Optimal. Leaf size=30 \[ -\frac {1}{2} e^5 \left (x+e^{-2 x} \left (e^{2 x}-x\right ) x^2\right )+\log (x) \]
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Rubi [A] time = 0.25, antiderivative size = 31, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 5, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {12, 6688, 2196, 2176, 2194} \begin {gather*} \frac {1}{2} e^{5-2 x} x^3-\frac {1}{8} e^5 (2 x+1)^2+\log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2176
Rule 2194
Rule 2196
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 35, normalized size = 1.17 \begin {gather*} -\frac {e^5 x}{2}-\frac {e^5 x^2}{2}+\frac {1}{2} e^{5-2 x} x^3+\log (x) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.75, size = 34, normalized size = 1.13 \begin {gather*} \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 \relax (x)\right )} e^{\left (-2 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.13, size = 26, normalized size = 0.87 \begin {gather*} \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 \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 27, normalized size = 0.90
method | result | size |
risch | \(-\frac {x^{2} {\mathrm e}^{5}}{2}-\frac {x \,{\mathrm e}^{5}}{2}+\ln \relax (x )+\frac {x^{3} {\mathrm e}^{-2 x +5}}{2}\) | \(27\) |
norman | \(\left (\frac {x^{3} {\mathrm e}^{5}}{2}-\frac {x \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{5} {\mathrm e}^{2 x} x^{2}}{2}\right ) {\mathrm e}^{-2 x}+\ln \relax (x )\) | \(37\) |
default | \(\ln \relax (x )-\frac {x^{2} {\mathrm e}^{5}}{2}+\frac {3 \,{\mathrm e}^{5} \left (-\frac {x^{2} {\mathrm e}^{-2 x}}{2}-\frac {x \,{\mathrm e}^{-2 x}}{2}-\frac {{\mathrm e}^{-2 x}}{4}\right )}{2}-{\mathrm e}^{5} \left (-\frac {x^{3} {\mathrm e}^{-2 x}}{2}-\frac {3 x^{2} {\mathrm e}^{-2 x}}{4}-\frac {3 x \,{\mathrm e}^{-2 x}}{4}-\frac {3 \,{\mathrm e}^{-2 x}}{8}\right )-\frac {x \,{\mathrm e}^{5}}{2}\) | \(79\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.37, size = 66, normalized size = 2.20 \begin {gather*} -\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 \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.13, size = 26, normalized size = 0.87 \begin {gather*} \ln \relax (x)-\frac {x\,{\mathrm {e}}^5}{2}-\frac {x^2\,{\mathrm {e}}^5}{2}+\frac {x^3\,{\mathrm {e}}^{5-2\,x}}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.16, size = 31, normalized size = 1.03 \begin {gather*} \frac {x^{3} e^{5} e^{- 2 x}}{2} - \frac {x^{2} e^{5}}{2} - \frac {x e^{5}}{2} + \log {\relax (x )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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