Optimal. Leaf size=30 \[ 2 x \left (x+\frac {e^x}{-2 x+x^2-\frac {\log (x)}{2 e^3 x^2}}\right ) \]
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Rubi [F] time = 3.61, 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 {e^6 \left (64 x^7-64 x^8+16 x^9\right )+e^x \left (4 e^3 x^2+e^6 \left (-24 x^6+8 x^7\right )\right )+\left (e^{3+x} \left (-12 x^2-4 x^3\right )+e^3 \left (32 x^4-16 x^5\right )\right ) \log (x)+4 x \log ^2(x)}{e^6 \left (16 x^6-16 x^7+4 x^8\right )+e^3 \left (8 x^3-4 x^4\right ) \log (x)+\log ^2(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 {4 x \left (e^{3+x} x+2 e^{6+x} (-3+x) x^5+4 e^6 (-2+x)^2 x^6-e^3 x \left (4 (-2+x) x^2+e^x (3+x)\right ) \log (x)+\log ^2(x)\right )}{\left (2 e^3 (-2+x) x^3-\log (x)\right )^2} \, dx\\ &=4 \int \frac {x \left (e^{3+x} x+2 e^{6+x} (-3+x) x^5+4 e^6 (-2+x)^2 x^6-e^3 x \left (4 (-2+x) x^2+e^x (3+x)\right ) \log (x)+\log ^2(x)\right )}{\left (2 e^3 (-2+x) x^3-\log (x)\right )^2} \, dx\\ &=4 \int \left (x+\frac {e^{3+x} x^2 \left (1-6 e^3 x^4+2 e^3 x^5-3 \log (x)-x \log (x)\right )}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2}\right ) \, dx\\ &=2 x^2+4 \int \frac {e^{3+x} x^2 \left (1-6 e^3 x^4+2 e^3 x^5-3 \log (x)-x \log (x)\right )}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2} \, dx\\ &=2 x^2+4 \int \left (\frac {e^{3+x} x^2 (3+x)}{-4 e^3 x^3+2 e^3 x^4-\log (x)}+\frac {e^{3+x} x^2 \left (1+12 e^3 x^3-8 e^3 x^4\right )}{\left (4 e^3 x^3-2 e^3 x^4+\log (x)\right )^2}\right ) \, dx\\ &=2 x^2+4 \int \frac {e^{3+x} x^2 (3+x)}{-4 e^3 x^3+2 e^3 x^4-\log (x)} \, dx+4 \int \frac {e^{3+x} x^2 \left (1+12 e^3 x^3-8 e^3 x^4\right )}{\left (4 e^3 x^3-2 e^3 x^4+\log (x)\right )^2} \, dx\\ &=2 x^2+4 \int \left (\frac {e^{3+x} x^2}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2}+\frac {12 e^{6+x} x^5}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2}-\frac {8 e^{6+x} x^6}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2}\right ) \, dx+4 \int \left (\frac {3 e^{3+x} x^2}{-4 e^3 x^3+2 e^3 x^4-\log (x)}+\frac {e^{3+x} x^3}{-4 e^3 x^3+2 e^3 x^4-\log (x)}\right ) \, dx\\ &=2 x^2+4 \int \frac {e^{3+x} x^2}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2} \, dx+4 \int \frac {e^{3+x} x^3}{-4 e^3 x^3+2 e^3 x^4-\log (x)} \, dx+12 \int \frac {e^{3+x} x^2}{-4 e^3 x^3+2 e^3 x^4-\log (x)} \, dx-32 \int \frac {e^{6+x} x^6}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2} \, dx+48 \int \frac {e^{6+x} x^5}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 33, normalized size = 1.10 \begin {gather*} 2 x^2 \left (1+\frac {2 e^{3+x} x}{2 e^3 (-2+x) x^3-\log (x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 52, normalized size = 1.73 \begin {gather*} \frac {2 \, {\left (2 \, x^{3} e^{\left (x + 3\right )} - x^{2} \log \relax (x) + 2 \, {\left (x^{6} - 2 \, x^{5}\right )} e^{3}\right )}}{2 \, {\left (x^{4} - 2 \, x^{3}\right )} e^{3} - \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 54, normalized size = 1.80 \begin {gather*} \frac {2 \, {\left (2 \, x^{6} e^{3} - 4 \, x^{5} e^{3} + 2 \, x^{3} e^{\left (x + 3\right )} - x^{2} \log \relax (x)\right )}}{2 \, x^{4} e^{3} - 4 \, x^{3} e^{3} - \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 37, normalized size = 1.23
method | result | size |
risch | \(2 x^{2}+\frac {4 x^{3} {\mathrm e}^{3+x}}{2 x^{4} {\mathrm e}^{3}-4 x^{3} {\mathrm e}^{3}-\ln \relax (x )}\) | \(37\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 54, normalized size = 1.80 \begin {gather*} \frac {2 \, {\left (2 \, x^{6} e^{3} - 4 \, x^{5} e^{3} + 2 \, x^{3} e^{\left (x + 3\right )} - x^{2} \log \relax (x)\right )}}{2 \, x^{4} e^{3} - 4 \, x^{3} e^{3} - \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {4\,x\,{\ln \relax (x)}^2+\left ({\mathrm {e}}^3\,\left (32\,x^4-16\,x^5\right )-{\mathrm {e}}^{x+3}\,\left (4\,x^3+12\,x^2\right )\right )\,\ln \relax (x)+{\mathrm {e}}^6\,\left (16\,x^9-64\,x^8+64\,x^7\right )-{\mathrm {e}}^x\,\left ({\mathrm {e}}^6\,\left (24\,x^6-8\,x^7\right )-4\,x^2\,{\mathrm {e}}^3\right )}{{\ln \relax (x)}^2+{\mathrm {e}}^3\,\left (8\,x^3-4\,x^4\right )\,\ln \relax (x)+{\mathrm {e}}^6\,\left (4\,x^8-16\,x^7+16\,x^6\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.35, size = 36, normalized size = 1.20 \begin {gather*} \frac {4 x^{3} e^{3} e^{x}}{2 x^{4} e^{3} - 4 x^{3} e^{3} - \log {\relax (x )}} + 2 x^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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