Optimal. Leaf size=29 \[ 2 x-x^2+\frac {-1+e^{x+x^2}}{-4+x+e^5 x} \]
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Rubi [B] time = 0.97, antiderivative size = 400, normalized size of antiderivative = 13.79, number of steps used = 17, number of rules used = 8, integrand size = 110, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {6, 6741, 27, 6742, 43, 77, 2289, 1850} \begin {gather*} -\frac {e^{10} x^2}{\left (1+e^5\right )^2}-\frac {2 e^5 x^2}{\left (1+e^5\right )^2}-\frac {x^2}{\left (1+e^5\right )^2}-\frac {e^{x^2+x}}{4-\left (1+e^5\right ) x}+\frac {18 x}{\left (1+e^5\right )^2}-\frac {2 e^{10} \left (7-e^5\right ) x}{\left (1+e^5\right )^3}-\frac {4 e^5 \left (3-5 e^5\right ) x}{\left (1+e^5\right )^3}-\frac {16 x}{\left (1+e^5\right )^3}+\frac {e^5 \left (1+195 e^5-61 e^{10}+e^{15}\right )}{\left (1+e^5\right )^4 \left (4-\left (1+e^5\right ) x\right )}+\frac {33}{\left (1+e^5\right ) \left (4-\left (1+e^5\right ) x\right )}-\frac {192}{\left (1+e^5\right )^2 \left (4-\left (1+e^5\right ) x\right )}+\frac {288}{\left (1+e^5\right )^3 \left (4-\left (1+e^5\right ) x\right )}-\frac {32 e^{10} \left (3-e^5\right )}{\left (1+e^5\right )^4 \left (4-\left (1+e^5\right ) x\right )}-\frac {128}{\left (1+e^5\right )^4 \left (4-\left (1+e^5\right ) x\right )}-\frac {16 e^5 \left (3-8 e^5+e^{10}\right ) \log \left (4-\left (1+e^5\right ) x\right )}{\left (1+e^5\right )^4}-\frac {48 \log \left (4-\left (1+e^5\right ) x\right )}{\left (1+e^5\right )^2}+\frac {144 \log \left (4-\left (1+e^5\right ) x\right )}{\left (1+e^5\right )^3}-\frac {16 e^{10} \left (5-e^5\right ) \log \left (4-\left (1+e^5\right ) x\right )}{\left (1+e^5\right )^4}-\frac {96 \log \left (4-\left (1+e^5\right ) x\right )}{\left (1+e^5\right )^4} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 27
Rule 43
Rule 77
Rule 1850
Rule 2289
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {33-48 x+18 x^2-2 x^3+e^5 \left (1-16 x+20 x^2-4 x^3\right )+e^{10} \left (2 x^2-2 x^3\right )+e^{x+x^2} \left (-5-7 x+2 x^2+e^5 \left (-1+x+2 x^2\right )\right )}{16-8 x+\left (1+e^{10}\right ) x^2+e^5 \left (-8 x+2 x^2\right )} \, dx\\ &=\int \frac {33-48 x+18 x^2-2 x^3+e^5 \left (1-16 x+20 x^2-4 x^3\right )+e^{10} \left (2 x^2-2 x^3\right )+e^{x+x^2} \left (-5-7 x+2 x^2+e^5 \left (-1+x+2 x^2\right )\right )}{16-8 \left (1+e^5\right ) x+\left (1+e^5\right )^2 x^2} \, dx\\ &=\int \frac {33-48 x+18 x^2-2 x^3+e^5 \left (1-16 x+20 x^2-4 x^3\right )+e^{10} \left (2 x^2-2 x^3\right )+e^{x+x^2} \left (-5-7 x+2 x^2+e^5 \left (-1+x+2 x^2\right )\right )}{\left (-4+x+e^5 x\right )^2} \, dx\\ &=\int \frac {33-48 x+18 x^2-2 x^3+e^5 \left (1-16 x+20 x^2-4 x^3\right )+e^{10} \left (2 x^2-2 x^3\right )+e^{x+x^2} \left (-5-7 x+2 x^2+e^5 \left (-1+x+2 x^2\right )\right )}{\left (-4+\left (1+e^5\right ) x\right )^2} \, dx\\ &=\int \left (\frac {33}{\left (4-\left (1+e^5\right ) x\right )^2}-\frac {48 x}{\left (4-\left (1+e^5\right ) x\right )^2}+\frac {18 x^2}{\left (4-\left (1+e^5\right ) x\right )^2}+\frac {2 e^{10} (1-x) x^2}{\left (4-\left (1+e^5\right ) x\right )^2}-\frac {2 x^3}{\left (4-\left (1+e^5\right ) x\right )^2}+\frac {e^{x+x^2} \left (-5-e^5-\left (7-e^5\right ) x+2 \left (1+e^5\right ) x^2\right )}{\left (4-\left (1+e^5\right ) x\right )^2}+\frac {e^5 \left (1-16 x+20 x^2-4 x^3\right )}{\left (4-\left (1+e^5\right ) x\right )^2}\right ) \, dx\\ &=\frac {33}{\left (1+e^5\right ) \left (4-\left (1+e^5\right ) x\right )}-2 \int \frac {x^3}{\left (4-\left (1+e^5\right ) x\right )^2} \, dx+18 \int \frac {x^2}{\left (4-\left (1+e^5\right ) x\right )^2} \, dx-48 \int \frac {x}{\left (4-\left (1+e^5\right ) x\right )^2} \, dx+e^5 \int \frac {1-16 x+20 x^2-4 x^3}{\left (4-\left (1+e^5\right ) x\right )^2} \, dx+\left (2 e^{10}\right ) \int \frac {(1-x) x^2}{\left (4-\left (1+e^5\right ) x\right )^2} \, dx+\int \frac {e^{x+x^2} \left (-5-e^5-\left (7-e^5\right ) x+2 \left (1+e^5\right ) x^2\right )}{\left (4-\left (1+e^5\right ) x\right )^2} \, dx\\ &=-\frac {e^{x+x^2}}{4-\left (1+e^5\right ) x}+\frac {33}{\left (1+e^5\right ) \left (4-\left (1+e^5\right ) x\right )}-2 \int \left (\frac {8}{\left (1+e^5\right )^3}+\frac {x}{\left (1+e^5\right )^2}+\frac {64}{\left (1+e^5\right )^3 \left (4-\left (1+e^5\right ) x\right )^2}+\frac {48}{\left (-1-e^5\right )^3 \left (4-\left (1+e^5\right ) x\right )}\right ) \, dx+18 \int \left (\frac {1}{\left (1+e^5\right )^2}+\frac {16}{\left (1+e^5\right )^2 \left (4-\left (1+e^5\right ) x\right )^2}+\frac {8}{\left (1+e^5\right )^2 \left (-4+\left (1+e^5\right ) x\right )}\right ) \, dx-48 \int \left (\frac {4}{\left (1+e^5\right ) \left (4-\left (1+e^5\right ) x\right )^2}+\frac {1}{\left (-1-e^5\right ) \left (4-\left (1+e^5\right ) x\right )}\right ) \, dx+e^5 \int \left (\frac {4 \left (-3+5 e^5\right )}{\left (1+e^5\right )^3}-\frac {4 x}{\left (1+e^5\right )^2}+\frac {1+195 e^5-61 e^{10}+e^{15}}{\left (1+e^5\right )^3 \left (4-\left (1+e^5\right ) x\right )^2}+\frac {16 \left (3-8 e^5+e^{10}\right )}{\left (1+e^5\right )^3 \left (4-\left (1+e^5\right ) x\right )}\right ) \, dx+\left (2 e^{10}\right ) \int \left (\frac {-7+e^5}{\left (1+e^5\right )^3}-\frac {x}{\left (1+e^5\right )^2}+\frac {16 \left (-3+e^5\right )}{\left (1+e^5\right )^3 \left (4-\left (1+e^5\right ) x\right )^2}+\frac {8 \left (5-e^5\right )}{\left (1+e^5\right )^3 \left (4-\left (1+e^5\right ) x\right )}\right ) \, dx\\ &=-\frac {16 x}{\left (1+e^5\right )^3}-\frac {4 e^5 \left (3-5 e^5\right ) x}{\left (1+e^5\right )^3}-\frac {2 e^{10} \left (7-e^5\right ) x}{\left (1+e^5\right )^3}+\frac {18 x}{\left (1+e^5\right )^2}-\frac {x^2}{\left (1+e^5\right )^2}-\frac {2 e^5 x^2}{\left (1+e^5\right )^2}-\frac {e^{10} x^2}{\left (1+e^5\right )^2}-\frac {e^{x+x^2}}{4-\left (1+e^5\right ) x}-\frac {128}{\left (1+e^5\right )^4 \left (4-\left (1+e^5\right ) x\right )}-\frac {32 e^{10} \left (3-e^5\right )}{\left (1+e^5\right )^4 \left (4-\left (1+e^5\right ) x\right )}+\frac {288}{\left (1+e^5\right )^3 \left (4-\left (1+e^5\right ) x\right )}-\frac {192}{\left (1+e^5\right )^2 \left (4-\left (1+e^5\right ) x\right )}+\frac {33}{\left (1+e^5\right ) \left (4-\left (1+e^5\right ) x\right )}+\frac {e^5 \left (1+195 e^5-61 e^{10}+e^{15}\right )}{\left (1+e^5\right )^4 \left (4-\left (1+e^5\right ) x\right )}-\frac {96 \log \left (4-\left (1+e^5\right ) x\right )}{\left (1+e^5\right )^4}-\frac {16 e^{10} \left (5-e^5\right ) \log \left (4-\left (1+e^5\right ) x\right )}{\left (1+e^5\right )^4}+\frac {144 \log \left (4-\left (1+e^5\right ) x\right )}{\left (1+e^5\right )^3}-\frac {48 \log \left (4-\left (1+e^5\right ) x\right )}{\left (1+e^5\right )^2}-\frac {16 e^5 \left (3-8 e^5+e^{10}\right ) \log \left (4-\left (1+e^5\right ) x\right )}{\left (1+e^5\right )^4}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.07, size = 29, normalized size = 1.00 \begin {gather*} 2 x-x^2+\frac {-1+e^{x+x^2}}{-4+x+e^5 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 44, normalized size = 1.52 \begin {gather*} -\frac {x^{3} - 6 \, x^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e^{5} + 8 \, x - e^{\left (x^{2} + x\right )} + 1}{x e^{5} + x - 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.59, size = 45, normalized size = 1.55 \begin {gather*} -\frac {x^{3} e^{5} + x^{3} - 2 \, x^{2} e^{5} - 6 \, x^{2} + 8 \, x - e^{\left (x^{2} + x\right )} + 1}{x e^{5} + x - 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.47, size = 37, normalized size = 1.28
method | result | size |
risch | \(-x^{2}+2 x -\frac {1}{x +x \,{\mathrm e}^{5}-4}+\frac {{\mathrm e}^{\left (x +1\right ) x}}{x +x \,{\mathrm e}^{5}-4}\) | \(37\) |
norman | \(\frac {\left (-{\mathrm e}^{5}-1\right ) x^{3}+\left (2 \,{\mathrm e}^{5}+6\right ) x^{2}+\left (-\frac {33}{4}-\frac {{\mathrm e}^{5}}{4}\right ) x +{\mathrm e}^{x^{2}+x}}{x +x \,{\mathrm e}^{5}-4}\) | \(46\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.43, size = 703, normalized size = 24.24 \begin {gather*} -{\left (\frac {x^{2} {\left (e^{5} + 1\right )} + 16 \, x}{e^{15} + 3 \, e^{10} + 3 \, e^{5} + 1} + \frac {96 \, \log \left (x {\left (e^{5} + 1\right )} - 4\right )}{e^{20} + 4 \, e^{15} + 6 \, e^{10} + 4 \, e^{5} + 1} - \frac {128}{x {\left (e^{25} + 5 \, e^{20} + 10 \, e^{15} + 10 \, e^{10} + 5 \, e^{5} + 1\right )} - 4 \, e^{20} - 16 \, e^{15} - 24 \, e^{10} - 16 \, e^{5} - 4}\right )} e^{10} + 2 \, {\left (\frac {x}{e^{10} + 2 \, e^{5} + 1} + \frac {8 \, \log \left (x {\left (e^{5} + 1\right )} - 4\right )}{e^{15} + 3 \, e^{10} + 3 \, e^{5} + 1} - \frac {16}{x {\left (e^{20} + 4 \, e^{15} + 6 \, e^{10} + 4 \, e^{5} + 1\right )} - 4 \, e^{15} - 12 \, e^{10} - 12 \, e^{5} - 4}\right )} e^{10} - 2 \, {\left (\frac {x^{2} {\left (e^{5} + 1\right )} + 16 \, x}{e^{15} + 3 \, e^{10} + 3 \, e^{5} + 1} + \frac {96 \, \log \left (x {\left (e^{5} + 1\right )} - 4\right )}{e^{20} + 4 \, e^{15} + 6 \, e^{10} + 4 \, e^{5} + 1} - \frac {128}{x {\left (e^{25} + 5 \, e^{20} + 10 \, e^{15} + 10 \, e^{10} + 5 \, e^{5} + 1\right )} - 4 \, e^{20} - 16 \, e^{15} - 24 \, e^{10} - 16 \, e^{5} - 4}\right )} e^{5} + 20 \, {\left (\frac {x}{e^{10} + 2 \, e^{5} + 1} + \frac {8 \, \log \left (x {\left (e^{5} + 1\right )} - 4\right )}{e^{15} + 3 \, e^{10} + 3 \, e^{5} + 1} - \frac {16}{x {\left (e^{20} + 4 \, e^{15} + 6 \, e^{10} + 4 \, e^{5} + 1\right )} - 4 \, e^{15} - 12 \, e^{10} - 12 \, e^{5} - 4}\right )} e^{5} - 16 \, {\left (\frac {\log \left (x {\left (e^{5} + 1\right )} - 4\right )}{e^{10} + 2 \, e^{5} + 1} - \frac {4}{x {\left (e^{15} + 3 \, e^{10} + 3 \, e^{5} + 1\right )} - 4 \, e^{10} - 8 \, e^{5} - 4}\right )} e^{5} - \frac {x^{2} {\left (e^{5} + 1\right )} + 16 \, x}{e^{15} + 3 \, e^{10} + 3 \, e^{5} + 1} + \frac {18 \, x}{e^{10} + 2 \, e^{5} + 1} - \frac {e^{5}}{x {\left (e^{10} + 2 \, e^{5} + 1\right )} - 4 \, e^{5} - 4} + \frac {e^{\left (x^{2} + x\right )}}{x {\left (e^{5} + 1\right )} - 4} - \frac {96 \, \log \left (x {\left (e^{5} + 1\right )} - 4\right )}{e^{20} + 4 \, e^{15} + 6 \, e^{10} + 4 \, e^{5} + 1} + \frac {144 \, \log \left (x {\left (e^{5} + 1\right )} - 4\right )}{e^{15} + 3 \, e^{10} + 3 \, e^{5} + 1} - \frac {48 \, \log \left (x {\left (e^{5} + 1\right )} - 4\right )}{e^{10} + 2 \, e^{5} + 1} + \frac {128}{x {\left (e^{25} + 5 \, e^{20} + 10 \, e^{15} + 10 \, e^{10} + 5 \, e^{5} + 1\right )} - 4 \, e^{20} - 16 \, e^{15} - 24 \, e^{10} - 16 \, e^{5} - 4} - \frac {288}{x {\left (e^{20} + 4 \, e^{15} + 6 \, e^{10} + 4 \, e^{5} + 1\right )} - 4 \, e^{15} - 12 \, e^{10} - 12 \, e^{5} - 4} + \frac {192}{x {\left (e^{15} + 3 \, e^{10} + 3 \, e^{5} + 1\right )} - 4 \, e^{10} - 8 \, e^{5} - 4} - \frac {33}{x {\left (e^{10} + 2 \, e^{5} + 1\right )} - 4 \, e^{5} - 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.24, size = 28, normalized size = 0.97 \begin {gather*} 2\,x+\frac {{\mathrm {e}}^{x^2+x}-1}{x\,\left ({\mathrm {e}}^5+1\right )-4}-x^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.44, size = 31, normalized size = 1.07 \begin {gather*} - x^{2} + 2 x + \frac {e^{x^{2} + x}}{x + x e^{5} - 4} - \frac {1}{x \left (1 + e^{5}\right ) - 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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