Optimal. Leaf size=36 \[ 4-x+\frac {4 \left (x-\frac {1}{4} x \log (4)\right )}{5 \left (e^{\frac {3 x}{\frac {4}{x}+x}}+x\right )} \]
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Rubi [F] time = 2.94, 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 {-80 x^2-40 x^4-5 x^6+e^{\frac {6 x^2}{4+x^2}} \left (-80-40 x^2-5 x^4\right )+e^{\frac {3 x^2}{4+x^2}} \left (64-160 x-64 x^2-80 x^3+4 x^4-10 x^5+\left (-16+16 x^2-x^4\right ) \log (4)\right )}{80 x^2+40 x^4+5 x^6+e^{\frac {6 x^2}{4+x^2}} \left (80+40 x^2+5 x^4\right )+e^{\frac {3 x^2}{4+x^2}} \left (160 x+80 x^3+10 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 {-5 e^{\frac {6 x^2}{4+x^2}} \left (4+x^2\right )^2-5 x^2 \left (4+x^2\right )^2-e^{\frac {3 x^2}{4+x^2}} \left (160 x+80 x^3+10 x^5+16 (-4+\log (4))-16 x^2 (-4+\log (4))+x^4 (-4+\log (4))\right )}{5 \left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {-5 e^{\frac {6 x^2}{4+x^2}} \left (4+x^2\right )^2-5 x^2 \left (4+x^2\right )^2-e^{\frac {3 x^2}{4+x^2}} \left (160 x+80 x^3+10 x^5+16 (-4+\log (4))-16 x^2 (-4+\log (4))+x^4 (-4+\log (4))\right )}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx\\ &=\frac {1}{5} \int \left (-5+\frac {x \left (16-16 x^2+x^4\right ) (-4+\log (4))}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2}-\frac {\left (16-16 x^2+x^4\right ) (-4+\log (4))}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2}\right ) \, dx\\ &=-x+\frac {1}{5} (4-\log (4)) \int \frac {16-16 x^2+x^4}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2} \, dx+\frac {1}{5} (-4+\log (4)) \int \frac {x \left (16-16 x^2+x^4\right )}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx\\ &=-x+\frac {1}{5} (4-\log (4)) \int \left (\frac {1}{e^{\frac {3 x^2}{4+x^2}}+x}+\frac {96}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2}-\frac {24}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )}\right ) \, dx+\frac {1}{5} (-4+\log (4)) \int \left (\frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2}+\frac {96 x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2}-\frac {24 x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )}\right ) \, dx\\ &=-x+\frac {1}{5} (4-\log (4)) \int \frac {1}{e^{\frac {3 x^2}{4+x^2}}+x} \, dx+\frac {1}{5} (24 (4-\log (4))) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )} \, dx-\frac {1}{5} (24 (4-\log (4))) \int \frac {1}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )} \, dx-\frac {1}{5} (96 (4-\log (4))) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx+\frac {1}{5} (96 (4-\log (4))) \int \frac {1}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2} \, dx+\frac {1}{5} (-4+\log (4)) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2} \, dx\\ &=-x+\frac {1}{5} (4-\log (4)) \int \frac {1}{e^{\frac {3 x^2}{4+x^2}}+x} \, dx+\frac {1}{5} (24 (4-\log (4))) \int \left (-\frac {1}{2 (2 i-x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2}+\frac {1}{2 (2 i+x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2}\right ) \, dx-\frac {1}{5} (24 (4-\log (4))) \int \left (\frac {i}{4 (2 i-x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )}+\frac {i}{4 (2 i+x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )}\right ) \, dx-\frac {1}{5} (96 (4-\log (4))) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx+\frac {1}{5} (96 (4-\log (4))) \int \frac {1}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2} \, dx+\frac {1}{5} (-4+\log (4)) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2} \, dx\\ &=-x-\frac {1}{5} (6 i (4-\log (4))) \int \frac {1}{(2 i-x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )} \, dx-\frac {1}{5} (6 i (4-\log (4))) \int \frac {1}{(2 i+x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )} \, dx+\frac {1}{5} (4-\log (4)) \int \frac {1}{e^{\frac {3 x^2}{4+x^2}}+x} \, dx-\frac {1}{5} (12 (4-\log (4))) \int \frac {1}{(2 i-x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2} \, dx+\frac {1}{5} (12 (4-\log (4))) \int \frac {1}{(2 i+x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2} \, dx-\frac {1}{5} (96 (4-\log (4))) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx+\frac {1}{5} (96 (4-\log (4))) \int \frac {1}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2} \, dx+\frac {1}{5} (-4+\log (4)) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.14, size = 32, normalized size = 0.89 \begin {gather*} \frac {1}{5} \left (-5 x-\frac {x (-4+\log (4))}{e^{3-\frac {12}{4+x^2}}+x}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 49, normalized size = 1.36 \begin {gather*} -\frac {5 \, x^{2} + 5 \, x e^{\left (\frac {3 \, x^{2}}{x^{2} + 4}\right )} + 2 \, x \log \relax (2) - 4 \, x}{5 \, {\left (x + e^{\left (\frac {3 \, x^{2}}{x^{2} + 4}\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 49, normalized size = 1.36 \begin {gather*} -\frac {5 \, x^{2} + 5 \, x e^{\left (\frac {3 \, x^{2}}{x^{2} + 4}\right )} + 2 \, x \log \relax (2) - 4 \, x}{5 \, {\left (x + e^{\left (\frac {3 \, x^{2}}{x^{2} + 4}\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.30, size = 29, normalized size = 0.81
method | result | size |
risch | \(-x -\frac {2 x \left (\ln \relax (2)-2\right )}{5 \left ({\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}+x \right )}\) | \(29\) |
norman | \(\frac {\left (-\frac {16}{5}+\frac {8 \ln \relax (2)}{5}\right ) {\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}+\left (-\frac {4}{5}+\frac {2 \ln \relax (2)}{5}\right ) x^{2} {\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}-4 x^{2}-x^{4}-4 x \,{\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}-{\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}} x^{3}}{x^{3}+x^{2} {\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}+4 x +4 \,{\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}}\) | \(131\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.66, size = 48, normalized size = 1.33 \begin {gather*} -\frac {5 \, x^{2} e^{\left (\frac {12}{x^{2} + 4}\right )} + 5 \, x e^{3} - 2 \, {\left (\log \relax (2) - 2\right )} e^{3}}{5 \, {\left (x e^{\left (\frac {12}{x^{2} + 4}\right )} + e^{3}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.51, size = 38, normalized size = 1.06 \begin {gather*} \frac {x^2-\frac {x\,\left (5\,x+2\,\ln \relax (2)-4\right )}{5}}{x+{\mathrm {e}}^{\frac {3\,x^2}{x^2+4}}}-x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.25, size = 27, normalized size = 0.75 \begin {gather*} - x + \frac {- 2 x \log {\relax (2 )} + 4 x}{5 x + 5 e^{\frac {3 x^{2}}{x^{2} + 4}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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