Optimal. Leaf size=32 \[ (3-x) \left (\frac {10 e^x}{5-x}+\frac {1}{2} e^{-\log ^2(3)} x\right ) \]
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Rubi [A] time = 0.28, antiderivative size = 46, normalized size of antiderivative = 1.44, number of steps used = 11, number of rules used = 7, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.127, Rules used = {12, 27, 6688, 2199, 2194, 2177, 2178} \begin {gather*} -\frac {1}{2} x^2 e^{-\log ^2(3)}+10 e^x-\frac {20 e^x}{5-x}+\frac {3}{2} x e^{-\log ^2(3)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 2177
Rule 2178
Rule 2194
Rule 2199
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=e^{-\log ^2(3)} \int \frac {75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )}{50-20 x+2 x^2} \, dx\\ &=e^{-\log ^2(3)} \int \frac {75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )}{2 (-5+x)^2} \, dx\\ &=\frac {1}{2} e^{-\log ^2(3)} \int \frac {75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )}{(-5+x)^2} \, dx\\ &=\frac {1}{2} e^{-\log ^2(3)} \int \left (3-2 x+\frac {20 e^{x+\log ^2(3)} \left (13-8 x+x^2\right )}{(-5+x)^2}\right ) \, dx\\ &=\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2+\left (10 e^{-\log ^2(3)}\right ) \int \frac {e^{x+\log ^2(3)} \left (13-8 x+x^2\right )}{(-5+x)^2} \, dx\\ &=\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2+\left (10 e^{-\log ^2(3)}\right ) \int \left (e^{x+\log ^2(3)}-\frac {2 e^{x+\log ^2(3)}}{(-5+x)^2}+\frac {2 e^{x+\log ^2(3)}}{-5+x}\right ) \, dx\\ &=\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2+\left (10 e^{-\log ^2(3)}\right ) \int e^{x+\log ^2(3)} \, dx-\left (20 e^{-\log ^2(3)}\right ) \int \frac {e^{x+\log ^2(3)}}{(-5+x)^2} \, dx+\left (20 e^{-\log ^2(3)}\right ) \int \frac {e^{x+\log ^2(3)}}{-5+x} \, dx\\ &=10 e^x-\frac {20 e^x}{5-x}+\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2+20 e^5 \text {Ei}(-5+x)-\left (20 e^{-\log ^2(3)}\right ) \int \frac {e^{x+\log ^2(3)}}{-5+x} \, dx\\ &=10 e^x-\frac {20 e^x}{5-x}+\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.07, size = 46, normalized size = 1.44 \begin {gather*} 10 e^x-\frac {20 e^x}{5-x}+\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2 \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 38, normalized size = 1.19 \begin {gather*} -\frac {{\left (x^{3} - 8 \, x^{2} - 20 \, {\left (x - 3\right )} e^{\left (\log \relax (3)^{2} + x\right )} + 15 \, x\right )} e^{\left (-\log \relax (3)^{2}\right )}}{2 \, {\left (x - 5\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.28, size = 45, normalized size = 1.41 \begin {gather*} -\frac {{\left (x^{3} - 8 \, x^{2} - 20 \, x e^{\left (\log \relax (3)^{2} + x\right )} + 15 \, x + 60 \, e^{\left (\log \relax (3)^{2} + x\right )}\right )} e^{\left (-\log \relax (3)^{2}\right )}}{2 \, {\left (x - 5\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.59, size = 36, normalized size = 1.12
method | result | size |
risch | \(-\frac {{\mathrm e}^{-\ln \relax (3)^{2}} x^{2}}{2}+\frac {3 x \,{\mathrm e}^{-\ln \relax (3)^{2}}}{2}+\frac {10 \left (x -3\right ) {\mathrm e}^{x}}{x -5}\) | \(36\) |
default | \(\frac {{\mathrm e}^{-\ln \relax (3)^{2}} \left (-x^{2}+3 x +\frac {40 \,{\mathrm e}^{\ln \relax (3)^{2}} {\mathrm e}^{x}}{x -5}+20 \,{\mathrm e}^{x} {\mathrm e}^{\ln \relax (3)^{2}}\right )}{2}\) | \(42\) |
norman | \(\frac {10 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{-\ln \relax (3)^{2}} x^{2}-\frac {{\mathrm e}^{-\ln \relax (3)^{2}} x^{3}}{2}-30 \,{\mathrm e}^{x}-\frac {75 \,{\mathrm e}^{-\ln \relax (3)^{2}}}{2}}{x -5}\) | \(50\) |
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*} -\frac {1}{2} \, {\left (x^{2} - 3 \, x - \frac {20 \, {\left (x^{2} e^{\left (\log \relax (3)^{2}\right )} - 8 \, x e^{\left (\log \relax (3)^{2}\right )}\right )} e^{x}}{x^{2} - 10 \, x + 25} + \frac {260 \, e^{\left (\log \relax (3)^{2} + 5\right )} E_{2}\left (-x + 5\right )}{x - 5} - \int \frac {40 \, {\left (x e^{\left (\log \relax (3)^{2}\right )} - 20 \, e^{\left (\log \relax (3)^{2}\right )}\right )} e^{x}}{x^{3} - 15 \, x^{2} + 75 \, x - 125}\,{d x}\right )} e^{\left (-\log \relax (3)^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.93, size = 36, normalized size = 1.12 \begin {gather*} \frac {{\mathrm {e}}^{-{\ln \relax (3)}^2}\,\left (x-3\right )\,\left (5\,x+20\,{\mathrm {e}}^{x+{\ln \relax (3)}^2}-x^2\right )}{2\,\left (x-5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.14, size = 34, normalized size = 1.06 \begin {gather*} - \frac {x^{2}}{2 e^{\log {\relax (3 )}^{2}}} + \frac {3 x}{2 e^{\log {\relax (3 )}^{2}}} + \frac {\left (10 x - 30\right ) e^{x}}{x - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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