Optimal. Leaf size=22 \[ -x+\frac {5}{x+\frac {e^{-x} x}{\sqrt {2}}} \]
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Rubi [F] time = 0.91, 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^{2-2 x} x^2+e^{1-x+\frac {1}{4} (4+\log (4))} \left (-5+5 x-2 x^2\right )+e^{\frac {1}{2} (4+\log (4))} \left (-5-x^2\right )}{e^{2-2 x} x^2+e^{\frac {1}{2} (4+\log (4))} x^2+2 e^{1-x+\frac {1}{4} (4+\log (4))} x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x^2+\sqrt {2} e^x \left (-5+5 x-2 x^2\right )-2 e^{2 x} \left (5+x^2\right )}{\left (x+\sqrt {2} e^x x\right )^2} \, dx\\ &=\int \left (-\frac {5}{\left (1+\sqrt {2} e^x\right )^2 x}+\frac {5 (1+x)}{\left (1+\sqrt {2} e^x\right ) x^2}+\frac {-5-x^2}{x^2}\right ) \, dx\\ &=-\left (5 \int \frac {1}{\left (1+\sqrt {2} e^x\right )^2 x} \, dx\right )+5 \int \frac {1+x}{\left (1+\sqrt {2} e^x\right ) x^2} \, dx+\int \frac {-5-x^2}{x^2} \, dx\\ &=5 \int \left (\frac {1}{\left (1+\sqrt {2} e^x\right ) x^2}+\frac {1}{\left (1+\sqrt {2} e^x\right ) x}\right ) \, dx-5 \int \frac {1}{\left (1+\sqrt {2} e^x\right )^2 x} \, dx+\int \left (-1-\frac {5}{x^2}\right ) \, dx\\ &=\frac {5}{x}-x+5 \int \frac {1}{\left (1+\sqrt {2} e^x\right ) x^2} \, dx-5 \int \frac {1}{\left (1+\sqrt {2} e^x\right )^2 x} \, dx+5 \int \frac {1}{\left (1+\sqrt {2} e^x\right ) x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.27, size = 34, normalized size = 1.55 \begin {gather*} -\frac {x^2+\sqrt {2} e^x \left (-5+x^2\right )}{x+\sqrt {2} e^x x} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.86, size = 50, normalized size = 2.27 \begin {gather*} -\frac {x^{2} e^{\left (-x + \frac {1}{2} \, \log \relax (2) + 2\right )} + {\left (x^{2} - 5\right )} e^{\left (\log \relax (2) + 2\right )}}{x e^{\left (-x + \frac {1}{2} \, \log \relax (2) + 2\right )} + x e^{\left (\log \relax (2) + 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {x^{2} e^{\left (-2 \, x + 2\right )} + {\left (2 \, x^{2} - 5 \, x + 5\right )} e^{\left (-x + \frac {1}{2} \, \log \relax (2) + 2\right )} + {\left (x^{2} + 5\right )} e^{\left (\log \relax (2) + 2\right )}}{2 \, x^{2} e^{\left (-x + \frac {1}{2} \, \log \relax (2) + 2\right )} + x^{2} e^{\left (-2 \, x + 2\right )} + x^{2} e^{\left (\log \relax (2) + 2\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.41, size = 29, normalized size = 1.32
method | result | size |
risch | \(-x +\frac {10 \,{\mathrm e}}{x \left (2 \,{\mathrm e}+\sqrt {2}\, {\mathrm e}^{1-x}\right )}\) | \(29\) |
norman | \(\frac {-x^{2} {\mathrm e}^{1-x}+5 \,{\mathrm e} \sqrt {2}-{\mathrm e} \sqrt {2}\, x^{2}}{x \left ({\mathrm e}^{1+\frac {\ln \relax (2)}{2}}+{\mathrm e}^{1-x}\right )}\) | \(50\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.69, size = 29, normalized size = 1.32 \begin {gather*} \frac {5\,\sqrt {2}\,\mathrm {e}}{x\,\left ({\mathrm {e}}^{1-x}+\sqrt {2}\,\mathrm {e}\right )}-x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.31, size = 26, normalized size = 1.18 \begin {gather*} - x + \frac {5 \sqrt {2} e}{x \left (e^{1 - x} + \sqrt {2} e\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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