Optimal. Leaf size=31 \[ \frac {4 x^2}{e^{2 x}-x}-\log \left (3+e^{1-x} x\right ) \]
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Rubi [F] time = 1.45, 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 {-12 x^3+e^{2 x} \left (24 x^2-24 x^3\right )+e^{1-x} x \left (e^{4 x} (-1+x)-x^2-3 x^3+e^{2 x} \left (2 x+6 x^2-8 x^3\right )\right )}{3 e^{4 x} x-6 e^{2 x} x^2+3 x^3+e^{1-x} x \left (e^{4 x} x-2 e^{2 x} x^2+x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{1+4 x} (-1+x)-24 e^{3 x} (-1+x) x-12 e^x x^2-e x^2 (1+3 x)-2 e^{1+2 x} x \left (-1-3 x+4 x^2\right )}{\left (e^{2 x}-x\right )^2 \left (3 e^x+e x\right )} \, dx\\ &=\int \left (-\frac {8 (-1+x) x}{e^{2 x}-x}-\frac {4 x^2 (-1+2 x)}{\left (e^{2 x}-x\right )^2}+\frac {e (-1+x)}{3 e^x+e x}\right ) \, dx\\ &=-\left (4 \int \frac {x^2 (-1+2 x)}{\left (e^{2 x}-x\right )^2} \, dx\right )-8 \int \frac {(-1+x) x}{e^{2 x}-x} \, dx+e \int \frac {-1+x}{3 e^x+e x} \, dx\\ &=-\left (4 \int \left (-\frac {x^2}{\left (e^{2 x}-x\right )^2}+\frac {2 x^3}{\left (e^{2 x}-x\right )^2}\right ) \, dx\right )-8 \int \left (-\frac {x}{e^{2 x}-x}+\frac {x^2}{e^{2 x}-x}\right ) \, dx+e \int \left (-\frac {1}{3 e^x+e x}+\frac {x}{3 e^x+e x}\right ) \, dx\\ &=4 \int \frac {x^2}{\left (e^{2 x}-x\right )^2} \, dx+8 \int \frac {x}{e^{2 x}-x} \, dx-8 \int \frac {x^2}{e^{2 x}-x} \, dx-8 \int \frac {x^3}{\left (e^{2 x}-x\right )^2} \, dx-e \int \frac {1}{3 e^x+e x} \, dx+e \int \frac {x}{3 e^x+e x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 30, normalized size = 0.97 \begin {gather*} x+\frac {4 x^2}{e^{2 x}-x}-\log \left (3 e^x+e x\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.69, size = 64, normalized size = 2.06 \begin {gather*} \frac {4 \, x e^{\left (-2 \, x + 2 \, \log \relax (x) + 2\right )} - {\left (x e^{2} - e^{\left (-2 \, x + 2 \, \log \relax (x) + 2\right )}\right )} \log \left (e^{\left (-x + \log \relax (x) + 1\right )} + 3\right )}{x e^{2} - e^{\left (-2 \, x + 2 \, \log \relax (x) + 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.75, size = 52, normalized size = 1.68 \begin {gather*} -\frac {7 \, x^{2} + x e^{\left (2 \, x\right )} + x \log \left (x e + 3 \, e^{x}\right ) - e^{\left (2 \, x\right )} \log \left (x e + 3 \, e^{x}\right )}{x - e^{\left (2 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.26, size = 29, normalized size = 0.94
method | result | size |
risch | \(x -\frac {4 x^{2}}{x -{\mathrm e}^{2 x}}-\ln \left (\frac {x \,{\mathrm e}}{3}+{\mathrm e}^{x}\right )\) | \(29\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.57, size = 36, normalized size = 1.16 \begin {gather*} -\frac {3 \, x^{2} + x e^{\left (2 \, x\right )}}{x - e^{\left (2 \, x\right )}} - \log \left (\frac {1}{3} \, x e + e^{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.35, size = 50, normalized size = 1.61 \begin {gather*} -\frac {x\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^{2\,x}\,\ln \left ({\mathrm {e}}^x+\frac {x\,\mathrm {e}}{3}\right )+x\,\ln \left ({\mathrm {e}}^x+\frac {x\,\mathrm {e}}{3}\right )+3\,x^2}{x-{\mathrm {e}}^{2\,x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.34, size = 37, normalized size = 1.19 \begin {gather*} - 4 x - \frac {4 x}{x e^{- 2 x} - 1} - \log {\relax (x )} - \log {\left (\frac {1}{\sqrt {e^{2 x}}} + \frac {3}{e x} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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