Optimal. Leaf size=30 \[ 5-e^{-\frac {2 e^2}{\left (5+\frac {x-\log (x)}{-2+x}\right )^2}} x^2 \]
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Rubi [F]
time = 9.48, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps
used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {}
\begin {gather*} \int \frac {\exp \left (-\frac {2 e^2 \left (4-4 x+x^2\right )}{100-120 x+36 x^2+(20-12 x) \log (x)+\log ^2(x)}\right ) \left (-2000 x+3600 x^2-2160 x^3+432 x^4+e^2 \left (-16 x+32 x^2-12 x^3\right )+\left (-600 x+720 x^2-216 x^3+e^2 \left (-8 x^2+4 x^3\right )\right ) \log (x)+\left (-60 x+36 x^2\right ) \log ^2(x)-2 x \log ^3(x)\right )}{1000-1800 x+1080 x^2-216 x^3+\left (300-360 x+108 x^2\right ) \log (x)+(30-18 x) \log ^2(x)+\log ^3(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x \left (8 (-5+3 x)^3-2 e^2 \left (4-8 x+3 x^2\right )+2 \left (-150-2 \left (-90+e^2\right ) x+\left (-54+e^2\right ) x^2\right ) \log (x)+6 (-5+3 x) \log ^2(x)-\log ^3(x)\right )}{(10-6 x+\log (x))^3} \, dx\\ &=2 \int \frac {e^{-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x \left (8 (-5+3 x)^3-2 e^2 \left (4-8 x+3 x^2\right )+2 \left (-150-2 \left (-90+e^2\right ) x+\left (-54+e^2\right ) x^2\right ) \log (x)+6 (-5+3 x) \log ^2(x)-\log ^3(x)\right )}{(10-6 x+\log (x))^3} \, dx\\ &=2 \int \left (-e^{-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x-\frac {2 e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} (-2+x)^2 x (-1+6 x)}{(-10+6 x-\log (x))^3}+\frac {2 e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} (-2+x) x^2}{(-10+6 x-\log (x))^2}\right ) \, dx\\ &=-\left (2 \int e^{-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x \, dx\right )-4 \int \frac {e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} (-2+x)^2 x (-1+6 x)}{(-10+6 x-\log (x))^3} \, dx+4 \int \frac {e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} (-2+x) x^2}{(-10+6 x-\log (x))^2} \, dx\\ &=-\left (2 \int e^{-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x \, dx\right )-4 \int \left (-\frac {4 e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x}{(-10+6 x-\log (x))^3}+\frac {28 e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^2}{(-10+6 x-\log (x))^3}-\frac {25 e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^3}{(-10+6 x-\log (x))^3}+\frac {6 e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^4}{(-10+6 x-\log (x))^3}\right ) \, dx+4 \int \left (-\frac {2 e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^2}{(-10+6 x-\log (x))^2}+\frac {e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^3}{(-10+6 x-\log (x))^2}\right ) \, dx\\ &=-\left (2 \int e^{-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x \, dx\right )+4 \int \frac {e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^3}{(-10+6 x-\log (x))^2} \, dx-8 \int \frac {e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^2}{(-10+6 x-\log (x))^2} \, dx+16 \int \frac {e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x}{(-10+6 x-\log (x))^3} \, dx-24 \int \frac {e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^4}{(-10+6 x-\log (x))^3} \, dx+100 \int \frac {e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^3}{(-10+6 x-\log (x))^3} \, dx-112 \int \frac {e^{2-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^2}{(-10+6 x-\log (x))^3} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.09, size = 26, normalized size = 0.87 \begin {gather*} -e^{-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^2 \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.28, size = 25, normalized size = 0.83
method | result | size |
risch | \(-x^{2} {\mathrm e}^{-\frac {2 \left (x -2\right )^{2} {\mathrm e}^{2}}{\left (10+\ln \left (x \right )-6 x \right )^{2}}}\) | \(25\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 43, normalized size = 1.43 \begin {gather*} -x^{2} e^{\left (-\frac {2 \, {\left (x^{2} - 4 \, x + 4\right )} e^{2}}{36 \, x^{2} - 4 \, {\left (3 \, x - 5\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 120 \, x + 100}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 21.60, size = 42, normalized size = 1.40 \begin {gather*} - x^{2} e^{- \frac {2 \left (x^{2} - 4 x + 4\right ) e^{2}}{36 x^{2} - 120 x + \left (20 - 12 x\right ) \log {\left (x \right )} + \log {\left (x \right )}^{2} + 100}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 71 vs.
\(2 (28) = 56\).
time = 1.06, size = 71, normalized size = 2.37 \begin {gather*} -x^{2} e^{\left (\frac {2 \, {\left (11 \, x^{2} e^{2} - 12 \, x e^{2} \log \left (x\right ) + e^{2} \log \left (x\right )^{2} - 20 \, x e^{2} + 20 \, e^{2} \log \left (x\right )\right )}}{25 \, {\left (36 \, x^{2} - 12 \, x \log \left (x\right ) + \log \left (x\right )^{2} - 120 \, x + 20 \, \log \left (x\right ) + 100\right )}} - \frac {2}{25} \, e^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.53, size = 50, normalized size = 1.67 \begin {gather*} -x^2\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^2\,x^2-8\,{\mathrm {e}}^2\,x+8\,{\mathrm {e}}^2}{36\,x^2-12\,x\,\ln \left (x\right )-120\,x+{\ln \left (x\right )}^2+20\,\ln \left (x\right )+100}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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