Optimal. Leaf size=30 \[ \frac {e^{x^8} x}{-1+e^2+\frac {e^{\frac {x}{5 \log (x)}}}{x}+x} \]
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Rubi [F]
time = 9.84, 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^{x^8} \left (-5 x^2-40 x^{10}+40 x^{11}+e^2 \left (5 x^2+40 x^{10}\right )\right ) \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (e^{x^8} x^2-e^{x^8} x^2 \log (x)+e^{x^8} \left (10 x+40 x^9\right ) \log ^2(x)\right )}{5 e^{\frac {2 x}{5 \log (x)}} \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (-10 x+10 e^2 x+10 x^2\right ) \log ^2(x)+\left (5 x^2+5 e^4 x^2-10 x^3+5 x^4+e^2 \left (-10 x^2+10 x^3\right )\right ) \log ^2(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 {e^{x^8} x \left (e^{\frac {x}{5 \log (x)}} x-e^{\frac {x}{5 \log (x)}} x \log (x)+5 \left (e^{\frac {x}{5 \log (x)}} \left (2+8 x^8\right )+e^2 \left (x+8 x^9\right )+x \left (-1-8 x^8+8 x^9\right )\right ) \log ^2(x)\right )}{5 \left (e^{\frac {x}{5 \log (x)}}+e^2 x+(-1+x) x\right )^2 \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \frac {e^{x^8} x \left (e^{\frac {x}{5 \log (x)}} x-e^{\frac {x}{5 \log (x)}} x \log (x)+5 \left (e^{\frac {x}{5 \log (x)}} \left (2+8 x^8\right )+e^2 \left (x+8 x^9\right )+x \left (-1-8 x^8+8 x^9\right )\right ) \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}+e^2 x+(-1+x) x\right )^2 \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^{x^8} x^2 \left (\left (1-e^2\right ) x-x^2-\left (1-e^2\right ) x \log (x)+x^2 \log (x)+5 \left (1-e^2\right ) \log ^2(x)-10 x \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)}+\frac {e^{x^8} x \left (x-x \log (x)+10 \log ^2(x)+40 x^8 \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right ) \log ^2(x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{x^8} x^2 \left (\left (1-e^2\right ) x-x^2-\left (1-e^2\right ) x \log (x)+x^2 \log (x)+5 \left (1-e^2\right ) \log ^2(x)-10 x \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)} \, dx+\frac {1}{5} \int \frac {e^{x^8} x \left (x-x \log (x)+10 \log ^2(x)+40 x^8 \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right ) \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {10 e^{x^8} x}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2}+\frac {40 e^{x^8} x^9}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2}+\frac {e^{x^8} x^2}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right ) \log ^2(x)}+\frac {e^{x^8} x^2}{\left (-e^{\frac {x}{5 \log (x)}}+\left (1-e^2\right ) x-x^2\right ) \log (x)}\right ) \, dx+\frac {1}{5} \int \frac {e^{x^8} x^2 \left (-x \left (-1+e^2+x\right )+x \left (-1+e^2+x\right ) \log (x)-5 \left (-1+e^2+2 x\right ) \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}+e^2 x+(-1+x) x\right )^2 \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {5 e^{x^8} \left (1-e^2\right ) x^2}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2}-\frac {10 e^{x^8} x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2}+\frac {e^{x^8} \left (1-e^2\right ) x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)}-\frac {e^{x^8} x^4}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)}-\frac {e^{x^8} \left (1-e^2\right ) x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log (x)}+\frac {e^{x^8} x^4}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log (x)}\right ) \, dx+\frac {1}{5} \int \frac {e^{x^8} x^2}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right ) \log ^2(x)} \, dx+\frac {1}{5} \int \frac {e^{x^8} x^2}{\left (-e^{\frac {x}{5 \log (x)}}+\left (1-e^2\right ) x-x^2\right ) \log (x)} \, dx+2 \int \frac {e^{x^8} x}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2} \, dx+8 \int \frac {e^{x^8} x^9}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2} \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{x^8} x^4}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)} \, dx\right )+\frac {1}{5} \int \frac {e^{x^8} x^2}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right ) \log ^2(x)} \, dx+\frac {1}{5} \int \frac {e^{x^8} x^2}{\left (-e^{\frac {x}{5 \log (x)}}+\left (1-e^2\right ) x-x^2\right ) \log (x)} \, dx+\frac {1}{5} \int \frac {e^{x^8} x^4}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log (x)} \, dx-2 \int \frac {e^{x^8} x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2} \, dx+2 \int \frac {e^{x^8} x}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2} \, dx+8 \int \frac {e^{x^8} x^9}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2} \, dx+\frac {1}{5} \left (1-e^2\right ) \int \frac {e^{x^8} x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)} \, dx+\left (1-e^2\right ) \int \frac {e^{x^8} x^2}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2} \, dx+\frac {1}{5} \left (-1+e^2\right ) \int \frac {e^{x^8} x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.15, size = 34, normalized size = 1.13 \begin {gather*} \frac {e^{x^8} x^2}{e^{\frac {x}{5 \log (x)}}-x+e^2 x+x^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.37, size = 30, normalized size = 1.00
method | result | size |
risch | \(\frac {x^{2} {\mathrm e}^{x^{8}}}{{\mathrm e}^{2} x +x^{2}+{\mathrm e}^{\frac {x}{5 \ln \left (x \right )}}-x}\) | \(30\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.38, size = 28, normalized size = 0.93 \begin {gather*} \frac {x^{2} e^{\left (x^{8}\right )}}{x^{2} + x {\left (e^{2} - 1\right )} + e^{\left (\frac {x}{5 \, \log \left (x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 29, normalized size = 0.97 \begin {gather*} \frac {x^{2} e^{\left (x^{8}\right )}}{x^{2} + x e^{2} - x + e^{\left (\frac {x}{5 \, \log \left (x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.17, size = 26, normalized size = 0.87 \begin {gather*} \frac {x^{2} e^{x^{8}}}{x^{2} - x + x e^{2} + e^{\frac {x}{5 \log {\left (x \right )}}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.49, size = 29, normalized size = 0.97 \begin {gather*} \frac {x^{2} e^{\left (x^{8}\right )}}{x^{2} + x e^{2} - x + e^{\left (\frac {x}{5 \, \log \left (x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {x}{5\,\ln \left (x\right )}}\,\left (x^2\,{\mathrm {e}}^{x^8}+{\mathrm {e}}^{x^8}\,{\ln \left (x\right )}^2\,\left (40\,x^9+10\,x\right )-x^2\,{\mathrm {e}}^{x^8}\,\ln \left (x\right )\right )+{\mathrm {e}}^{x^8}\,{\ln \left (x\right )}^2\,\left ({\mathrm {e}}^2\,\left (40\,x^{10}+5\,x^2\right )-5\,x^2-40\,x^{10}+40\,x^{11}\right )}{5\,{\mathrm {e}}^{\frac {2\,x}{5\,\ln \left (x\right )}}\,{\ln \left (x\right )}^2+{\ln \left (x\right )}^2\,\left (5\,x^2\,{\mathrm {e}}^4-{\mathrm {e}}^2\,\left (10\,x^2-10\,x^3\right )+5\,x^2-10\,x^3+5\,x^4\right )+{\mathrm {e}}^{\frac {x}{5\,\ln \left (x\right )}}\,{\ln \left (x\right )}^2\,\left (10\,x\,{\mathrm {e}}^2-10\,x+10\,x^2\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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