Optimal. Leaf size=25 \[ -2+\frac {x}{-\frac {5}{-4+x}+\frac {e^{e^x+x}}{x^2}} \]
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Rubi [F] time = 6.56, 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 {20 x^4-10 x^5+e^{e^x} \left (e^{2 x} \left (-16 x^3+8 x^4-x^5\right )+e^x \left (48 x^2-40 x^3+11 x^4-x^5\right )\right )}{25 x^4+e^{2 e^x+2 x} \left (16-8 x+x^2\right )+e^{e^x+x} \left (40 x^2-10 x^3\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 {x^2 \left (-e^{e^x+x} (-4+x)^2 (-3+x)-e^{e^x+2 x} (-4+x)^2 x-10 (-2+x) x^2\right )}{\left (e^{e^x+x} (-4+x)-5 x^2\right )^2} \, dx\\ &=\int \left (-e^{-e^x} x^3-\frac {5 e^{-e^x} x^4 \left (8 e^{e^x}-5 e^{e^x} x+e^{e^x} x^2+5 x^3\right )}{\left (-4 e^{e^x+x}+e^{e^x+x} x-5 x^2\right )^2}-\frac {e^{-e^x} x^2 \left (12 e^{e^x}-7 e^{e^x} x+e^{e^x} x^2+10 x^3\right )}{-4 e^{e^x+x}+e^{e^x+x} x-5 x^2}\right ) \, dx\\ &=-\left (5 \int \frac {e^{-e^x} x^4 \left (8 e^{e^x}-5 e^{e^x} x+e^{e^x} x^2+5 x^3\right )}{\left (-4 e^{e^x+x}+e^{e^x+x} x-5 x^2\right )^2} \, dx\right )-\int e^{-e^x} x^3 \, dx-\int \frac {e^{-e^x} x^2 \left (12 e^{e^x}-7 e^{e^x} x+e^{e^x} x^2+10 x^3\right )}{-4 e^{e^x+x}+e^{e^x+x} x-5 x^2} \, dx\\ &=-\left (5 \int \frac {e^{-e^x} \left (5 x^7+e^{e^x} x^4 \left (8-5 x+x^2\right )\right )}{\left (e^{e^x+x} (-4+x)-5 x^2\right )^2} \, dx\right )-\int e^{-e^x} x^3 \, dx-\int \frac {e^{-e^x} \left (10 x^5+e^{e^x} x^2 \left (12-7 x+x^2\right )\right )}{e^{e^x+x} (-4+x)-5 x^2} \, dx\\ &=-\left (5 \int \left (\frac {8 x^4}{\left (-4 e^{e^x+x}+e^{e^x+x} x-5 x^2\right )^2}-\frac {5 x^5}{\left (-4 e^{e^x+x}+e^{e^x+x} x-5 x^2\right )^2}+\frac {x^6}{\left (-4 e^{e^x+x}+e^{e^x+x} x-5 x^2\right )^2}+\frac {5 e^{-e^x} x^7}{\left (4 e^{e^x+x}-e^{e^x+x} x+5 x^2\right )^2}\right ) \, dx\right )-\int e^{-e^x} x^3 \, dx-\int \left (\frac {12 x^2}{-4 e^{e^x+x}+e^{e^x+x} x-5 x^2}-\frac {7 x^3}{-4 e^{e^x+x}+e^{e^x+x} x-5 x^2}+\frac {x^4}{-4 e^{e^x+x}+e^{e^x+x} x-5 x^2}-\frac {10 e^{-e^x} x^5}{4 e^{e^x+x}-e^{e^x+x} x+5 x^2}\right ) \, dx\\ &=-\left (5 \int \frac {x^6}{\left (-4 e^{e^x+x}+e^{e^x+x} x-5 x^2\right )^2} \, dx\right )+7 \int \frac {x^3}{-4 e^{e^x+x}+e^{e^x+x} x-5 x^2} \, dx+10 \int \frac {e^{-e^x} x^5}{4 e^{e^x+x}-e^{e^x+x} x+5 x^2} \, dx-12 \int \frac {x^2}{-4 e^{e^x+x}+e^{e^x+x} x-5 x^2} \, dx+25 \int \frac {x^5}{\left (-4 e^{e^x+x}+e^{e^x+x} x-5 x^2\right )^2} \, dx-25 \int \frac {e^{-e^x} x^7}{\left (4 e^{e^x+x}-e^{e^x+x} x+5 x^2\right )^2} \, dx-40 \int \frac {x^4}{\left (-4 e^{e^x+x}+e^{e^x+x} x-5 x^2\right )^2} \, dx-\int e^{-e^x} x^3 \, dx-\int \frac {x^4}{-4 e^{e^x+x}+e^{e^x+x} x-5 x^2} \, dx\\ &=-\left (5 \int \frac {x^6}{\left (e^{e^x+x} (-4+x)-5 x^2\right )^2} \, dx\right )+7 \int \frac {x^3}{e^{e^x+x} (-4+x)-5 x^2} \, dx+10 \int \frac {e^{-e^x} x^5}{-e^{e^x+x} (-4+x)+5 x^2} \, dx-12 \int \frac {x^2}{e^{e^x+x} (-4+x)-5 x^2} \, dx+25 \int \frac {x^5}{\left (e^{e^x+x} (-4+x)-5 x^2\right )^2} \, dx-25 \int \frac {e^{-e^x} x^7}{\left (e^{e^x+x} (-4+x)-5 x^2\right )^2} \, dx-40 \int \frac {x^4}{\left (e^{e^x+x} (-4+x)-5 x^2\right )^2} \, dx-\int e^{-e^x} x^3 \, dx-\int \frac {x^4}{e^{e^x+x} (-4+x)-5 x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.88, size = 28, normalized size = 1.12 \begin {gather*} -\frac {(-4+x) x^3}{-e^{e^x+x} (-4+x)+5 x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 29, normalized size = 1.16 \begin {gather*} -\frac {x^{4} - 4 \, x^{3}}{5 \, x^{2} - {\left (x - 4\right )} e^{\left (x + e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 34, normalized size = 1.36 \begin {gather*} -\frac {x^{4} - 4 \, x^{3}}{5 \, x^{2} - x e^{\left (x + e^{x}\right )} + 4 \, e^{\left (x + e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 32, normalized size = 1.28
method | result | size |
risch | \(-\frac {\left (x -4\right ) x^{3}}{-x \,{\mathrm e}^{{\mathrm e}^{x}+x}+5 x^{2}+4 \,{\mathrm e}^{{\mathrm e}^{x}+x}}\) | \(32\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.53, size = 29, normalized size = 1.16 \begin {gather*} -\frac {x^{4} - 4 \, x^{3}}{5 \, x^{2} - {\left (x - 4\right )} e^{\left (x + e^{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}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^{2\,x}\,\left (x^5-8\,x^4+16\,x^3\right )-{\mathrm {e}}^x\,\left (-x^5+11\,x^4-40\,x^3+48\,x^2\right )\right )-20\,x^4+10\,x^5}{25\,x^4+{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (x^2-8\,x+16\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^x\,\left (40\,x^2-10\,x^3\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.28, size = 27, normalized size = 1.08 \begin {gather*} \frac {x^{4} - 4 x^{3}}{- 5 x^{2} + \left (x e^{x} - 4 e^{x}\right ) e^{e^{x}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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