Optimal. Leaf size=34 \[ 2+\frac {1}{4} \left (2 x+\frac {e^3 \log \left (2 e^{-2 x}\right )}{e^2-\log ^2(x)}\right ) \]
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Rubi [F]
time = 0.74, 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 {e^4 x-e^5 x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(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 {\left (e^4-e^5\right ) x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx\\ &=\int \frac {\left (e^4-e^5\right ) x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 x \left (e^2-\log ^2(x)\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {\left (e^4-e^5\right ) x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{x \left (e^2-\log ^2(x)\right )^2} \, dx\\ &=\frac {1}{2} \int \left (1+\frac {e^2 \log \left (2 e^{-2 x}\right )}{4 x (e-\log (x))^2}-\frac {e^2}{2 (e-\log (x))}-\frac {e^2 \log \left (2 e^{-2 x}\right )}{4 x (e+\log (x))^2}-\frac {e^2}{2 (e+\log (x))}\right ) \, dx\\ &=\frac {x}{2}+\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e-\log (x))^2} \, dx-\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e+\log (x))^2} \, dx-\frac {1}{4} e^2 \int \frac {1}{e-\log (x)} \, dx-\frac {1}{4} e^2 \int \frac {1}{e+\log (x)} \, dx\\ &=\frac {x}{2}+\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e-\log (x))^2} \, dx-\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e+\log (x))^2} \, dx-\frac {1}{4} e^2 \text {Subst}\left (\int \frac {e^x}{e-x} \, dx,x,\log (x)\right )-\frac {1}{4} e^2 \text {Subst}\left (\int \frac {e^x}{e+x} \, dx,x,\log (x)\right )\\ &=\frac {x}{2}+\frac {1}{4} e^{2+e} \text {Ei}(-e+\log (x))-\frac {1}{4} e^{2-e} \text {Ei}(e+\log (x))+\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e-\log (x))^2} \, dx-\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.07, size = 33, normalized size = 0.97 \begin {gather*} \frac {1}{2} \left (x+\frac {e^3 \log \left (2 e^{-2 x}\right )}{2 \left (e^2-\log ^2(x)\right )}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 0.23, size = 423, normalized size = 12.44
method | result | size |
risch | \(-\frac {{\mathrm e}^{3} \ln \left ({\mathrm e}^{x}\right )}{2 \left ({\mathrm e}^{2}-\ln \left (x \right )^{2}\right )}+\frac {-i \pi \,{\mathrm e}^{3} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+2 i \pi \,{\mathrm e}^{3} \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-i \pi \,{\mathrm e}^{3} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-2 \,{\mathrm e}^{3} \ln \left (2\right )-4 \,{\mathrm e}^{2} x +4 x \ln \left (x \right )^{2}}{-8 \,{\mathrm e}^{2}+8 \ln \left (x \right )^{2}}\) | \(112\) |
default | \(-\frac {{\mathrm e}^{3} x}{2 \left ({\mathrm e}^{2}-\ln \left (x \right )^{2}\right )}-\frac {{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )+{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{3} {\mathrm e}^{-{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )-{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )+{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{2} {\mathrm e}^{-{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )-{\mathrm e}\right )}{8}+\frac {x}{2}+\frac {{\mathrm e}^{3} {\mathrm e}^{-1} {\mathrm e}^{{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )+{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{3} {\mathrm e}^{-1} {\mathrm e}^{-{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )-{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{4} {\mathrm e}^{-2} {\mathrm e}^{{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )+{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{4} {\mathrm e}^{-2} {\mathrm e}^{-{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )-{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{2} {\mathrm e}^{-1} {\mathrm e}^{{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )+{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{2} {\mathrm e}^{-1} {\mathrm e}^{-{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )-{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{2} {\mathrm e}^{3} {\mathrm e}^{-2} {\mathrm e}^{-1} {\mathrm e}^{-{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )-{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{2} {\mathrm e}^{3} {\mathrm e}^{-2} {\mathrm e}^{-1} {\mathrm e}^{{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )+{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{2} {\mathrm e}^{3} {\mathrm e}^{-2} {\mathrm e}^{{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )+{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{2} {\mathrm e}^{3} {\mathrm e}^{-2} {\mathrm e}^{-{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )-{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{4} {\mathrm e}^{-2} {\mathrm e}^{-1} {\mathrm e}^{{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )+{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{4} {\mathrm e}^{-2} {\mathrm e}^{-1} {\mathrm e}^{-{\mathrm e}} \expIntegral \left (1, -\ln \left (x \right )-{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{3} \left (\ln \left (2 \,{\mathrm e}^{-2 x}\right )+2 \ln \left ({\mathrm e}^{x}\right )\right )}{4 \,{\mathrm e}^{2}-4 \ln \left (x \right )^{2}}-\frac {{\mathrm e}^{3} \left (\ln \left ({\mathrm e}^{x}\right )-x \right )}{2 \left ({\mathrm e}^{2}-\ln \left (x \right )^{2}\right )}\) | \(423\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.49, size = 37, normalized size = 1.09 \begin {gather*} \frac {2 \, x \log \left (x\right )^{2} + 2 \, x {\left (e^{3} - e^{2}\right )} - e^{3} \log \left (2\right )}{4 \, {\left (\log \left (x\right )^{2} - e^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 37, normalized size = 1.09 \begin {gather*} \frac {2 \, x \log \left (x\right )^{2} + 2 \, x e^{3} - 2 \, x e^{2} - e^{3} \log \left (2\right )}{4 \, {\left (\log \left (x\right )^{2} - e^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.07, size = 27, normalized size = 0.79 \begin {gather*} \frac {x}{2} + \frac {2 x e^{3} - e^{3} \log {\left (2 \right )}}{4 \log {\left (x \right )}^{2} - 4 e^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.43, size = 37, normalized size = 1.09 \begin {gather*} \frac {2 \, x \log \left (x\right )^{2} + 2 \, x e^{3} - 2 \, x e^{2} - e^{3} \log \left (2\right )}{4 \, {\left (\log \left (x\right )^{2} - e^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 8.07, size = 29, normalized size = 0.85 \begin {gather*} \frac {x}{2}-\frac {{\mathrm {e}}^3\,\left (2\,x-\ln \left (2\right )\right )}{4\,\left ({\mathrm {e}}^2-{\ln \left (x\right )}^2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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