Optimal. Leaf size=29 \[ x \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right ) \]
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Rubi [F] time = 3.95, 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 {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} \left (-75+e^8 \left (4 x^2-4 x^3\right )\right )+\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )\right )}{\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\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 {125+15 e^{2 x^2}+e^{3 x^2}-e^{x^2} \left (-75-4 e^8 (-1+x) x^2\right )-\left (5+e^{x^2}\right )^3 (-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )}{\left (5+e^{x^2}\right )^3 (1-x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\\ &=\int \left (\frac {20 e^8 x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}-\frac {4 e^8 x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}+\frac {-1-\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )+x \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}\right ) \, dx\\ &=-\left (\left (4 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\right )+\left (20 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx+\int \frac {-1-\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )+x \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\\ &=-\left (\left (4 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\right )+\left (20 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx+\int \left (-\frac {1}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}+\log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )\right ) \, dx\\ &=-\left (\left (4 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\right )+\left (20 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx-\int \frac {1}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx+\int \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right ) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.28, size = 29, normalized size = 1.00 \begin {gather*} x \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.63, size = 32, normalized size = 1.10 \begin {gather*} x \log \left (\log \left (\frac {x e^{\left (\frac {e^{8}}{e^{\left (2 \, x^{2}\right )} + 10 \, e^{\left (x^{2}\right )} + 25}\right )}}{4 \, {\left (x - 1\right )}}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.66, size = 302, normalized size = 10.41
method | result | size |
risch | \(\ln \left (-2 \ln \relax (2)+\ln \relax (x )+\ln \left ({\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}\right )-\ln \left (x -1\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right )+\mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right )+\mathrm {csgn}\left (\frac {i}{x -1}\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right ) \left (-\mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right )+\mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right )\right )}{2}\right ) x\) | \(302\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.85, size = 76, normalized size = 2.62 \begin {gather*} x \log \left (-{\left (2 \, \log \relax (2) + \log \left (x - 1\right )\right )} e^{\left (2 \, x^{2}\right )} - 10 \, {\left (2 \, \log \relax (2) + \log \left (x - 1\right )\right )} e^{\left (x^{2}\right )} + {\left (e^{\left (2 \, x^{2}\right )} + 10 \, e^{\left (x^{2}\right )} + 25\right )} \log \relax (x) + e^{8} - 50 \, \log \relax (2) - 25 \, \log \left (x - 1\right )\right ) - 2 \, x \log \left (e^{\left (x^{2}\right )} + 5\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 {15\,{\mathrm {e}}^{2\,x^2}+{\mathrm {e}}^{3\,x^2}-{\mathrm {e}}^{x^2}\,\left ({\mathrm {e}}^8\,\left (4\,x^2-4\,x^3\right )-75\right )-\ln \left (\ln \left (\frac {x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^8}{10\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{2\,x^2}+25}}}{4\,x-4}\right )\right )\,\ln \left (\frac {x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^8}{10\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{2\,x^2}+25}}}{4\,x-4}\right )\,\left (125\,x+{\mathrm {e}}^{2\,x^2}\,\left (15\,x-15\right )+{\mathrm {e}}^{3\,x^2}\,\left (x-1\right )+{\mathrm {e}}^{x^2}\,\left (75\,x-75\right )-125\right )+125}{\ln \left (\frac {x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^8}{10\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{2\,x^2}+25}}}{4\,x-4}\right )\,\left (125\,x+{\mathrm {e}}^{2\,x^2}\,\left (15\,x-15\right )+{\mathrm {e}}^{3\,x^2}\,\left (x-1\right )+{\mathrm {e}}^{x^2}\,\left (75\,x-75\right )-125\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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