Optimal. Leaf size=29 \[ \frac {x}{4+e^x-x+4 \left (16-x^2\right )+\frac {1}{4 \log (5)}} \]
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Rubi [F] time = 1.41, 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 {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\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 {4 \log (5) \left (1+272 \log (5)-4 e^x (-1+x) \log (5)+16 x^2 \log (5)\right )}{\left (1+272 \log (5)+4 e^x \log (5)-4 x \log (5)-16 x^2 \log (5)\right )^2} \, dx\\ &=(4 \log (5)) \int \frac {1+272 \log (5)-4 e^x (-1+x) \log (5)+16 x^2 \log (5)}{\left (1+272 \log (5)+4 e^x \log (5)-4 x \log (5)-16 x^2 \log (5)\right )^2} \, dx\\ &=(4 \log (5)) \int \left (-\frac {x \left (-1-276 \log (5)-28 x \log (5)+16 x^2 \log (5)\right )}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2}+\frac {-1+x}{-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)}\right ) \, dx\\ &=-\left ((4 \log (5)) \int \frac {x \left (-1-276 \log (5)-28 x \log (5)+16 x^2 \log (5)\right )}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2} \, dx\right )+(4 \log (5)) \int \frac {-1+x}{-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)} \, dx\\ &=-\left ((4 \log (5)) \int \left (-\frac {28 x^2 \log (5)}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2}+\frac {16 x^3 \log (5)}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2}-\frac {x (1+276 \log (5))}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2}\right ) \, dx\right )+(4 \log (5)) \int \left (\frac {1}{1+272 \log (5)+4 e^x \log (5)-4 x \log (5)-16 x^2 \log (5)}+\frac {x}{-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)}\right ) \, dx\\ &=(4 \log (5)) \int \frac {1}{1+272 \log (5)+4 e^x \log (5)-4 x \log (5)-16 x^2 \log (5)} \, dx+(4 \log (5)) \int \frac {x}{-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)} \, dx-\left (64 \log ^2(5)\right ) \int \frac {x^3}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2} \, dx+\left (112 \log ^2(5)\right ) \int \frac {x^2}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2} \, dx+(4 \log (5) (1+276 \log (5))) \int \frac {x}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.38, size = 32, normalized size = 1.10 \begin {gather*} \frac {4 x \log (5)}{1+272 \log (5)+4 e^x \log (5)-4 x \log (5)-16 x^2 \log (5)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.93, size = 27, normalized size = 0.93 \begin {gather*} -\frac {4 \, x \log \relax (5)}{4 \, {\left (4 \, x^{2} + x - 68\right )} \log \relax (5) - 4 \, e^{x} \log \relax (5) - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 31, normalized size = 1.07 \begin {gather*} -\frac {4 \, x \log \relax (5)}{16 \, x^{2} \log \relax (5) + 4 \, x \log \relax (5) - 4 \, e^{x} \log \relax (5) - 272 \, \log \relax (5) - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.31, size = 32, normalized size = 1.10
method | result | size |
risch | \(-\frac {4 \ln \relax (5) x}{16 x^{2} \ln \relax (5)+4 x \ln \relax (5)-4 \,{\mathrm e}^{x} \ln \relax (5)-272 \ln \relax (5)-1}\) | \(32\) |
norman | \(\frac {-4 \,{\mathrm e}^{x} \ln \relax (5)+16 x^{2} \ln \relax (5)-1-272 \ln \relax (5)}{16 x^{2} \ln \relax (5)+4 x \ln \relax (5)-4 \,{\mathrm e}^{x} \ln \relax (5)-272 \ln \relax (5)-1}\) | \(47\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 31, normalized size = 1.07 \begin {gather*} -\frac {4 \, x \log \relax (5)}{16 \, x^{2} \log \relax (5) + 4 \, x \log \relax (5) - 4 \, e^{x} \log \relax (5) - 272 \, \log \relax (5) - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.52, size = 31, normalized size = 1.07 \begin {gather*} \frac {4\,x\,\ln \relax (5)}{272\,\ln \relax (5)-4\,x\,\ln \relax (5)-16\,x^2\,\ln \relax (5)+4\,{\mathrm {e}}^x\,\ln \relax (5)+1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.19, size = 36, normalized size = 1.24 \begin {gather*} \frac {4 x \log {\relax (5 )}}{- 16 x^{2} \log {\relax (5 )} - 4 x \log {\relax (5 )} + 4 e^{x} \log {\relax (5 )} + 1 + 272 \log {\relax (5 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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