Optimal. Leaf size=29 \[ \frac {x}{x^2-\log \left (\frac {1}{2 \left (2-e^{-x}-2 x\right )^2}\right )} \]
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Rubi [F] time = 3.13, 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 x-x^2+e^x \left (-4 x+2 x^2-2 x^3\right )+\left (-1+e^x (2-2 x)\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )}{x^4+e^x \left (-2 x^4+2 x^5\right )+\left (-2 x^2+e^x \left (4 x^2-4 x^3\right )\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )+\left (1+e^x (-2+2 x)\right ) \log ^2\left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\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 \left (-2+x+2 e^x \left (2-x+x^2\right )\right )+\left (-1-2 e^x (-1+x)\right ) \log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )}{\left (1+2 e^x (-1+x)\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx\\ &=\int \left (\frac {2 x^2}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {-2 x+x^2-x^3+\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )-x \log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}\right ) \, dx\\ &=2 \int \frac {x^2}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+\int \frac {-2 x+x^2-x^3+\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )-x \log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx\\ &=2 \int \left (\frac {1}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {1}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {x}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}\right ) \, dx+\int \frac {-x \left (-2+x-x^2\right )+(-1+x) \log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )}{(1-x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx\\ &=2 \int \frac {1}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {1}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {x}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+\int \left (-\frac {2 x \left (1-x+x^2\right )}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {1}{x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )}\right ) \, dx\\ &=2 \int \frac {1}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {1}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {x}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx-2 \int \frac {x \left (1-x+x^2\right )}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+\int \frac {1}{x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )} \, dx\\ &=-\left (2 \int \left (\frac {1}{\left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {1}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {x^2}{\left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}\right ) \, dx\right )+2 \int \frac {1}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {1}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {x}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+\int \frac {1}{x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )} \, dx\\ &=-\left (2 \int \frac {1}{\left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx\right )-2 \int \frac {1}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx-2 \int \frac {x^2}{\left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {1}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {1}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {x}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+\int \frac {1}{x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 33, normalized size = 1.14 \begin {gather*} -\frac {x}{-x^2+\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.83, size = 42, normalized size = 1.45 \begin {gather*} \frac {x}{x^{2} - \log \left (\frac {e^{\left (2 \, x\right )}}{2 \, {\left (4 \, {\left (x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + 4 \, {\left (x - 1\right )} e^{x} + 1\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 10.39, size = 45, normalized size = 1.55 \begin {gather*} \frac {x}{x^{2} - 2 \, x + \log \left (8 \, x^{2} e^{\left (2 \, x\right )} - 16 \, x e^{\left (2 \, x\right )} + 8 \, x e^{x} + 8 \, e^{\left (2 \, x\right )} - 8 \, e^{x} + 2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.70, size = 332, normalized size = 11.45
method | result | size |
risch | \(\frac {2 x}{i \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )^{2}-i \pi \,\mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )^{2}+2 i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )^{3}-i \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-i \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}\right )^{3}+i \pi \,\mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )+i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+2 x^{2}+6 \ln \relax (2)+4 \ln \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )-4 \ln \left ({\mathrm e}^{x}\right )}\) | \(332\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.98, size = 25, normalized size = 0.86 \begin {gather*} \frac {x}{x^{2} - 2 \, x + \log \relax (2) + 2 \, \log \left (2 \, {\left (x - 1\right )} e^{x} + 1\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 = -0.03 \begin {gather*} \int -\frac {\ln \left (\frac {{\mathrm {e}}^{2\,x}}{{\mathrm {e}}^{2\,x}\,\left (8\,x^2-16\,x+8\right )+{\mathrm {e}}^x\,\left (8\,x-8\right )+2}\right )\,\left ({\mathrm {e}}^x\,\left (2\,x-2\right )+1\right )-2\,x+x^2+{\mathrm {e}}^x\,\left (2\,x^3-2\,x^2+4\,x\right )}{\ln \left (\frac {{\mathrm {e}}^{2\,x}}{{\mathrm {e}}^{2\,x}\,\left (8\,x^2-16\,x+8\right )+{\mathrm {e}}^x\,\left (8\,x-8\right )+2}\right )\,\left ({\mathrm {e}}^x\,\left (4\,x^2-4\,x^3\right )-2\,x^2\right )-{\mathrm {e}}^x\,\left (2\,x^4-2\,x^5\right )+x^4+{\ln \left (\frac {{\mathrm {e}}^{2\,x}}{{\mathrm {e}}^{2\,x}\,\left (8\,x^2-16\,x+8\right )+{\mathrm {e}}^x\,\left (8\,x-8\right )+2}\right )}^2\,\left ({\mathrm {e}}^x\,\left (2\,x-2\right )+1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.34, size = 37, normalized size = 1.28 \begin {gather*} - \frac {x}{- x^{2} + \log {\left (\frac {e^{2 x}}{\left (8 x - 8\right ) e^{x} + \left (8 x^{2} - 16 x + 8\right ) e^{2 x} + 2} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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