln \left (x\right )-\frac {x\,{\mathrm {e}}^5}{2}-\frac {x^2\,{\mathrm {e}}^5}{2}+\frac {x^3\,{\mathrm {e}}^{5-2\,x}}{2} \]
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