Optimal. Leaf size=25 \[ 4+e^{2 x}+\frac {e^5}{(1-2 x)^2}+2 x-\log (x) \]
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Rubi [B] time = 0.71, antiderivative size = 61, normalized size of antiderivative = 2.44, number of steps used = 12, number of rules used = 7, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6, 6741, 6742, 2194, 44, 37, 43} \begin {gather*} -\frac {12 x^2}{(1-2 x)^2}+2 x+e^{2 x}-\frac {5}{1-2 x}+\frac {1}{2 x-1}+\frac {2+e^5}{(1-2 x)^2}+\frac {1}{(1-2 x)^2}-\log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 37
Rule 43
Rule 44
Rule 2194
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1+\left (-8-4 e^5\right ) x+24 x^2-32 x^3+16 x^4+e^{2 x} \left (-2 x+12 x^2-24 x^3+16 x^4\right )}{-x+6 x^2-12 x^3+8 x^4} \, dx\\ &=\int \frac {-1-\left (-8-4 e^5\right ) x-24 x^2+32 x^3-16 x^4-e^{2 x} \left (-2 x+12 x^2-24 x^3+16 x^4\right )}{x \left (1-6 x+12 x^2-8 x^3\right )} \, dx\\ &=\int \left (2 e^{2 x}-\frac {4 \left (2+e^5\right )}{(-1+2 x)^3}+\frac {1}{x (-1+2 x)^3}+\frac {24 x}{(-1+2 x)^3}-\frac {32 x^2}{(-1+2 x)^3}+\frac {16 x^3}{(-1+2 x)^3}\right ) \, dx\\ &=\frac {2+e^5}{(1-2 x)^2}+2 \int e^{2 x} \, dx+16 \int \frac {x^3}{(-1+2 x)^3} \, dx+24 \int \frac {x}{(-1+2 x)^3} \, dx-32 \int \frac {x^2}{(-1+2 x)^3} \, dx+\int \frac {1}{x (-1+2 x)^3} \, dx\\ &=e^{2 x}+\frac {2+e^5}{(1-2 x)^2}-\frac {12 x^2}{(1-2 x)^2}+16 \int \left (\frac {1}{8}+\frac {1}{8 (-1+2 x)^3}+\frac {3}{8 (-1+2 x)^2}+\frac {3}{8 (-1+2 x)}\right ) \, dx-32 \int \left (\frac {1}{4 (-1+2 x)^3}+\frac {1}{2 (-1+2 x)^2}+\frac {1}{4 (-1+2 x)}\right ) \, dx+\int \left (-\frac {1}{x}+\frac {2}{(-1+2 x)^3}-\frac {2}{(-1+2 x)^2}+\frac {2}{-1+2 x}\right ) \, dx\\ &=e^{2 x}+\frac {1}{(1-2 x)^2}+\frac {2+e^5}{(1-2 x)^2}-\frac {5}{1-2 x}+2 x-\frac {12 x^2}{(1-2 x)^2}+\frac {1}{-1+2 x}-\log (x)\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.19, size = 25, normalized size = 1.00 \begin {gather*} -3+e^{2 x}+\frac {e^5}{(1-2 x)^2}+2 x-\log (x) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.87, size = 58, normalized size = 2.32 \begin {gather*} \frac {8 \, x^{3} - 8 \, x^{2} + {\left (4 \, x^{2} - 4 \, x + 1\right )} e^{\left (2 \, x\right )} - {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \relax (x) + 2 \, x + e^{5}}{4 \, x^{2} - 4 \, x + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.30, size = 65, normalized size = 2.60 \begin {gather*} \frac {8 \, x^{3} + 4 \, x^{2} e^{\left (2 \, x\right )} - 4 \, x^{2} \log \relax (x) - 8 \, x^{2} - 4 \, x e^{\left (2 \, x\right )} + 4 \, x \log \relax (x) + 2 \, x + e^{5} + e^{\left (2 \, x\right )} - \log \relax (x)}{4 \, x^{2} - 4 \, x + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 25, normalized size = 1.00
method | result | size |
derivativedivides | \(-\ln \left (2 x \right )+2 x +{\mathrm e}^{2 x}+\frac {{\mathrm e}^{5}}{\left (2 x -1\right )^{2}}\) | \(25\) |
default | \(-\ln \left (2 x \right )+2 x +{\mathrm e}^{2 x}+\frac {{\mathrm e}^{5}}{\left (2 x -1\right )^{2}}\) | \(25\) |
risch | \(2 x +\frac {{\mathrm e}^{5}}{4 x^{2}-4 x +1}-\ln \relax (x )+{\mathrm e}^{2 x}\) | \(27\) |
norman | \(\frac {-6 x +8 x^{3}-4 x \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{2 x} x^{2}+2+{\mathrm e}^{5}+{\mathrm e}^{2 x}}{\left (2 x -1\right )^{2}}-\ln \relax (x )\) | \(46\) |
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*} 2 \, x + \frac {2 \, {\left (4 \, x^{3} - 6 \, x^{2} + 3 \, x\right )} e^{\left (2 \, x\right )}}{8 \, x^{3} - 12 \, x^{2} + 6 \, x - 1} - \frac {12 \, x - 5}{2 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {2 \, {\left (8 \, x - 3\right )}}{4 \, x^{2} - 4 \, x + 1} - \frac {3 \, {\left (4 \, x - 1\right )}}{4 \, x^{2} - 4 \, x + 1} + \frac {4 \, x - 3}{2 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {e^{5}}{4 \, x^{2} - 4 \, x + 1} + \frac {e E_{3}\left (-2 \, x + 1\right )}{{\left (2 \, x - 1\right )}^{2}} + \frac {2}{4 \, x^{2} - 4 \, x + 1} + 6 \, \int \frac {e^{\left (2 \, x\right )}}{16 \, x^{4} - 32 \, x^{3} + 24 \, x^{2} - 8 \, x + 1}\,{d x} - \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.17, size = 22, normalized size = 0.88 \begin {gather*} 2\,x+{\mathrm {e}}^{2\,x}-\ln \relax (x)+\frac {{\mathrm {e}}^5}{{\left (2\,x-1\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.27, size = 24, normalized size = 0.96 \begin {gather*} 2 x + e^{2 x} - \log {\relax (x )} + \frac {e^{5}}{4 x^{2} - 4 x + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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