Optimal. Leaf size=23 \[ (2+x) \log \left (-1-e^4+e^x+\frac {2}{5 x}+x\right ) \]
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Rubi [F] time = 2.16, 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 {-4-2 x+10 x^2+5 x^3+e^x \left (10 x^2+5 x^3\right )+\left (2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3\right ) \log \left (\frac {2-5 x-5 e^4 x+5 e^x x+5 x^2}{5 x}\right )}{2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4-2 x+10 x^2+5 x^3+e^x \left (10 x^2+5 x^3\right )+\left (2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3\right ) \log \left (\frac {2-5 x-5 e^4 x+5 e^x x+5 x^2}{5 x}\right )}{2 x+5 e^x x^2+\left (-5-5 e^4\right ) x^2+5 x^3} \, dx\\ &=\int \left (\frac {(2+x) \left (-2+5 \left (1+e^x\right ) x^2\right )}{x \left (2+5 \left (-1-e^4+e^x\right ) x+5 x^2\right )}+\log \left (-1-e^4+e^x+\frac {2}{5 x}+x\right )\right ) \, dx\\ &=\int \frac {(2+x) \left (-2+5 \left (1+e^x\right ) x^2\right )}{x \left (2+5 \left (-1-e^4+e^x\right ) x+5 x^2\right )} \, dx+\int \log \left (-1-e^4+e^x+\frac {2}{5 x}+x\right ) \, dx\\ &=x \log \left (-1-e^4+e^x+\frac {2}{5 x}+x\right )-\int \frac {-2+5 \left (1+e^x\right ) x^2}{2+5 \left (-1-e^4+e^x\right ) x+5 x^2} \, dx+\int \left (2+x+\frac {-4-6 x+2 \left (9+5 e^4\right ) x^2+5 e^4 x^3-5 x^4}{x \left (2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2\right )}\right ) \, dx\\ &=2 x+\frac {x^2}{2}+x \log \left (-1-e^4+e^x+\frac {2}{5 x}+x\right )+\int \frac {-4-6 x+2 \left (9+5 e^4\right ) x^2+5 e^4 x^3-5 x^4}{x \left (2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2\right )} \, dx-\int \left (x+\frac {-2-2 x+5 \left (2+e^4\right ) x^2-5 x^3}{2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2}\right ) \, dx\\ &=2 x+x \log \left (-1-e^4+e^x+\frac {2}{5 x}+x\right )-\int \frac {-2-2 x+5 \left (2+e^4\right ) x^2-5 x^3}{2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2} \, dx+\int \left (\frac {6}{-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2}+\frac {4}{x \left (-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2\right )}+\frac {5 x^3}{-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2}+\frac {2 \left (9+5 e^4\right ) x}{2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2}+\frac {5 e^4 x^2}{2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2}\right ) \, dx\\ &=2 x+x \log \left (-1-e^4+e^x+\frac {2}{5 x}+x\right )+4 \int \frac {1}{x \left (-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2\right )} \, dx+5 \int \frac {x^3}{-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2} \, dx+6 \int \frac {1}{-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2} \, dx+\left (5 e^4\right ) \int \frac {x^2}{2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2} \, dx+\left (2 \left (9+5 e^4\right )\right ) \int \frac {x}{2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2} \, dx-\int \left (\frac {2}{-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2}+\frac {2 x}{-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2}+\frac {5 x^3}{-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2}+\frac {5 \left (2+e^4\right ) x^2}{2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2}\right ) \, dx\\ &=2 x+x \log \left (-1-e^4+e^x+\frac {2}{5 x}+x\right )-2 \int \frac {1}{-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2} \, dx-2 \int \frac {x}{-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2} \, dx+4 \int \frac {1}{x \left (-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2\right )} \, dx+6 \int \frac {1}{-2-5 e^x x+5 \left (1+e^4\right ) x-5 x^2} \, dx+\left (5 e^4\right ) \int \frac {x^2}{2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2} \, dx-\left (5 \left (2+e^4\right )\right ) \int \frac {x^2}{2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2} \, dx+\left (2 \left (9+5 e^4\right )\right ) \int \frac {x}{2+5 e^x x-5 \left (1+e^4\right ) x+5 x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.59, size = 51, normalized size = 2.22 \begin {gather*} -2 \log (x)+x \log \left (-1-e^4+e^x+\frac {2}{5 x}+x\right )+2 \log \left (2-5 x-5 e^4 x+5 e^x x+5 x^2\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.72, size = 30, normalized size = 1.30 \begin {gather*} {\left (x + 2\right )} \log \left (\frac {5 \, x^{2} - 5 \, x e^{4} + 5 \, x e^{x} - 5 \, x + 2}{5 \, x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.22, size = 56, normalized size = 2.43 \begin {gather*} x \log \left (\frac {5 \, x^{2} - 5 \, x e^{4} + 5 \, x e^{x} - 5 \, x + 2}{5 \, x}\right ) + 2 \, \log \left (-5 \, x^{2} + 5 \, x e^{4} - 5 \, x e^{x} + 5 \, x - 2\right ) - 2 \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.27, size = 58, normalized size = 2.52
method | result | size |
norman | \(x \ln \left (\frac {5 \,{\mathrm e}^{x} x -5 x \,{\mathrm e}^{4}+5 x^{2}-5 x +2}{5 x}\right )+2 \ln \left (\frac {5 \,{\mathrm e}^{x} x -5 x \,{\mathrm e}^{4}+5 x^{2}-5 x +2}{5 x}\right )\) | \(58\) |
risch | \(x \ln \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )-x \ln \relax (x )+\frac {i \pi x \,\mathrm {csgn}\left (i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )}{x}\right )^{2}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right )}{2}+\frac {i \pi x \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )}{x}\right )^{3}}{2}-i \pi x \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )}{x}\right )^{2}+\frac {i \pi x \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )}{2}+i \pi x +2 \ln \left ({\mathrm e}^{x}-\frac {5 x \,{\mathrm e}^{4}-5 x^{2}+5 x -2}{5 x}\right )\) | \(283\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.50, size = 60, normalized size = 2.61 \begin {gather*} -x \log \relax (5) + x \log \left (5 \, x^{2} - 5 \, x {\left (e^{4} + 1\right )} + 5 \, x e^{x} + 2\right ) - x \log \relax (x) + 2 \, \log \left (\frac {5 \, x^{2} - 5 \, x {\left (e^{4} + 1\right )} + 5 \, x e^{x} + 2}{5 \, x}\right ) \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 = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^x\,\left (5\,x^3+10\,x^2\right )-2\,x+\ln \left (\frac {x\,{\mathrm {e}}^x-x\,{\mathrm {e}}^4-x+x^2+\frac {2}{5}}{x}\right )\,\left (2\,x+5\,x^2\,{\mathrm {e}}^x-5\,x^2\,{\mathrm {e}}^4-5\,x^2+5\,x^3\right )+10\,x^2+5\,x^3-4}{2\,x+5\,x^2\,{\mathrm {e}}^x-5\,x^2\,{\mathrm {e}}^4-5\,x^2+5\,x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.65, size = 49, normalized size = 2.13 \begin {gather*} x \log {\left (\frac {x^{2} + x e^{x} - x e^{4} - x + \frac {2}{5}}{x} \right )} + 2 \log {\left (e^{x} + \frac {5 x^{2} - 5 x e^{4} - 5 x + 2}{5 x} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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