Optimal. Leaf size=32 \[ \frac {\frac {2}{x}-\frac {2 \left (5-e^5\right )}{x-\left (e^x+x\right )^2}}{e^2} \]
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Rubi [F] time = 1.85, 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 e^{4 x}-8 e^{3 x} x+8 x^2-16 x^3-2 x^4+e^{2 x} \left (4 x-32 x^2+4 e^5 x^2\right )+e^5 \left (-2 x^2+4 x^3\right )+e^x \left (-12 x^2-28 x^3+e^5 \left (4 x^2+4 x^3\right )\right )}{e^{2+4 x} x^2+4 e^{2+3 x} x^3+e^{2+2 x} \left (-2 x^3+6 x^4\right )+e^{2+x} \left (-4 x^4+4 x^5\right )+e^2 \left (x^4-2 x^5+x^6\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 {2 \left (-e^{4 x}-4 e^{3 x} x+2 e^{5+2 x} x^2+2 e^{5+x} x^2 (1+x)-2 e^x x^2 (3+7 x)-2 e^{2 x} x (-1+8 x)-x^2 \left (-4+8 x+x^2\right )-e^5 \left (x^2-2 x^3\right )\right )}{e^2 x^2 \left (e^{2 x}+2 e^x x+(-1+x) x\right )^2} \, dx\\ &=\frac {2 \int \frac {-e^{4 x}-4 e^{3 x} x+2 e^{5+2 x} x^2+2 e^{5+x} x^2 (1+x)-2 e^x x^2 (3+7 x)-2 e^{2 x} x (-1+8 x)-x^2 \left (-4+8 x+x^2\right )-e^5 \left (x^2-2 x^3\right )}{x^2 \left (e^{2 x}+2 e^x x+(-1+x) x\right )^2} \, dx}{e^2}\\ &=\frac {2 \int \left (-\frac {1}{x^2}+\frac {2 \left (-5+e^5\right )}{e^{2 x}-x+2 e^x x+x^2}-\frac {\left (-5+e^5\right ) \left (1-2 e^x-4 x+2 e^x x+2 x^2\right )}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}\right ) \, dx}{e^2}\\ &=\frac {2}{e^2 x}+\frac {\left (2 \left (5-e^5\right )\right ) \int \frac {1-2 e^x-4 x+2 e^x x+2 x^2}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}-\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {1}{e^{2 x}-x+2 e^x x+x^2} \, dx}{e^2}\\ &=\frac {2}{e^2 x}+\frac {\left (2 \left (5-e^5\right )\right ) \int \left (\frac {1}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}-\frac {2 e^x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}-\frac {4 x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}+\frac {2 e^x x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}+\frac {2 x^2}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}\right ) \, dx}{e^2}-\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {1}{e^{2 x}-x+2 e^x x+x^2} \, dx}{e^2}\\ &=\frac {2}{e^2 x}+\frac {\left (2 \left (5-e^5\right )\right ) \int \frac {1}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}-\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {e^x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}+\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {e^x x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}+\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {x^2}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}-\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {1}{e^{2 x}-x+2 e^x x+x^2} \, dx}{e^2}-\frac {\left (8 \left (5-e^5\right )\right ) \int \frac {x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 37, normalized size = 1.16 \begin {gather*} -\frac {2 \left (-\frac {1}{x}+\frac {-5+e^5}{e^{2 x}-x+2 e^x x+x^2}\right )}{e^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.82, size = 65, normalized size = 2.03 \begin {gather*} -\frac {2 \, {\left (x e^{9} - {\left (x^{2} + 4 \, x\right )} e^{4} - 2 \, x e^{\left (x + 4\right )} - e^{\left (2 \, x + 4\right )}\right )}}{2 \, x^{2} e^{\left (x + 6\right )} + {\left (x^{3} - x^{2}\right )} e^{6} + x e^{\left (2 \, x + 6\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.48, size = 65, normalized size = 2.03 \begin {gather*} \frac {2 \, {\left (x^{2} e^{2} - 2 \, x e^{7} + 9 \, x e^{2} + 2 \, x e^{\left (x + 2\right )} + e^{\left (2 \, x + 2\right )}\right )}}{x^{3} e^{4} - x^{2} e^{4} + 2 \, x^{2} e^{\left (x + 4\right )} + x e^{\left (2 \, x + 4\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 55, normalized size = 1.72
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{-2}}{x}-\frac {2 \,{\mathrm e}^{-2} {\mathrm e}^{5}}{{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}-x}+\frac {10 \,{\mathrm e}^{-2}}{{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}-x}\) | \(55\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.46, size = 55, normalized size = 1.72 \begin {gather*} \frac {2 \, {\left (x^{2} - x {\left (e^{5} - 4\right )} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}}{x^{3} e^{2} - x^{2} e^{2} + 2 \, x^{2} e^{\left (x + 2\right )} + x e^{\left (2 \, x + 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.76, size = 58, normalized size = 1.81 \begin {gather*} \frac {2\,{\mathrm {e}}^{2\,x}+x\,\left (4\,{\mathrm {e}}^x-2\,{\mathrm {e}}^5+8\right )+2\,x^2}{x\,{\mathrm {e}}^{2\,x+2}+2\,x^2\,{\mathrm {e}}^{x+2}-x^2\,{\mathrm {e}}^2+x^3\,{\mathrm {e}}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.21, size = 42, normalized size = 1.31 \begin {gather*} \frac {10 - 2 e^{5}}{x^{2} e^{2} + 2 x e^{2} e^{x} - x e^{2} + e^{2} e^{2 x}} + \frac {2}{x e^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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