Optimal. Leaf size=32 \[ \left (\frac {x}{8}+e^5 \left (-4+e^{\left (1+x^2\right )^2}-\frac {\log (3)}{1+x}\right )\right )^2 \]
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Rubi [F] time = 0.96, 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 {x+3 x^2+3 x^3+x^4+e^5 \left (-32-96 x-96 x^2-32 x^3\right )+e^{12+4 x^2+2 x^4} \left (256 x+768 x^2+1024 x^3+1024 x^4+768 x^5+256 x^6\right )+\left (e^{10} (-256-256 x)+e^5 (-8-8 x)\right ) \log (3)-64 e^{10} \log ^2(3)+e^{1+2 x^2+x^4} \left (e^{10} \left (-1024 x-3072 x^2-4096 x^3-4096 x^4-3072 x^5-1024 x^6\right )+e^5 \left (8+24 x+56 x^2+104 x^3+128 x^4+128 x^5+96 x^6+32 x^7\right )+e^{10} \left (64-192 x-512 x^2-512 x^3-512 x^4-256 x^5\right ) \log (3)\right )}{32+96 x+96 x^2+32 x^3} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {x}{32}+8 e^{12+4 x^2+2 x^4} x \left (1+x^2\right )+\frac {1}{4} e^{6+2 x^2+x^4} \left (1+2 x^2\right )^2-\frac {2 e^{10} \log (3) (4+4 x+\log (3))}{(1+x)^3}-\frac {e^5 \left (4+8 x+4 x^2+\log (3)\right )}{4 (1+x)^2}-\frac {2 e^{11+2 x^2+x^4} \left (16 x^5-\log (3)+4 x (4+\log (3))+4 x^2 (8+\log (3))+4 x^3 (8+\log (3))+4 x^4 (8+\log (3))\right )}{(1+x)^2}\right ) \, dx\\ &=\frac {x^2}{64}+\frac {1}{4} \int e^{6+2 x^2+x^4} \left (1+2 x^2\right )^2 \, dx-2 \int \frac {e^{11+2 x^2+x^4} \left (16 x^5-\log (3)+4 x (4+\log (3))+4 x^2 (8+\log (3))+4 x^3 (8+\log (3))+4 x^4 (8+\log (3))\right )}{(1+x)^2} \, dx+8 \int e^{12+4 x^2+2 x^4} x \left (1+x^2\right ) \, dx-\frac {1}{4} e^5 \int \frac {4+8 x+4 x^2+\log (3)}{(1+x)^2} \, dx-\left (2 e^{10} \log (3)\right ) \int \frac {4+4 x+\log (3)}{(1+x)^3} \, dx\\ &=e^{12+4 x^2+2 x^4}+\frac {x^2}{64}+\frac {e^{10} (4+4 x+\log (3))^2}{(1+x)^2}-\frac {2 e^{11+2 x^2+x^4} \left (4 x^5+x (4+\log (3))+x^2 (8+\log (3))+x^3 (8+\log (3))+x^4 (8+\log (3))\right )}{(1+x)^2 \left (x+x^3\right )}+\frac {1}{4} \int \left (e^{6+2 x^2+x^4}+4 e^{6+2 x^2+x^4} x^2+4 e^{6+2 x^2+x^4} x^4\right ) \, dx-\frac {1}{4} e^5 \int \left (4+\frac {\log (3)}{(1+x)^2}\right ) \, dx\\ &=e^{12+4 x^2+2 x^4}-e^5 x+\frac {x^2}{64}+\frac {e^5 \log (3)}{4 (1+x)}+\frac {e^{10} (4+4 x+\log (3))^2}{(1+x)^2}-\frac {2 e^{11+2 x^2+x^4} \left (4 x^5+x (4+\log (3))+x^2 (8+\log (3))+x^3 (8+\log (3))+x^4 (8+\log (3))\right )}{(1+x)^2 \left (x+x^3\right )}+\frac {1}{4} \int e^{6+2 x^2+x^4} \, dx+\int e^{6+2 x^2+x^4} x^2 \, dx+\int e^{6+2 x^2+x^4} x^4 \, dx\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.42, size = 109, normalized size = 3.41 \begin {gather*} e^{2 \left (6+2 x^2+x^4\right )}+\frac {1}{4} e^{6+2 x^2+x^4} x+\frac {x^2}{64}-\frac {e^5 \left (4+8 x+4 x^2-\log (3)\right )}{4 (1+x)}-\frac {2 e^{11+2 x^2+x^4} (4+4 x+\log (3))}{1+x}+\frac {e^{10} \log (3) (8+8 x+\log (3))}{(1+x)^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.76, size = 136, normalized size = 4.25 \begin {gather*} \frac {x^{4} + 2 \, x^{3} + 64 \, e^{10} \log \relax (3)^{2} + x^{2} - 64 \, {\left (x^{3} + 2 \, x^{2} + x\right )} e^{5} + 64 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, x^{4} + 4 \, x^{2} + 12\right )} - 16 \, {\left (8 \, {\left (x + 1\right )} e^{10} \log \relax (3) + 32 \, {\left (x^{2} + 2 \, x + 1\right )} e^{10} - {\left (x^{3} + 2 \, x^{2} + x\right )} e^{5}\right )} e^{\left (x^{4} + 2 \, x^{2} + 1\right )} + 16 \, {\left (32 \, {\left (x + 1\right )} e^{10} + {\left (x + 1\right )} e^{5}\right )} \log \relax (3)}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.22, size = 246, normalized size = 7.69 \begin {gather*} \frac {x^{4} - 64 \, x^{3} e^{5} + 16 \, x^{3} e^{\left (x^{4} + 2 \, x^{2} + 6\right )} + 2 \, x^{3} - 128 \, x^{2} e^{5} + 64 \, x^{2} e^{\left (2 \, x^{4} + 4 \, x^{2} + 12\right )} - 512 \, x^{2} e^{\left (x^{4} + 2 \, x^{2} + 11\right )} + 32 \, x^{2} e^{\left (x^{4} + 2 \, x^{2} + 6\right )} + 512 \, x e^{10} \log \relax (3) + 16 \, x e^{5} \log \relax (3) - 128 \, x e^{\left (x^{4} + 2 \, x^{2} + 11\right )} \log \relax (3) + 64 \, e^{10} \log \relax (3)^{2} + x^{2} - 64 \, x e^{5} + 128 \, x e^{\left (2 \, x^{4} + 4 \, x^{2} + 12\right )} - 1024 \, x e^{\left (x^{4} + 2 \, x^{2} + 11\right )} + 16 \, x e^{\left (x^{4} + 2 \, x^{2} + 6\right )} + 512 \, e^{10} \log \relax (3) + 16 \, e^{5} \log \relax (3) - 128 \, e^{\left (x^{4} + 2 \, x^{2} + 11\right )} \log \relax (3) + 64 \, e^{\left (2 \, x^{4} + 4 \, x^{2} + 12\right )} - 512 \, e^{\left (x^{4} + 2 \, x^{2} + 11\right )}}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.26, size = 114, normalized size = 3.56
method | result | size |
risch | \(\frac {x^{2}}{64}-x \,{\mathrm e}^{5}+\frac {\frac {\left (256 \,{\mathrm e}^{10} \ln \relax (3)+8 \,{\mathrm e}^{5} \ln \relax (3)\right ) x}{32}+{\mathrm e}^{10} \ln \relax (3)^{2}+8 \,{\mathrm e}^{10} \ln \relax (3)+\frac {{\mathrm e}^{5} \ln \relax (3)}{4}}{x^{2}+2 x +1}+{\mathrm e}^{2 x^{4}+4 x^{2}+12}-\frac {\left (8 \,{\mathrm e}^{5} \ln \relax (3)+32 x \,{\mathrm e}^{5}-x^{2}+32 \,{\mathrm e}^{5}-x \right ) {\mathrm e}^{x^{4}+2 x^{2}+6}}{4 \left (x +1\right )}\) | \(114\) |
norman | \(\frac {\left (\frac {1}{32}-{\mathrm e}^{5}\right ) x^{3}+\left (-2 \,{\mathrm e}^{10} \ln \relax (3)-8 \,{\mathrm e}^{10}\right ) {\mathrm e}^{x^{4}+2 x^{2}+1}+\left (-\frac {1}{32}+3 \,{\mathrm e}^{5}+8 \,{\mathrm e}^{10} \ln \relax (3)+\frac {{\mathrm e}^{5} \ln \relax (3)}{4}\right ) x +{\mathrm e}^{10} {\mathrm e}^{2 x^{4}+4 x^{2}+2}+\left (\frac {{\mathrm e}^{5}}{2}-8 \,{\mathrm e}^{10}\right ) x^{2} {\mathrm e}^{x^{4}+2 x^{2}+1}+\left (-2 \,{\mathrm e}^{10} \ln \relax (3)-16 \,{\mathrm e}^{10}+\frac {{\mathrm e}^{5}}{4}\right ) x \,{\mathrm e}^{x^{4}+2 x^{2}+1}+{\mathrm e}^{10} {\mathrm e}^{2 x^{4}+4 x^{2}+2} x^{2}+\frac {x^{4}}{64}+\frac {{\mathrm e}^{5} {\mathrm e}^{x^{4}+2 x^{2}+1} x^{3}}{4}+2 \,{\mathrm e}^{10} {\mathrm e}^{2 x^{4}+4 x^{2}+2} x -\frac {1}{64}+2 \,{\mathrm e}^{5}+8 \,{\mathrm e}^{10} \ln \relax (3)+\frac {{\mathrm e}^{5} \ln \relax (3)}{4}+{\mathrm e}^{10} \ln \relax (3)^{2}}{\left (x +1\right )^{2}}\) | \(235\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.56, size = 323, normalized size = 10.09 \begin {gather*} \frac {1}{64} \, x^{2} - \frac {1}{2} \, {\left (2 \, x - \frac {6 \, x + 5}{x^{2} + 2 \, x + 1} - 6 \, \log \left (x + 1\right )\right )} e^{5} - \frac {3}{2} \, {\left (\frac {4 \, x + 3}{x^{2} + 2 \, x + 1} + 2 \, \log \left (x + 1\right )\right )} e^{5} + \frac {4 \, {\left (2 \, x + 1\right )} e^{10} \log \relax (3)}{x^{2} + 2 \, x + 1} + \frac {{\left (2 \, x + 1\right )} e^{5} \log \relax (3)}{8 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {e^{10} \log \relax (3)^{2}}{x^{2} + 2 \, x + 1} + \frac {3 \, {\left (2 \, x + 1\right )} e^{5}}{2 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {4 \, e^{10} \log \relax (3)}{x^{2} + 2 \, x + 1} + \frac {e^{5} \log \relax (3)}{8 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {8 \, x + 7}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} - \frac {3 \, {\left (6 \, x + 5\right )}}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {3 \, {\left (4 \, x + 3\right )}}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} - \frac {2 \, x + 1}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {4 \, {\left (x e^{12} + e^{12}\right )} e^{\left (2 \, x^{4} + 4 \, x^{2}\right )} + {\left (x^{2} e^{6} - x {\left (32 \, e^{11} - e^{6}\right )} - 8 \, {\left (\log \relax (3) + 4\right )} e^{11}\right )} e^{\left (x^{4} + 2 \, x^{2}\right )}}{4 \, {\left (x + 1\right )}} + \frac {e^{5}}{2 \, {\left (x^{2} + 2 \, x + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.23, size = 218, normalized size = 6.81 \begin {gather*} -\frac {x^2\,\left (64\,{\mathrm {e}}^5+8\,{\mathrm {e}}^5\,\ln \relax (3)+256\,{\mathrm {e}}^{10}\,\ln \relax (3)+32\,{\mathrm {e}}^{10}\,{\ln \relax (3)}^2-\frac {1}{2}\right )-64\,x\,{\mathrm {e}}^{2\,x^4+4\,x^2+12}-8\,x^3\,{\mathrm {e}}^{x^4+2\,x^2+6}-32\,{\mathrm {e}}^{2\,x^4+4\,x^2+12}+x^3\,\left (32\,{\mathrm {e}}^5-1\right )+{\mathrm {e}}^{x^4+2\,x^2+11}\,\left (64\,\ln \relax (3)+256\right )-32\,x^2\,{\mathrm {e}}^{2\,x^4+4\,x^2+12}-\frac {x^4}{2}+x\,\left (32\,{\mathrm {e}}^5+8\,{\mathrm {e}}^5\,\ln \relax (3)+256\,{\mathrm {e}}^{10}\,\ln \relax (3)+64\,{\mathrm {e}}^{10}\,{\ln \relax (3)}^2\right )+8\,x\,{\mathrm {e}}^{x^4+2\,x^2+6}\,\left (64\,{\mathrm {e}}^5+8\,{\mathrm {e}}^5\,\ln \relax (3)-1\right )+16\,x^2\,{\mathrm {e}}^{x^4+2\,x^2+6}\,\left (16\,{\mathrm {e}}^5-1\right )}{32\,x^2+64\,x+32} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.66, size = 136, normalized size = 4.25 \begin {gather*} \frac {x^{2}}{64} - x e^{5} + \frac {x \left (e^{5} \log {\relax (3 )} + 32 e^{10} \log {\relax (3 )}\right ) + e^{5} \log {\relax (3 )} + 4 e^{10} \log {\relax (3 )}^{2} + 32 e^{10} \log {\relax (3 )}}{4 x^{2} + 8 x + 4} + \frac {\left (4 x e^{10} + 4 e^{10}\right ) e^{2 x^{4} + 4 x^{2} + 2} + \left (x^{2} e^{5} - 32 x e^{10} + x e^{5} - 32 e^{10} - 8 e^{10} \log {\relax (3 )}\right ) e^{x^{4} + 2 x^{2} + 1}}{4 x + 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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