Optimal. Leaf size=31 \[ \frac {e^5 \left (5+2 x-e^{-2 x} (-2+x) x^2-\log \left (x^2\right )\right )}{x} \]
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Rubi [A] time = 0.31, antiderivative size = 46, normalized size of antiderivative = 1.48, number of steps used = 13, number of rules used = 7, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.137, Rules used = {6688, 12, 14, 2196, 2194, 2176, 2304} \begin {gather*} -e^{5-2 x} x^2+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}+2 e^{5-2 x} x-\frac {2 e^5}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 2176
Rule 2194
Rule 2196
Rule 2304
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^5 \left (-7+2 e^{-2 x} x^2 \left (1-3 x+x^2\right )+\log \left (x^2\right )\right )}{x^2} \, dx\\ &=e^5 \int \frac {-7+2 e^{-2 x} x^2 \left (1-3 x+x^2\right )+\log \left (x^2\right )}{x^2} \, dx\\ &=e^5 \int \left (2 e^{-2 x} \left (1-3 x+x^2\right )+\frac {-7+\log \left (x^2\right )}{x^2}\right ) \, dx\\ &=e^5 \int \frac {-7+\log \left (x^2\right )}{x^2} \, dx+\left (2 e^5\right ) \int e^{-2 x} \left (1-3 x+x^2\right ) \, dx\\ &=-\frac {2 e^5}{x}+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}+\left (2 e^5\right ) \int \left (e^{-2 x}-3 e^{-2 x} x+e^{-2 x} x^2\right ) \, dx\\ &=-\frac {2 e^5}{x}+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}+\left (2 e^5\right ) \int e^{-2 x} \, dx+\left (2 e^5\right ) \int e^{-2 x} x^2 \, dx-\left (6 e^5\right ) \int e^{-2 x} x \, dx\\ &=-e^{5-2 x}-\frac {2 e^5}{x}+3 e^{5-2 x} x-e^{5-2 x} x^2+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}+\left (2 e^5\right ) \int e^{-2 x} x \, dx-\left (3 e^5\right ) \int e^{-2 x} \, dx\\ &=\frac {1}{2} e^{5-2 x}-\frac {2 e^5}{x}+2 e^{5-2 x} x-e^{5-2 x} x^2+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}+e^5 \int e^{-2 x} \, dx\\ &=-\frac {2 e^5}{x}+2 e^{5-2 x} x-e^{5-2 x} x^2+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.28, size = 26, normalized size = 0.84 \begin {gather*} -\frac {e^5 \left (-5+e^{-2 x} (-2+x) x^2+\log \left (x^2\right )\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.89, size = 43, normalized size = 1.39 \begin {gather*} -\frac {{\left ({\left (x^{3} - 2 \, x^{2}\right )} e^{10} + e^{\left (2 \, x + 10\right )} \log \left (x^{2}\right ) - 5 \, e^{\left (2 \, x + 10\right )}\right )} e^{\left (-2 \, x - 5\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 38, normalized size = 1.23 \begin {gather*} -\frac {x^{3} e^{\left (-2 \, x + 5\right )} - 2 \, x^{2} e^{\left (-2 \, x + 5\right )} + e^{5} \log \left (x^{2}\right ) - 5 \, e^{5}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 44, normalized size = 1.42
method | result | size |
default | \(-\frac {{\mathrm e}^{5} \ln \left (x^{2}\right )}{x}+\frac {5 \,{\mathrm e}^{5}}{x}+2 \,{\mathrm e}^{5} \left (-\frac {x^{2} {\mathrm e}^{-2 x}}{2}+x \,{\mathrm e}^{-2 x}\right )\) | \(44\) |
norman | \(\frac {\left (2 x^{2} {\mathrm e}^{5}-x^{3} {\mathrm e}^{5}+5 \,{\mathrm e}^{5} {\mathrm e}^{2 x}-{\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )\right ) {\mathrm e}^{-2 x}}{x}\) | \(46\) |
risch | \(-\frac {2 \,{\mathrm e}^{5} \ln \relax (x )}{x}-\frac {\left (-i {\mathrm e}^{2 x} \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i {\mathrm e}^{2 x} \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i {\mathrm e}^{2 x} \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 x^{3}-4 x^{2}-10 \,{\mathrm e}^{2 x}\right ) {\mathrm e}^{-2 x +5}}{2 x}\) | \(100\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.35, size = 69, normalized size = 2.23 \begin {gather*} -{\left (\frac {\log \left (x^{2}\right )}{x} + \frac {2}{x}\right )} e^{5} - \frac {1}{2} \, {\left (2 \, x^{2} e^{5} + 2 \, x e^{5} + e^{5}\right )} e^{\left (-2 \, x\right )} + \frac {3}{2} \, {\left (2 \, x e^{5} + e^{5}\right )} e^{\left (-2 \, x\right )} + \frac {7 \, e^{5}}{x} - e^{\left (-2 \, x + 5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.41, size = 39, normalized size = 1.26 \begin {gather*} 2\,x\,{\mathrm {e}}^{5-2\,x}+\frac {5\,{\mathrm {e}}^5}{x}-x^2\,{\mathrm {e}}^{5-2\,x}-\frac {\ln \left (x^2\right )\,{\mathrm {e}}^5}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.37, size = 34, normalized size = 1.10 \begin {gather*} \left (- x^{2} e^{5} + 2 x e^{5}\right ) e^{- 2 x} - \frac {e^{5} \log {\left (x^{2} \right )}}{x} + \frac {5 e^{5}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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