Optimal. Leaf size=25 \[ \log \left (-5-2 x+\frac {1}{2} e^{3+x-3 (1+x)} x-\log (x)\right ) \]
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Rubi [F]
time = 1.68, 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+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1+2 x}{x (5+2 x+\log (x))}+\frac {-4+10 x+4 x^2-\log (x)+2 x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}\right ) \, dx\\ &=\int \frac {1+2 x}{x (5+2 x+\log (x))} \, dx+\int \frac {-4+10 x+4 x^2-\log (x)+2 x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx\\ &=\log (5+2 x+\log (x))+\int \left (-\frac {4}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}+\frac {10 x}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}+\frac {4 x^2}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}-\frac {\log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}+\frac {2 x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}\right ) \, dx\\ &=\log (5+2 x+\log (x))+2 \int \frac {x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx-4 \int \frac {1}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx+4 \int \frac {x^2}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx+10 \int \frac {x}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx-\int \frac {\log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.38, size = 18, normalized size = 0.72 \begin {gather*} \log \left (10+\left (4-e^{-2 x}\right ) x+2 \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.02, size = 28, normalized size = 1.12
method | result | size |
risch | \(\ln \left (\ln \left (x \right )+\frac {\left (4 x \,{\mathrm e}^{2 x}+10 \,{\mathrm e}^{2 x}-x \right ) {\mathrm e}^{-2 x}}{2}\right )\) | \(28\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 41 vs.
\(2 (17) = 34\).
time = 0.40, size = 41, normalized size = 1.64 \begin {gather*} -2 \, x + \log \left (2 \, x + \log \left (x\right ) + 5\right ) + \log \left (\frac {2 \, {\left (2 \, x + \log \left (x\right ) + 5\right )} e^{\left (2 \, x\right )} - x}{2 \, {\left (2 \, x + \log \left (x\right ) + 5\right )}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 17, normalized size = 0.68 \begin {gather*} \log \left (-x e^{\left (-2 \, x\right )} + 4 \, x + 2 \, \log \left (x\right ) + 10\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.13, size = 24, normalized size = 0.96 \begin {gather*} \log {\left (x \right )} + \log {\left (e^{- 2 x} + \frac {- 4 x - 2 \log {\left (x \right )} - 10}{x} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 16, normalized size = 0.64 \begin {gather*} \log \left (x e^{\left (-2 \, x\right )} - 4 \, x - 2 \, \log \left (x\right ) - 10\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.18, size = 15, normalized size = 0.60 \begin {gather*} \ln \left (2\,x+\ln \left (x\right )-\frac {x\,{\mathrm {e}}^{-2\,x}}{2}+5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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