Optimal. Leaf size=32 \[ -x+\frac {\frac {5+x}{2+e^{e+x^2}}-\log (x)}{3+e^3} \]
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Rubi [F] time = 1.48, 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 {-4-10 x-4 e^3 x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{12 x+4 e^3 x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\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 {-4-10 x-4 e^3 x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{\left (12+4 e^3\right ) x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx\\ &=\int \frac {-4+\left (-10-4 e^3\right ) x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{\left (12+4 e^3\right ) x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx\\ &=\int \frac {-4+\left (-10-4 e^3\right ) x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{\left (3+e^3\right ) \left (2+e^{e+x^2}\right )^2 x} \, dx\\ &=\frac {\int \frac {-4+\left (-10-4 e^3\right ) x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{\left (2+e^{e+x^2}\right )^2 x} \, dx}{3+e^3}\\ &=\frac {\int \left (\frac {4 x (5+x)}{\left (2+e^{e+x^2}\right )^2}+\frac {-1-\left (3+e^3\right ) x}{x}-\frac {-1+10 x+2 x^2}{2+e^{e+x^2}}\right ) \, dx}{3+e^3}\\ &=\frac {\int \frac {-1-\left (3+e^3\right ) x}{x} \, dx}{3+e^3}-\frac {\int \frac {-1+10 x+2 x^2}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {4 \int \frac {x (5+x)}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}\\ &=\frac {\int \left (-3-e^3-\frac {1}{x}\right ) \, dx}{3+e^3}-\frac {\int \left (-\frac {1}{2+e^{e+x^2}}+\frac {10 x}{2+e^{e+x^2}}+\frac {2 x^2}{2+e^{e+x^2}}\right ) \, dx}{3+e^3}+\frac {4 \int \left (\frac {5 x}{\left (2+e^{e+x^2}\right )^2}+\frac {x^2}{\left (2+e^{e+x^2}\right )^2}\right ) \, dx}{3+e^3}\\ &=-x-\frac {\log (x)}{3+e^3}+\frac {\int \frac {1}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {2 \int \frac {x^2}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {4 \int \frac {x^2}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}-\frac {10 \int \frac {x}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {20 \int \frac {x}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}\\ &=-x-\frac {\log (x)}{3+e^3}+\frac {\int \frac {1}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {2 \int \frac {x^2}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {4 \int \frac {x^2}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{2+e^{e+x}} \, dx,x,x^2\right )}{3+e^3}+\frac {10 \operatorname {Subst}\left (\int \frac {1}{\left (2+e^{e+x}\right )^2} \, dx,x,x^2\right )}{3+e^3}\\ &=-x-\frac {\log (x)}{3+e^3}+\frac {\int \frac {1}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {2 \int \frac {x^2}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {4 \int \frac {x^2}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{x (2+x)} \, dx,x,e^{e+x^2}\right )}{3+e^3}+\frac {10 \operatorname {Subst}\left (\int \frac {1}{x (2+x)^2} \, dx,x,e^{e+x^2}\right )}{3+e^3}\\ &=-x-\frac {\log (x)}{3+e^3}+\frac {\int \frac {1}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {2 \int \frac {x^2}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{e+x^2}\right )}{2 \left (3+e^3\right )}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{2+x} \, dx,x,e^{e+x^2}\right )}{2 \left (3+e^3\right )}+\frac {4 \int \frac {x^2}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}+\frac {10 \operatorname {Subst}\left (\int \left (\frac {1}{4 x}-\frac {1}{2 (2+x)^2}-\frac {1}{4 (2+x)}\right ) \, dx,x,e^{e+x^2}\right )}{3+e^3}\\ &=\frac {5}{\left (3+e^3\right ) \left (2+e^{e+x^2}\right )}-x-\frac {\log (x)}{3+e^3}+\frac {\int \frac {1}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {2 \int \frac {x^2}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {4 \int \frac {x^2}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 36, normalized size = 1.12 \begin {gather*} -\frac {\frac {-5-x}{2+e^{e+x^2}}+\left (3+e^3\right ) x+\log (x)}{3+e^3} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.87, size = 60, normalized size = 1.88 \begin {gather*} -\frac {2 \, x e^{3} + {\left (x e^{3} + 3 \, x\right )} e^{\left (x^{2} + e\right )} + {\left (e^{\left (x^{2} + e\right )} + 2\right )} \log \relax (x) + 5 \, x - 5}{{\left (e^{3} + 3\right )} e^{\left (x^{2} + e\right )} + 2 \, e^{3} + 6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.22, size = 71, normalized size = 2.22 \begin {gather*} -\frac {2 \, x e^{3} + x e^{\left (x^{2} + e + 3\right )} + 3 \, x e^{\left (x^{2} + e\right )} + e^{\left (x^{2} + e\right )} \log \relax (x) + 5 \, x + 2 \, \log \relax (x) - 5}{2 \, e^{3} + e^{\left (x^{2} + e + 3\right )} + 3 \, e^{\left (x^{2} + e\right )} + 6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.13, size = 36, normalized size = 1.12
method | result | size |
risch | \(-x -\frac {\ln \relax (x )}{{\mathrm e}^{3}+3}+\frac {5+x}{\left ({\mathrm e}^{3}+3\right ) \left (2+{\mathrm e}^{{\mathrm e}+x^{2}}\right )}\) | \(36\) |
norman | \(\frac {-x \,{\mathrm e}^{{\mathrm e}+x^{2}}-\frac {\left (5+2 \,{\mathrm e}^{3}\right ) x}{{\mathrm e}^{3}+3}+\frac {5}{{\mathrm e}^{3}+3}}{2+{\mathrm e}^{{\mathrm e}+x^{2}}}-\frac {\ln \relax (x )}{{\mathrm e}^{3}+3}\) | \(58\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.52, size = 64, normalized size = 2.00 \begin {gather*} -\frac {x {\left (e^{\left (e + 3\right )} + 3 \, e^{e}\right )} e^{\left (x^{2}\right )} + x {\left (2 \, e^{3} + 5\right )} - 5}{{\left (e^{\left (e + 3\right )} + 3 \, e^{e}\right )} e^{\left (x^{2}\right )} + 2 \, e^{3} + 6} - \frac {\log \relax (x)}{e^{3} + 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.76, size = 76, normalized size = 2.38 \begin {gather*} -\frac {3\,x+\ln \relax (x)+x\,{\mathrm {e}}^3}{{\mathrm {e}}^3+3}-\frac {5\,x+2\,\ln \relax (x)+2\,x\,{\mathrm {e}}^3-\frac {\left (2\,{\mathrm {e}}^3+6\right )\,\left (3\,x+\ln \relax (x)+x\,{\mathrm {e}}^3+\frac {5}{2}\right )}{{\mathrm {e}}^3+3}}{\left ({\mathrm {e}}^{x^2+\mathrm {e}}+2\right )\,\left ({\mathrm {e}}^3+3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.19, size = 39, normalized size = 1.22 \begin {gather*} \frac {x + 5}{\left (3 + e^{3}\right ) e^{x^{2} + e} + 6 + 2 e^{3}} + \frac {- x \left (3 + e^{3}\right ) - \log {\relax (x )}}{3 + e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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