Optimal. Leaf size=24 \[ e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x \]
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Rubi [F] time = 20.11, 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^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{25+e^4+20 x+4 x^2+e^2 (10+4 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{\left (5+e^2\right )^2+4 \left (5+e^2\right ) x+4 x^2} \, dx\\ &=\int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{\left (5+e^2+2 x\right )^2} \, dx\\ &=\int \left (\frac {25 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2}-\frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x} (-1+x)}{\left (5+e^2+2 x\right )^2}-\frac {5 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x}{\left (5+e^2+2 x\right )^2}+\frac {3 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}} \left (5+e^2\right ) x}{\left (5+e^2+2 x\right )^2}-\frac {16 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x^2}{\left (5+e^2+2 x\right )^2}-\frac {4 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x^3}{\left (5+e^2+2 x\right )^2}-\frac {2 e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x} (-1+x) (5+2 x)}{\left (5+e^2+2 x\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x} (-1+x) (5+2 x)}{\left (5+e^2+2 x\right )^2} \, dx\right )-4 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x^3}{\left (5+e^2+2 x\right )^2} \, dx-5 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x}{\left (5+e^2+2 x\right )^2} \, dx-16 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x^2}{\left (5+e^2+2 x\right )^2} \, dx+25 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx+\left (3 \left (5+e^2\right )\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}} x}{\left (5+e^2+2 x\right )^2} \, dx-\int \frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x} (-1+x)}{\left (5+e^2+2 x\right )^2} \, dx\\ &=-\left (2 \int \left (\frac {1}{2} e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x}+\frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (7+e^2\right )}{2 \left (5+e^2+2 x\right )^2}+\frac {e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (-7-2 e^2\right )}{2 \left (5+e^2+2 x\right )}\right ) \, dx\right )-4 \int \left (\frac {1}{4} e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (-5-e^2\right )+\frac {1}{4} e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x-\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (5+e^2\right )^3}{8 \left (5+e^2+2 x\right )^2}+\frac {3 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (5+e^2\right )^2}{8 \left (5+e^2+2 x\right )}\right ) \, dx-5 \int \left (\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (-5-e^2\right )}{2 \left (5+e^2+2 x\right )^2}+\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{2 \left (5+e^2+2 x\right )}\right ) \, dx-16 \int \left (\frac {1}{4} e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}+\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (5+e^2\right )^2}{4 \left (5+e^2+2 x\right )^2}+\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (-5-e^2\right )}{2 \left (5+e^2+2 x\right )}\right ) \, dx+25 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx+\left (3 \left (5+e^2\right )\right ) \int \left (\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}} \left (-5-e^2\right )}{2 \left (5+e^2+2 x\right )^2}+\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}}}{2 \left (5+e^2+2 x\right )}\right ) \, dx-\int \left (\frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (-7-e^2\right )}{2 \left (5+e^2+2 x\right )^2}+\frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x}}{2 \left (5+e^2+2 x\right )}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x}}{5+e^2+2 x} \, dx\right )-\frac {5}{2} \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{5+e^2+2 x} \, dx-4 \int e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \, dx+25 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx-\left (-7-2 e^2\right ) \int \frac {e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x}}{5+e^2+2 x} \, dx-\frac {1}{2} \left (-7-e^2\right ) \int \frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx-\left (-5-e^2\right ) \int e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \, dx+\frac {1}{2} \left (3 \left (5+e^2\right )\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}}}{5+e^2+2 x} \, dx+\frac {1}{2} \left (5 \left (5+e^2\right )\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx+\left (8 \left (5+e^2\right )\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{5+e^2+2 x} \, dx-\frac {1}{2} \left (3 \left (5+e^2\right )^2\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}}}{\left (5+e^2+2 x\right )^2} \, dx-\frac {1}{2} \left (3 \left (5+e^2\right )^2\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{5+e^2+2 x} \, dx-\left (4 \left (5+e^2\right )^2\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx+\frac {1}{2} \left (5+e^2\right )^3 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx-\left (7+e^2\right ) \int \frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx-\int e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x} \, dx-\int e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.32, size = 24, normalized size = 1.00 \begin {gather*} e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 21, normalized size = 0.88 \begin {gather*} x e^{\left (-x + e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5\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 {{\left (4 \, x^{3} + 16 \, x^{2} + {\left (x - 1\right )} e^{4} + 2 \, {\left (2 \, x^{2} + 3 \, x - 5\right )} e^{2} - 3 \, {\left (x e^{2} + 5 \, x\right )} e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5 \, x - 25\right )} e^{\left (-x + e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5\right )}}{4 \, x^{2} + 2 \, {\left (2 \, x + 5\right )} e^{2} + 20 \, x + e^{4} + 25}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.33, size = 22, normalized size = 0.92
method | result | size |
risch | \(x \,{\mathrm e}^{{\mathrm e}^{\frac {3 x}{{\mathrm e}^{2}+5+2 x}}-x +5}\) | \(22\) |
norman | \(\frac {\left (\left ({\mathrm e}^{2}+5\right ) x +2 x^{2}\right ) {\mathrm e}^{{\mathrm e}^{\frac {3 x}{{\mathrm e}^{2}+5+2 x}}-x +5}}{{\mathrm e}^{2}+5+2 x}\) | \(44\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (4 \, x^{3} + 16 \, x^{2} + {\left (x - 1\right )} e^{4} + 2 \, {\left (2 \, x^{2} + 3 \, x - 5\right )} e^{2} - 3 \, {\left (x e^{2} + 5 \, x\right )} e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5 \, x - 25\right )} e^{\left (-x + e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5\right )}}{4 \, x^{2} + 2 \, {\left (2 \, x + 5\right )} e^{2} + 20 \, x + e^{4} + 25}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.55, size = 22, normalized size = 0.92 \begin {gather*} x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {3\,x}{2\,x+{\mathrm {e}}^2+5}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 36.81, size = 19, normalized size = 0.79 \begin {gather*} x e^{- x + e^{\frac {3 x}{2 x + 5 + e^{2}}} + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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