Optimal. Leaf size=29 \[ \frac {10 e^{-e^3+i \pi +x} \log (1+x)}{-5+x+\log (-5+x)} \]
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Rubi [F] time = 11.71, 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^{-e^3+i \pi +x} \left (125-50 x+5 x^2\right )+2 e^{-e^3+i \pi +x} (-25+5 x) \log (-5+x)+\left (2 e^{-e^3+i \pi +x} \left (145+90 x-50 x^2+5 x^3\right )+2 e^{-e^3+i \pi +x} \left (-25-20 x+5 x^2\right ) \log (-5+x)\right ) \log (1+x)}{-125-50 x+60 x^2-14 x^3+x^4+\left (50+30 x-18 x^2+2 x^3\right ) \log (-5+x)+\left (-5-4 x+x^2\right ) \log ^2(-5+x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 e^{-e^3+x} \left ((-5+x)^2+\left (29+18 x-10 x^2+x^3\right ) \log (1+x)+(-5+x) \log (-5+x) (1+(1+x) \log (1+x))\right )}{\left (5+4 x-x^2\right ) (5-x-\log (-5+x))^2} \, dx\\ &=10 \int \frac {e^{-e^3+x} \left ((-5+x)^2+\left (29+18 x-10 x^2+x^3\right ) \log (1+x)+(-5+x) \log (-5+x) (1+(1+x) \log (1+x))\right )}{\left (5+4 x-x^2\right ) (5-x-\log (-5+x))^2} \, dx\\ &=10 \int \left (-\frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))}-\frac {e^{-e^3+x} \left (29-11 x+x^2-5 \log (-5+x)+x \log (-5+x)\right ) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}\right ) \, dx\\ &=-\left (10 \int \frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))} \, dx\right )-10 \int \frac {e^{-e^3+x} \left (29-11 x+x^2-5 \log (-5+x)+x \log (-5+x)\right ) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx\\ &=-\left (10 \int \frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))} \, dx\right )-10 \int \left (\frac {29 e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}-\frac {11 e^{-e^3+x} x \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}+\frac {e^{-e^3+x} x^2 \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}-\frac {5 e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}+\frac {e^{-e^3+x} x \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}\right ) \, dx\\ &=-\left (10 \int \frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))} \, dx\right )-10 \int \frac {e^{-e^3+x} x^2 \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx-10 \int \frac {e^{-e^3+x} x \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx+50 \int \frac {e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx+110 \int \frac {e^{-e^3+x} x \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx-290 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx\\ &=-\left (10 \int \frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))} \, dx\right )-10 \int \left (\frac {5 e^{-e^3+x} \log (1+x)}{(-5+x+\log (-5+x))^2}+\frac {25 e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}+\frac {e^{-e^3+x} x \log (1+x)}{(-5+x+\log (-5+x))^2}\right ) \, dx-10 \int \left (\frac {e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x+\log (-5+x))^2}+\frac {5 e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}\right ) \, dx+50 \int \frac {e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx+110 \int \left (\frac {e^{-e^3+x} \log (1+x)}{(-5+x+\log (-5+x))^2}+\frac {5 e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}\right ) \, dx-290 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx\\ &=-\left (10 \int \frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))} \, dx\right )-10 \int \frac {e^{-e^3+x} x \log (1+x)}{(-5+x+\log (-5+x))^2} \, dx-10 \int \frac {e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x+\log (-5+x))^2} \, dx-50 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x+\log (-5+x))^2} \, dx+110 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x+\log (-5+x))^2} \, dx-250 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx-290 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx+550 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.67, size = 24, normalized size = 0.83 \begin {gather*} -\frac {10 e^{-e^3+x} \log (1+x)}{-5+x+\log (-5+x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.75, size = 24, normalized size = 0.83 \begin {gather*} -\frac {5 \, e^{\left (x - e^{3} + \log \relax (2)\right )} \log \left (x + 1\right )}{x + \log \left (x - 5\right ) - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 22, normalized size = 0.76 \begin {gather*} -\frac {10 \, e^{\left (x - e^{3}\right )} \log \left (x + 1\right )}{x + \log \left (x - 5\right ) - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 23, normalized size = 0.79
method | result | size |
risch | \(-\frac {10 \,{\mathrm e}^{x -{\mathrm e}^{3}} \ln \left (x +1\right )}{-5+x +\ln \left (x -5\right )}\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.58, size = 29, normalized size = 1.00 \begin {gather*} -\frac {10 \, e^{x} \log \left (x + 1\right )}{x e^{\left (e^{3}\right )} + e^{\left (e^{3}\right )} \log \left (x - 5\right ) - 5 \, e^{\left (e^{3}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.44, size = 22, normalized size = 0.76 \begin {gather*} -\frac {10\,\ln \left (x+1\right )\,{\mathrm {e}}^{-{\mathrm {e}}^3}\,{\mathrm {e}}^x}{x+\ln \left (x-5\right )-5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.55, size = 22, normalized size = 0.76 \begin {gather*} - \frac {10 e^{x - e^{3}} \log {\left (x + 1 \right )}}{x + \log {\left (x - 5 \right )} - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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