Optimal. Leaf size=36 \[ -\left (e^x x-\frac {2}{e^x+x}\right )^2+\log \left (-x+\frac {1}{5} \left (-\frac {2}{x}+x\right )\right ) \]
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Rubi [F] time = 2.39, 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 {8 x+15 x^3+2 x^5+e^{5 x} \left (-2 x^2-2 x^3-4 x^4-4 x^5\right )+e^{4 x} \left (-6 x^3-6 x^4-12 x^5-12 x^6\right )+e^x \left (8 x-3 x^2+16 x^3+10 x^4+8 x^6\right )+e^{3 x} \left (-1+4 x+2 x^2+8 x^3-6 x^4-6 x^5-12 x^6-12 x^7\right )+e^{2 x} \left (-3 x+4 x^2+10 x^3+8 x^4+6 x^5-2 x^6-4 x^7-4 x^8\right )}{x^4+2 x^6+e^{3 x} \left (x+2 x^3\right )+e^{2 x} \left (3 x^2+6 x^4\right )+e^x \left (3 x^3+6 x^5\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 {8 x+15 x^3+2 x^5+e^{5 x} \left (-2 x^2-2 x^3-4 x^4-4 x^5\right )+e^{4 x} \left (-6 x^3-6 x^4-12 x^5-12 x^6\right )+e^x \left (8 x-3 x^2+16 x^3+10 x^4+8 x^6\right )+e^{3 x} \left (-1+4 x+2 x^2+8 x^3-6 x^4-6 x^5-12 x^6-12 x^7\right )+e^{2 x} \left (-3 x+4 x^2+10 x^3+8 x^4+6 x^5-2 x^6-4 x^7-4 x^8\right )}{x \left (e^x+x\right )^3 \left (1+2 x^2\right )} \, dx\\ &=\int \left (-2 e^{2 x} x (1+x)-\frac {8 (-1+x)}{\left (e^x+x\right )^3}+\frac {4 (-2+x) x}{e^x+x}-\frac {4 \left (-2-x^2+x^3\right )}{\left (e^x+x\right )^2}+\frac {-1+4 x+2 x^2+8 x^3}{x \left (1+2 x^2\right )}\right ) \, dx\\ &=-\left (2 \int e^{2 x} x (1+x) \, dx\right )+4 \int \frac {(-2+x) x}{e^x+x} \, dx-4 \int \frac {-2-x^2+x^3}{\left (e^x+x\right )^2} \, dx-8 \int \frac {-1+x}{\left (e^x+x\right )^3} \, dx+\int \frac {-1+4 x+2 x^2+8 x^3}{x \left (1+2 x^2\right )} \, dx\\ &=-\left (2 \int \left (e^{2 x} x+e^{2 x} x^2\right ) \, dx\right )-4 \int \left (-\frac {2}{\left (e^x+x\right )^2}-\frac {x^2}{\left (e^x+x\right )^2}+\frac {x^3}{\left (e^x+x\right )^2}\right ) \, dx+4 \int \left (-\frac {2 x}{e^x+x}+\frac {x^2}{e^x+x}\right ) \, dx-8 \int \left (-\frac {1}{\left (e^x+x\right )^3}+\frac {x}{\left (e^x+x\right )^3}\right ) \, dx+\int \left (4-\frac {1}{x}+\frac {4 x}{1+2 x^2}\right ) \, dx\\ &=4 x-\log (x)-2 \int e^{2 x} x \, dx-2 \int e^{2 x} x^2 \, dx+4 \int \frac {x^2}{\left (e^x+x\right )^2} \, dx-4 \int \frac {x^3}{\left (e^x+x\right )^2} \, dx+4 \int \frac {x^2}{e^x+x} \, dx+4 \int \frac {x}{1+2 x^2} \, dx+8 \int \frac {1}{\left (e^x+x\right )^3} \, dx-8 \int \frac {x}{\left (e^x+x\right )^3} \, dx+8 \int \frac {1}{\left (e^x+x\right )^2} \, dx-8 \int \frac {x}{e^x+x} \, dx\\ &=4 x-e^{2 x} x-e^{2 x} x^2-\frac {4}{\left (e^x+x\right )^2}-\log (x)+\log \left (1+2 x^2\right )+2 \int e^{2 x} x \, dx+4 \int \frac {x^2}{\left (e^x+x\right )^2} \, dx-4 \int \frac {x^3}{\left (e^x+x\right )^2} \, dx+4 \int \frac {x^2}{e^x+x} \, dx-8 \int \frac {x}{e^x+x} \, dx+\int e^{2 x} \, dx\\ &=\frac {e^{2 x}}{2}+4 x-e^{2 x} x^2-\frac {4}{\left (e^x+x\right )^2}-\log (x)+\log \left (1+2 x^2\right )+4 \int \frac {x^2}{\left (e^x+x\right )^2} \, dx-4 \int \frac {x^3}{\left (e^x+x\right )^2} \, dx+4 \int \frac {x^2}{e^x+x} \, dx-8 \int \frac {x}{e^x+x} \, dx-\int e^{2 x} \, dx\\ &=4 x-e^{2 x} x^2-\frac {4}{\left (e^x+x\right )^2}-\log (x)+\log \left (1+2 x^2\right )+4 \int \frac {x^2}{\left (e^x+x\right )^2} \, dx-4 \int \frac {x^3}{\left (e^x+x\right )^2} \, dx+4 \int \frac {x^2}{e^x+x} \, dx-8 \int \frac {x}{e^x+x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 40, normalized size = 1.11 \begin {gather*} -\frac {\left (-2+e^{2 x} x+e^x x^2\right )^2}{\left (e^x+x\right )^2}-\log (x)+\log \left (1+2 x^2\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.65, size = 94, normalized size = 2.61 \begin {gather*} -\frac {2 \, x^{3} e^{\left (3 \, x\right )} + x^{2} e^{\left (4 \, x\right )} - 4 \, x^{2} e^{x} + {\left (x^{4} - 4 \, x\right )} e^{\left (2 \, x\right )} - {\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )} \log \left (2 \, x^{2} + 1\right ) + {\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )} \log \relax (x) + 4}{x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.28, size = 118, normalized size = 3.28 \begin {gather*} -\frac {x^{4} e^{\left (2 \, x\right )} + 2 \, x^{3} e^{\left (3 \, x\right )} + x^{2} e^{\left (4 \, x\right )} - 4 \, x^{2} e^{x} - x^{2} \log \left (2 \, x^{2} + 1\right ) - 2 \, x e^{x} \log \left (2 \, x^{2} + 1\right ) + x^{2} \log \relax (x) + 2 \, x e^{x} \log \relax (x) - 4 \, x e^{\left (2 \, x\right )} - e^{\left (2 \, x\right )} \log \left (2 \, x^{2} + 1\right ) + e^{\left (2 \, x\right )} \log \relax (x) + 4}{x^{2} + 2 \, x e^{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 = 0.10, size = 45, normalized size = 1.25
method | result | size |
risch | \(4 x -\ln \relax (x )+\ln \left (2 x^{2}+1\right )-{\mathrm e}^{2 x} x^{2}-\frac {4 \left (x^{3}+{\mathrm e}^{x} x^{2}+1\right )}{\left ({\mathrm e}^{x}+x \right )^{2}}\) | \(45\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.60, size = 68, normalized size = 1.89 \begin {gather*} -\frac {2 \, x^{3} e^{\left (3 \, x\right )} + x^{2} e^{\left (4 \, x\right )} - 4 \, x^{2} e^{x} + {\left (x^{4} - 4 \, x\right )} e^{\left (2 \, x\right )} + 4}{x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}} + \log \left (2 \, x^{2} + 1\right ) - \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.45, size = 62, normalized size = 1.72 \begin {gather*} 4\,x+\ln \left (x^2+\frac {1}{2}\right )-\ln \relax (x)-x^2\,{\mathrm {e}}^{2\,x}-\frac {4}{{\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^x+x^2}+\frac {4\,\left (x^2-x^3\right )}{\left (x+{\mathrm {e}}^x\right )\,\left (x-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.28, size = 54, normalized size = 1.50 \begin {gather*} - x^{2} e^{2 x} + 4 x - \log {\relax (x )} + \log {\left (2 x^{2} + 1 \right )} + \frac {- 4 x^{3} - 4 x^{2} e^{x} - 4}{x^{2} + 2 x e^{x} + e^{2 x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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