Optimal. Leaf size=25 \[ \frac {3 e^{-5+\left (-4+e^{10}\right ) (-2+x) x+x^2}}{x+\log (3)} \]
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Rubi [B] time = 0.38, antiderivative size = 90, normalized size of antiderivative = 3.60, number of steps used = 2, number of rules used = 2, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {27, 2288} \begin {gather*} \frac {3 e^{-3 x^2-e^{10} \left (2 x-x^2\right )+8 x-5} \left (-3 x^2-e^{10} \left (x-x^2\right )+4 x+\left (-e^{10} (1-x)-3 x+4\right ) \log (3)\right )}{\left (-e^{10} (1-x)-3 x+4\right ) (x+\log (3))^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 27
Rule 2288
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-5+8 x-3 x^2-e^{10} \left (2 x-x^2\right )} \left (-3+24 x-18 x^2+e^{10} \left (-6 x+6 x^2\right )+\left (24-18 x+e^{10} (-6+6 x)\right ) \log (3)\right )}{(x+\log (3))^2} \, dx\\ &=\frac {3 e^{-5+8 x-3 x^2-e^{10} \left (2 x-x^2\right )} \left (4 x-3 x^2-e^{10} \left (x-x^2\right )+\left (4-e^{10} (1-x)-3 x\right ) \log (3)\right )}{\left (4-e^{10} (1-x)-3 x\right ) (x+\log (3))^2}\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.22, size = 85, normalized size = 3.40 \begin {gather*} \frac {3 e^{-5+8 x+e^{10} (-2+x) x-3 x^2} \left (2 \left (-3+e^{10}\right ) x^2+8 \log (3)+x \left (8-6 \log (3)+e^{10} (-2+\log (9))\right )-e^{10} \log (9)\right )}{\left (8+e^{10} (-2+x)-6 x+e^{10} x\right ) (x+\log (3))^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.93, size = 29, normalized size = 1.16 \begin {gather*} \frac {3 \, e^{\left (-3 \, x^{2} + {\left (x^{2} - 2 \, x\right )} e^{10} + 8 \, x - 5\right )}}{x + \log \relax (3)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {3 \, {\left (6 \, x^{2} - 2 \, {\left (x^{2} - x\right )} e^{10} - 2 \, {\left ({\left (x - 1\right )} e^{10} - 3 \, x + 4\right )} \log \relax (3) - 8 \, x + 1\right )} e^{\left (-3 \, x^{2} + {\left (x^{2} - 2 \, x\right )} e^{10} + 8 \, x - 5\right )}}{x^{2} + 2 \, x \log \relax (3) + \log \relax (3)^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.63, size = 31, normalized size = 1.24
method | result | size |
risch | \(\frac {3 \,{\mathrm e}^{x^{2} {\mathrm e}^{10}-2 x \,{\mathrm e}^{10}-3 x^{2}+8 x -5}}{\ln \relax (3)+x}\) | \(31\) |
norman | \(\frac {3 \,{\mathrm e}^{-\left (-x^{2}+2 x \right ) {\mathrm e}^{10}-3 x^{2}+8 x -5}}{\ln \relax (3)+x}\) | \(36\) |
gosper | \(\frac {3 \,{\mathrm e}^{x^{2} {\mathrm e}^{10}-2 x \,{\mathrm e}^{10}-3 x^{2}+8 x -5}}{\ln \relax (3)+x}\) | \(38\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 35, normalized size = 1.40 \begin {gather*} \frac {3 \, e^{\left (x^{2} e^{10} - 3 \, x^{2} - 2 \, x e^{10} + 8 \, x\right )}}{x e^{5} + e^{5} \log \relax (3)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.81, size = 33, normalized size = 1.32 \begin {gather*} \frac {3\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{10}}\,{\mathrm {e}}^{8\,x}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-3\,x^2}\,{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^{10}}}{x+\ln \relax (3)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.24, size = 27, normalized size = 1.08 \begin {gather*} \frac {3 e^{- 3 x^{2} + 8 x - \left (- x^{2} + 2 x\right ) e^{10} - 5}}{x + \log {\relax (3 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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