Optimal. Leaf size=28 \[ e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}+x \]
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Rubi [F] time = 21.48, 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 {2+8 x+8 x^2+e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x} \left (2+8 x+8 x^2+e^{-e+e^x+\frac {e^{-e+e^x}}{2+4 x}} \left (-2+e^x (1+2 x)\right )\right )}{2+8 x+8 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 {2+8 x+8 x^2+e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x} \left (2+8 x+8 x^2+e^{-e+e^x+\frac {e^{-e+e^x}}{2+4 x}} \left (-2+e^x (1+2 x)\right )\right )}{2 (1+2 x)^2} \, dx\\ &=\frac {1}{2} \int \frac {2+8 x+8 x^2+e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x} \left (2+8 x+8 x^2+e^{-e+e^x+\frac {e^{-e+e^x}}{2+4 x}} \left (-2+e^x (1+2 x)\right )\right )}{(1+2 x)^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {2}{(1+2 x)^2}+\frac {2 e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}}{(1+2 x)^2}+\frac {8 x}{(1+2 x)^2}+\frac {8 e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x} x}{(1+2 x)^2}+\frac {8 x^2}{(1+2 x)^2}+\frac {8 e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x} x^2}{(1+2 x)^2}+\frac {\exp \left (e^x+e^{\frac {e^{-e+e^x}}{2+4 x}}-e \left (1-e^{24}\right )+x+\frac {e^{-e+e^x}}{2+4 x}\right ) \left (-2+e^x+2 e^x x\right )}{(1+2 x)^2}\right ) \, dx\\ &=-\frac {1}{2 (1+2 x)}+\frac {1}{2} \int \frac {\exp \left (e^x+e^{\frac {e^{-e+e^x}}{2+4 x}}-e \left (1-e^{24}\right )+x+\frac {e^{-e+e^x}}{2+4 x}\right ) \left (-2+e^x+2 e^x x\right )}{(1+2 x)^2} \, dx+4 \int \frac {x}{(1+2 x)^2} \, dx+4 \int \frac {e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x} x}{(1+2 x)^2} \, dx+4 \int \frac {x^2}{(1+2 x)^2} \, dx+4 \int \frac {e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x} x^2}{(1+2 x)^2} \, dx+\int \frac {e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}}{(1+2 x)^2} \, dx\\ &=-\frac {1}{2 (1+2 x)}+\frac {1}{2} \int \left (-\frac {2 \exp \left (e^x+e^{\frac {e^{-e+e^x}}{2+4 x}}-e \left (1-e^{24}\right )+x+\frac {e^{-e+e^x}}{2+4 x}\right )}{(1+2 x)^2}+\frac {\exp \left (e^x+e^{\frac {e^{-e+e^x}}{2+4 x}}-e \left (1-e^{24}\right )+2 x+\frac {e^{-e+e^x}}{2+4 x}\right )}{1+2 x}\right ) \, dx+4 \int \left (\frac {1}{4}+\frac {1}{4 (1+2 x)^2}-\frac {1}{2 (1+2 x)}\right ) \, dx+4 \int \left (-\frac {1}{2 (1+2 x)^2}+\frac {1}{2 (1+2 x)}\right ) \, dx+4 \int \left (\frac {1}{4} e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}+\frac {e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}}{4 (1+2 x)^2}-\frac {e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}}{2 (1+2 x)}\right ) \, dx+4 \int \left (-\frac {e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}}{2 (1+2 x)^2}+\frac {e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}}{2 (1+2 x)}\right ) \, dx+\int \frac {e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}}{(1+2 x)^2} \, dx\\ &=x+\frac {1}{2} \int \frac {\exp \left (e^x+e^{\frac {e^{-e+e^x}}{2+4 x}}-e \left (1-e^{24}\right )+2 x+\frac {e^{-e+e^x}}{2+4 x}\right )}{1+2 x} \, dx-2 \int \frac {e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}}{(1+2 x)^2} \, dx+\int e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x} \, dx+2 \int \frac {e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}}{(1+2 x)^2} \, dx-\int \frac {\exp \left (e^x+e^{\frac {e^{-e+e^x}}{2+4 x}}-e \left (1-e^{24}\right )+x+\frac {e^{-e+e^x}}{2+4 x}\right )}{(1+2 x)^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.74, size = 28, normalized size = 1.00 \begin {gather*} e^{e^{25}+e^{\frac {e^{-e+e^x}}{2+4 x}}+x}+x \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.90, size = 65, normalized size = 2.32 \begin {gather*} x + e^{\left ({\left ({\left (x + e^{25}\right )} e^{\left (-e + e^{x}\right )} + e^{\left (-\frac {2 \, {\left (2 \, x + 1\right )} e - 2 \, {\left (2 \, x + 1\right )} e^{x} - e^{\left (-e + e^{x}\right )}}{2 \, {\left (2 \, x + 1\right )}}\right )}\right )} e^{\left (e - e^{x}\right )}\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 {8 \, x^{2} + {\left (8 \, x^{2} + {\left ({\left (2 \, x + 1\right )} e^{x} - 2\right )} e^{\left (\frac {e^{\left (-e + e^{x}\right )}}{2 \, {\left (2 \, x + 1\right )}} - e + e^{x}\right )} + 8 \, x + 2\right )} e^{\left (x + e^{25} + e^{\left (\frac {e^{\left (-e + e^{x}\right )}}{2 \, {\left (2 \, x + 1\right )}}\right )}\right )} + 8 \, x + 2}{2 \, {\left (4 \, x^{2} + 4 \, x + 1\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.32, size = 26, normalized size = 0.93
method | result | size |
risch | \({\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{x}-{\mathrm e}}}{4 x +2}}+{\mathrm e}^{25}+x}+x\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.61, size = 25, normalized size = 0.89 \begin {gather*} x + e^{\left (x + e^{25} + e^{\left (\frac {e^{\left (e^{x}\right )}}{2 \, {\left (2 \, x e^{e} + e^{e}\right )}}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.08, size = 26, normalized size = 0.93 \begin {gather*} x+{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^{-\mathrm {e}}\,{\mathrm {e}}^{{\mathrm {e}}^x}}{4\,x+2}}}\,{\mathrm {e}}^{{\mathrm {e}}^{25}}\,{\mathrm {e}}^x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 6.40, size = 22, normalized size = 0.79 \begin {gather*} x + e^{x + e^{\frac {e^{e^{x} - e}}{4 x + 2}} + e^{25}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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