Optimal. Leaf size=36 \[ \frac {2-\frac {(4-x) \left (-e^x+e^{-x^2}+2 x\right )}{e^5+x}}{x} \]
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Rubi [F] time = 4.32, 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^{-x^2} \left (8 x-x^2+8 x^3-2 x^4+e^5 \left (4+8 x^2-2 x^3\right )+e^{x^2} \left (-2 e^{10}+6 x^2+e^5 \left (-4 x+2 x^2\right )+e^x \left (-8 x+5 x^2-x^3+e^5 \left (-4+4 x-x^2\right )\right )\right )\right )}{e^{10} x^2+2 e^5 x^3+x^4} \, 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^{-x^2} \left (8 x-x^2+8 x^3-2 x^4+e^5 \left (4+8 x^2-2 x^3\right )+e^{x^2} \left (-2 e^{10}+6 x^2+e^5 \left (-4 x+2 x^2\right )+e^x \left (-8 x+5 x^2-x^3+e^5 \left (-4+4 x-x^2\right )\right )\right )\right )}{x^2 \left (e^{10}+2 e^5 x+x^2\right )} \, dx\\ &=\int \frac {e^{-x^2} \left (8 x-x^2+8 x^3-2 x^4+e^5 \left (4+8 x^2-2 x^3\right )+e^{x^2} \left (-2 e^{10}+6 x^2+e^5 \left (-4 x+2 x^2\right )+e^x \left (-8 x+5 x^2-x^3+e^5 \left (-4+4 x-x^2\right )\right )\right )\right )}{x^2 \left (e^5+x\right )^2} \, dx\\ &=\int \left (-\frac {e^{-x^2}}{\left (e^5+x\right )^2}+\frac {8 e^{-x^2}}{x \left (e^5+x\right )^2}+\frac {8 e^{-x^2} x}{\left (e^5+x\right )^2}-\frac {2 e^{-x^2} x^2}{\left (e^5+x\right )^2}-\frac {2 e^{5-x^2} \left (-2-4 x^2+x^3\right )}{x^2 \left (e^5+x\right )^2}+\frac {-2 e^{10}-4 e^{5+x}-4 e^5 x-8 e^x \left (1-\frac {e^5}{2}\right ) x+5 e^x \left (1-\frac {e^5}{5}\right ) x^2+6 \left (1+\frac {e^5}{3}\right ) x^2-e^x x^3}{x^2 \left (e^5+x\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{-x^2} x^2}{\left (e^5+x\right )^2} \, dx\right )-2 \int \frac {e^{5-x^2} \left (-2-4 x^2+x^3\right )}{x^2 \left (e^5+x\right )^2} \, dx+8 \int \frac {e^{-x^2}}{x \left (e^5+x\right )^2} \, dx+8 \int \frac {e^{-x^2} x}{\left (e^5+x\right )^2} \, dx-\int \frac {e^{-x^2}}{\left (e^5+x\right )^2} \, dx+\int \frac {-2 e^{10}-4 e^{5+x}-4 e^5 x-8 e^x \left (1-\frac {e^5}{2}\right ) x+5 e^x \left (1-\frac {e^5}{5}\right ) x^2+6 \left (1+\frac {e^5}{3}\right ) x^2-e^x x^3}{x^2 \left (e^5+x\right )^2} \, dx\\ &=\frac {e^{-x^2}}{e^5+x}+2 \int e^{-x^2} \, dx-2 \int \left (e^{-x^2}+\frac {e^{10-x^2}}{\left (e^5+x\right )^2}-\frac {2 e^{5-x^2}}{e^5+x}\right ) \, dx-2 \int \left (-\frac {2 e^{-5-x^2}}{x^2}+\frac {4 e^{-10-x^2}}{x}+\frac {e^{-5-x^2} \left (-2-4 e^{10}-e^{15}\right )}{\left (e^5+x\right )^2}+\frac {e^{-10-x^2} \left (-4+e^{15}\right )}{e^5+x}\right ) \, dx+8 \int \left (-\frac {e^{5-x^2}}{\left (e^5+x\right )^2}+\frac {e^{-x^2}}{e^5+x}\right ) \, dx+8 \int \left (\frac {e^{-10-x^2}}{x}-\frac {e^{-5-x^2}}{\left (e^5+x\right )^2}-\frac {e^{-10-x^2}}{e^5+x}\right ) \, dx-\left (2 e^5\right ) \int \frac {e^{-x^2}}{e^5+x} \, dx+\int \frac {-2 e^{10}-e^{5+x} (-2+x)^2+2 e^5 (-2+x) x+6 x^2-e^x x \left (8-5 x+x^2\right )}{x^2 \left (e^5+x\right )^2} \, dx\\ &=\frac {e^{-x^2}}{e^5+x}+\sqrt {\pi } \text {erf}(x)-2 \int e^{-x^2} \, dx-2 \int \frac {e^{10-x^2}}{\left (e^5+x\right )^2} \, dx+4 \int \frac {e^{-5-x^2}}{x^2} \, dx+4 \int \frac {e^{5-x^2}}{e^5+x} \, dx-8 \int \frac {e^{-5-x^2}}{\left (e^5+x\right )^2} \, dx-8 \int \frac {e^{5-x^2}}{\left (e^5+x\right )^2} \, dx+8 \int \frac {e^{-x^2}}{e^5+x} \, dx-8 \int \frac {e^{-10-x^2}}{e^5+x} \, dx-\left (2 e^5\right ) \int \frac {e^{-x^2}}{e^5+x} \, dx+\left (2 \left (4-e^{15}\right )\right ) \int \frac {e^{-10-x^2}}{e^5+x} \, dx+\left (2 \left (2+4 e^{10}+e^{15}\right )\right ) \int \frac {e^{-5-x^2}}{\left (e^5+x\right )^2} \, dx+\int \left (\frac {2 \left (-e^{10}-2 e^5 x+\left (3+e^5\right ) x^2\right )}{x^2 \left (e^5+x\right )^2}+\frac {e^x \left (-4 e^5-4 \left (2-e^5\right ) x+\left (5-e^5\right ) x^2-x^3\right )}{x^2 \left (e^5+x\right )^2}\right ) \, dx\\ &=-\frac {4 e^{-5-x^2}}{x}+\frac {e^{-x^2}}{e^5+x}+\frac {8 e^{-5-x^2}}{e^5+x}+\frac {8 e^{5-x^2}}{e^5+x}+\frac {2 e^{10-x^2}}{e^5+x}-\frac {2 e^{-5-x^2} \left (2+4 e^{10}+e^{15}\right )}{e^5+x}+2 \int \frac {-e^{10}-2 e^5 x+\left (3+e^5\right ) x^2}{x^2 \left (e^5+x\right )^2} \, dx+4 \int e^{10-x^2} \, dx+4 \int \frac {e^{5-x^2}}{e^5+x} \, dx-8 \int e^{-5-x^2} \, dx+8 \int \frac {e^{-x^2}}{e^5+x} \, dx-8 \int \frac {e^{-10-x^2}}{e^5+x} \, dx+16 \int e^{-5-x^2} \, dx+16 \int e^{5-x^2} \, dx-\left (2 e^5\right ) \int \frac {e^{-x^2}}{e^5+x} \, dx-\left (4 e^5\right ) \int \frac {e^{10-x^2}}{e^5+x} \, dx-\left (16 e^5\right ) \int \frac {e^{-5-x^2}}{e^5+x} \, dx-\left (16 e^5\right ) \int \frac {e^{5-x^2}}{e^5+x} \, dx+\left (2 \left (4-e^{15}\right )\right ) \int \frac {e^{-10-x^2}}{e^5+x} \, dx-\left (4 \left (2+4 e^{10}+e^{15}\right )\right ) \int e^{-5-x^2} \, dx+\left (4 e^5 \left (2+4 e^{10}+e^{15}\right )\right ) \int \frac {e^{-5-x^2}}{e^5+x} \, dx+\int \frac {e^x \left (-4 e^5-4 \left (2-e^5\right ) x+\left (5-e^5\right ) x^2-x^3\right )}{x^2 \left (e^5+x\right )^2} \, dx\\ &=-\frac {4 e^{-5-x^2}}{x}+\frac {e^{-x^2}}{e^5+x}+\frac {8 e^{-5-x^2}}{e^5+x}+\frac {8 e^{5-x^2}}{e^5+x}+\frac {2 e^{10-x^2}}{e^5+x}-\frac {2 e^{-5-x^2} \left (2+4 e^{10}+e^{15}\right )}{e^5+x}+\frac {4 \sqrt {\pi } \text {erf}(x)}{e^5}+8 e^5 \sqrt {\pi } \text {erf}(x)+2 e^{10} \sqrt {\pi } \text {erf}(x)-\frac {2 \left (2+4 e^{10}+e^{15}\right ) \sqrt {\pi } \text {erf}(x)}{e^5}+2 \int \left (-\frac {1}{x^2}+\frac {4+e^5}{\left (e^5+x\right )^2}\right ) \, dx+4 \int \frac {e^{5-x^2}}{e^5+x} \, dx+8 \int \frac {e^{-x^2}}{e^5+x} \, dx-8 \int \frac {e^{-10-x^2}}{e^5+x} \, dx-\left (2 e^5\right ) \int \frac {e^{-x^2}}{e^5+x} \, dx-\left (4 e^5\right ) \int \frac {e^{10-x^2}}{e^5+x} \, dx-\left (16 e^5\right ) \int \frac {e^{-5-x^2}}{e^5+x} \, dx-\left (16 e^5\right ) \int \frac {e^{5-x^2}}{e^5+x} \, dx+\left (2 \left (4-e^{15}\right )\right ) \int \frac {e^{-10-x^2}}{e^5+x} \, dx+\left (4 e^5 \left (2+4 e^{10}+e^{15}\right )\right ) \int \frac {e^{-5-x^2}}{e^5+x} \, dx+\int \left (-\frac {4 e^{-5+x}}{x^2}+\frac {4 e^{-5+x}}{x}+\frac {e^{-5+x} \left (4+e^5\right )}{\left (e^5+x\right )^2}+\frac {e^{-5+x} \left (-4-e^5\right )}{e^5+x}\right ) \, dx\\ &=\frac {2}{x}-\frac {4 e^{-5-x^2}}{x}+\frac {e^{-x^2}}{e^5+x}+\frac {8 e^{-5-x^2}}{e^5+x}+\frac {8 e^{5-x^2}}{e^5+x}+\frac {2 e^{10-x^2}}{e^5+x}-\frac {2 \left (4+e^5\right )}{e^5+x}-\frac {2 e^{-5-x^2} \left (2+4 e^{10}+e^{15}\right )}{e^5+x}+\frac {4 \sqrt {\pi } \text {erf}(x)}{e^5}+8 e^5 \sqrt {\pi } \text {erf}(x)+2 e^{10} \sqrt {\pi } \text {erf}(x)-\frac {2 \left (2+4 e^{10}+e^{15}\right ) \sqrt {\pi } \text {erf}(x)}{e^5}-4 \int \frac {e^{-5+x}}{x^2} \, dx+4 \int \frac {e^{-5+x}}{x} \, dx+4 \int \frac {e^{5-x^2}}{e^5+x} \, dx+8 \int \frac {e^{-x^2}}{e^5+x} \, dx-8 \int \frac {e^{-10-x^2}}{e^5+x} \, dx-\left (2 e^5\right ) \int \frac {e^{-x^2}}{e^5+x} \, dx-\left (4 e^5\right ) \int \frac {e^{10-x^2}}{e^5+x} \, dx-\left (16 e^5\right ) \int \frac {e^{-5-x^2}}{e^5+x} \, dx-\left (16 e^5\right ) \int \frac {e^{5-x^2}}{e^5+x} \, dx+\left (-4-e^5\right ) \int \frac {e^{-5+x}}{e^5+x} \, dx+\left (4+e^5\right ) \int \frac {e^{-5+x}}{\left (e^5+x\right )^2} \, dx+\left (2 \left (4-e^{15}\right )\right ) \int \frac {e^{-10-x^2}}{e^5+x} \, dx+\left (4 e^5 \left (2+4 e^{10}+e^{15}\right )\right ) \int \frac {e^{-5-x^2}}{e^5+x} \, dx\\ &=\frac {2}{x}+\frac {4 e^{-5+x}}{x}-\frac {4 e^{-5-x^2}}{x}+\frac {e^{-x^2}}{e^5+x}+\frac {8 e^{-5-x^2}}{e^5+x}+\frac {8 e^{5-x^2}}{e^5+x}+\frac {2 e^{10-x^2}}{e^5+x}-\frac {2 \left (4+e^5\right )}{e^5+x}-\frac {e^{-5+x} \left (4+e^5\right )}{e^5+x}-\frac {2 e^{-5-x^2} \left (2+4 e^{10}+e^{15}\right )}{e^5+x}+\frac {4 \sqrt {\pi } \text {erf}(x)}{e^5}+8 e^5 \sqrt {\pi } \text {erf}(x)+2 e^{10} \sqrt {\pi } \text {erf}(x)-\frac {2 \left (2+4 e^{10}+e^{15}\right ) \sqrt {\pi } \text {erf}(x)}{e^5}+\frac {4 \text {Ei}(x)}{e^5}-e^{-5-e^5} \left (4+e^5\right ) \text {Ei}\left (e^5+x\right )-4 \int \frac {e^{-5+x}}{x} \, dx+4 \int \frac {e^{5-x^2}}{e^5+x} \, dx+8 \int \frac {e^{-x^2}}{e^5+x} \, dx-8 \int \frac {e^{-10-x^2}}{e^5+x} \, dx-\left (2 e^5\right ) \int \frac {e^{-x^2}}{e^5+x} \, dx-\left (4 e^5\right ) \int \frac {e^{10-x^2}}{e^5+x} \, dx-\left (16 e^5\right ) \int \frac {e^{-5-x^2}}{e^5+x} \, dx-\left (16 e^5\right ) \int \frac {e^{5-x^2}}{e^5+x} \, dx+\left (4+e^5\right ) \int \frac {e^{-5+x}}{e^5+x} \, dx+\left (2 \left (4-e^{15}\right )\right ) \int \frac {e^{-10-x^2}}{e^5+x} \, dx+\left (4 e^5 \left (2+4 e^{10}+e^{15}\right )\right ) \int \frac {e^{-5-x^2}}{e^5+x} \, dx\\ &=\frac {2}{x}+\frac {4 e^{-5+x}}{x}-\frac {4 e^{-5-x^2}}{x}+\frac {e^{-x^2}}{e^5+x}+\frac {8 e^{-5-x^2}}{e^5+x}+\frac {8 e^{5-x^2}}{e^5+x}+\frac {2 e^{10-x^2}}{e^5+x}-\frac {2 \left (4+e^5\right )}{e^5+x}-\frac {e^{-5+x} \left (4+e^5\right )}{e^5+x}-\frac {2 e^{-5-x^2} \left (2+4 e^{10}+e^{15}\right )}{e^5+x}+\frac {4 \sqrt {\pi } \text {erf}(x)}{e^5}+8 e^5 \sqrt {\pi } \text {erf}(x)+2 e^{10} \sqrt {\pi } \text {erf}(x)-\frac {2 \left (2+4 e^{10}+e^{15}\right ) \sqrt {\pi } \text {erf}(x)}{e^5}+4 \int \frac {e^{5-x^2}}{e^5+x} \, dx+8 \int \frac {e^{-x^2}}{e^5+x} \, dx-8 \int \frac {e^{-10-x^2}}{e^5+x} \, dx-\left (2 e^5\right ) \int \frac {e^{-x^2}}{e^5+x} \, dx-\left (4 e^5\right ) \int \frac {e^{10-x^2}}{e^5+x} \, dx-\left (16 e^5\right ) \int \frac {e^{-5-x^2}}{e^5+x} \, dx-\left (16 e^5\right ) \int \frac {e^{5-x^2}}{e^5+x} \, dx+\left (2 \left (4-e^{15}\right )\right ) \int \frac {e^{-10-x^2}}{e^5+x} \, dx+\left (4 e^5 \left (2+4 e^{10}+e^{15}\right )\right ) \int \frac {e^{-5-x^2}}{e^5+x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 53, normalized size = 1.47 \begin {gather*} \frac {e^{-x^2} \left (-4-e^{x+x^2} (-4+x)-2 e^{5+x^2} (-1+x)+x-6 e^{x^2} x\right )}{x \left (e^5+x\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.65, size = 45, normalized size = 1.25 \begin {gather*} -\frac {{\left ({\left (2 \, {\left (x - 1\right )} e^{5} + {\left (x - 4\right )} e^{x} + 6 \, x\right )} e^{\left (x^{2}\right )} - x + 4\right )} e^{\left (-x^{2}\right )}}{x^{2} + x e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.21, size = 66, normalized size = 1.83 \begin {gather*} -\frac {x e^{\left (x^{2} + x\right )} + 2 \, x e^{\left (x^{2} + 5\right )} + 6 \, x e^{\left (x^{2}\right )} - x - 4 \, e^{\left (x^{2} + x\right )} - 2 \, e^{\left (x^{2} + 5\right )} + 4}{x^{2} e^{\left (x^{2}\right )} + x e^{\left (x^{2} + 5\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.39, size = 57, normalized size = 1.58
method | result | size |
norman | \(\frac {\left (-4+x +\left (-2 \,{\mathrm e}^{5}-6\right ) x \,{\mathrm e}^{x^{2}}+2 \,{\mathrm e}^{5} {\mathrm e}^{x^{2}}+4 \,{\mathrm e}^{x} {\mathrm e}^{x^{2}}-x \,{\mathrm e}^{x} {\mathrm e}^{x^{2}}\right ) {\mathrm e}^{-x^{2}}}{x \left ({\mathrm e}^{5}+x \right )}\) | \(57\) |
risch | \(\frac {\left (-2 \,{\mathrm e}^{5}-6\right ) x +2 \,{\mathrm e}^{5}}{\left ({\mathrm e}^{5}+x \right ) x}-\frac {\left (x -4\right ) {\mathrm e}^{x}}{x \left ({\mathrm e}^{5}+x \right )}+\frac {\left (x -4\right ) {\mathrm e}^{-x^{2}}}{x \left ({\mathrm e}^{5}+x \right )}\) | \(60\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 41, normalized size = 1.14 \begin {gather*} -\frac {2 \, x {\left (e^{5} + 3\right )} - {\left (x - 4\right )} e^{\left (-x^{2}\right )} + {\left (x - 4\right )} e^{x} - 2 \, e^{5}}{x^{2} + x e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.75, size = 49, normalized size = 1.36 \begin {gather*} -\frac {6\,x-2\,{\mathrm {e}}^5+4\,{\mathrm {e}}^{-x^2}-4\,{\mathrm {e}}^x+2\,x\,{\mathrm {e}}^5-x\,{\mathrm {e}}^{-x^2}+x\,{\mathrm {e}}^x}{x\,\left (x+{\mathrm {e}}^5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.58, size = 54, normalized size = 1.50 \begin {gather*} \frac {\left (4 - x\right ) e^{x}}{x^{2} + x e^{5}} + \frac {\left (x - 4\right ) e^{- x^{2}}}{x^{2} + x e^{5}} + \frac {x \left (- 2 e^{5} - 6\right ) + 2 e^{5}}{x^{2} + x e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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