Optimal. Leaf size=27 \[ \left (-3+x^2\right ) \log \left (1+\left (1+e^5+e^{1-x}\right )^2 x^2\right ) \]
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Rubi [F] time = 96.42, 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 {-6 x+2 x^3+e^{10} \left (-6 x+2 x^3\right )+e^5 \left (-12 x+4 x^3\right )+e^{2-2 x} \left (-6 x+6 x^2+2 x^3-2 x^4\right )+e^{1-x} \left (-12 x+6 x^2+4 x^3-2 x^4+e^5 \left (-12 x+6 x^2+4 x^3-2 x^4\right )\right )+\left (2 x+2 x^3+4 e^5 x^3+2 e^{10} x^3+2 e^{2-2 x} x^3+e^{1-x} \left (4 x^3+4 e^5 x^3\right )\right ) \log \left (1+x^2+2 e^5 x^2+e^{10} x^2+e^{2-2 x} x^2+e^{1-x} \left (2 x^2+2 e^5 x^2\right )\right )}{1+x^2+2 e^5 x^2+e^{10} x^2+e^{2-2 x} x^2+e^{1-x} \left (2 x^2+2 e^5 x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-6 x+2 x^3+e^{10} \left (-6 x+2 x^3\right )+e^5 \left (-12 x+4 x^3\right )+e^{2-2 x} \left (-6 x+6 x^2+2 x^3-2 x^4\right )+e^{1-x} \left (-12 x+6 x^2+4 x^3-2 x^4+e^5 \left (-12 x+6 x^2+4 x^3-2 x^4\right )\right )+\left (2 x+2 x^3+4 e^5 x^3+2 e^{10} x^3+2 e^{2-2 x} x^3+e^{1-x} \left (4 x^3+4 e^5 x^3\right )\right ) \log \left (1+x^2+2 e^5 x^2+e^{10} x^2+e^{2-2 x} x^2+e^{1-x} \left (2 x^2+2 e^5 x^2\right )\right )}{1+e^{10} x^2+e^{2-2 x} x^2+\left (1+2 e^5\right ) x^2+e^{1-x} \left (2 x^2+2 e^5 x^2\right )} \, dx\\ &=\int \frac {-6 x+2 x^3+e^{10} \left (-6 x+2 x^3\right )+e^5 \left (-12 x+4 x^3\right )+e^{2-2 x} \left (-6 x+6 x^2+2 x^3-2 x^4\right )+e^{1-x} \left (-12 x+6 x^2+4 x^3-2 x^4+e^5 \left (-12 x+6 x^2+4 x^3-2 x^4\right )\right )+\left (2 x+2 x^3+4 e^5 x^3+2 e^{10} x^3+2 e^{2-2 x} x^3+e^{1-x} \left (4 x^3+4 e^5 x^3\right )\right ) \log \left (1+x^2+2 e^5 x^2+e^{10} x^2+e^{2-2 x} x^2+e^{1-x} \left (2 x^2+2 e^5 x^2\right )\right )}{1+e^{2-2 x} x^2+\left (1+2 e^5+e^{10}\right ) x^2+e^{1-x} \left (2 x^2+2 e^5 x^2\right )} \, dx\\ &=\int \frac {2 x \left (-3+x^2+2 e^5 \left (1+\frac {e^5}{2}\right ) \left (-3+x^2\right )+e^{2-2 x} \left (-3+3 x+x^2-x^3\right )-e^{1-x} \left (1+e^5\right ) \left (6-3 x-2 x^2+x^3\right )+e^{-2 x} \left (e^2 x^2+2 e^{1+x} x^2+e^{2 (5+x)} x^2+2 e^{6+x} x^2+2 e^{5+2 x} x^2+e^{2 x} \left (1+x^2\right )\right ) \log \left (1+e^{-2 x} \left (e+e^x+e^{5+x}\right )^2 x^2\right )\right )}{1+e^{-2 x} \left (e+e^x+e^{5+x}\right )^2 x^2} \, dx\\ &=2 \int \frac {x \left (-3+x^2+2 e^5 \left (1+\frac {e^5}{2}\right ) \left (-3+x^2\right )+e^{2-2 x} \left (-3+3 x+x^2-x^3\right )-e^{1-x} \left (1+e^5\right ) \left (6-3 x-2 x^2+x^3\right )+e^{-2 x} \left (e^2 x^2+2 e^{1+x} x^2+e^{2 (5+x)} x^2+2 e^{6+x} x^2+2 e^{5+2 x} x^2+e^{2 x} \left (1+x^2\right )\right ) \log \left (1+e^{-2 x} \left (e+e^x+e^{5+x}\right )^2 x^2\right )\right )}{1+e^{-2 x} \left (e+e^x+e^{5+x}\right )^2 x^2} \, dx\\ &=2 \int \left (\frac {e x \left (3-x^2\right ) \left (-e-2 e^x \left (1+e^5\right )+e x+e^x \left (1+e^5\right ) x+e \left (1+e^5 \left (2+e^5\right )\right ) x^3+e^x \left (1+e^5 \left (3+3 e^5+e^{10}\right )\right ) x^3\right )}{\left (1+\left (1+e^5\right )^2 x^2\right ) \left (e^{2 x}+e^2 x^2+2 e^{1+x} \left (1+e^5\right ) x^2+e^{2 x} \left (1+e^5 \left (2+e^5\right )\right ) x^2\right )}+\frac {x \left (-3 \left (1+e^5 \left (2+e^5\right )\right )+\left (1+e^5 \left (2+e^5\right )\right ) x^2+\log \left (1+e^{-2 x} \left (e+e^x+e^{5+x}\right )^2 x^2\right )+\left (1+e^5 \left (2+e^5\right )\right ) x^2 \log \left (1+e^{-2 x} \left (e+e^x+e^{5+x}\right )^2 x^2\right )\right )}{1+\left (1+e^5\right )^2 x^2}\right ) \, dx\\ &=2 \int \frac {x \left (-3 \left (1+e^5 \left (2+e^5\right )\right )+\left (1+e^5 \left (2+e^5\right )\right ) x^2+\log \left (1+e^{-2 x} \left (e+e^x+e^{5+x}\right )^2 x^2\right )+\left (1+e^5 \left (2+e^5\right )\right ) x^2 \log \left (1+e^{-2 x} \left (e+e^x+e^{5+x}\right )^2 x^2\right )\right )}{1+\left (1+e^5\right )^2 x^2} \, dx+(2 e) \int \frac {x \left (3-x^2\right ) \left (-e-2 e^x \left (1+e^5\right )+e x+e^x \left (1+e^5\right ) x+e \left (1+e^5 \left (2+e^5\right )\right ) x^3+e^x \left (1+e^5 \left (3+3 e^5+e^{10}\right )\right ) x^3\right )}{\left (1+\left (1+e^5\right )^2 x^2\right ) \left (e^{2 x}+e^2 x^2+2 e^{1+x} \left (1+e^5\right ) x^2+e^{2 x} \left (1+e^5 \left (2+e^5\right )\right ) x^2\right )} \, dx\\ &=2 \int x \left (\frac {\left (1+e^5\right )^2 \left (-3+x^2\right )}{1+\left (1+e^5\right )^2 x^2}+\log \left (1+e^{-2 x} \left (e+e^x+e^{5+x}\right )^2 x^2\right )\right ) \, dx+(2 e) \int \left (\frac {x \left (e+2 e^x \left (1+e^5\right )-e x-e^x \left (1+e^5\right ) x-e \left (1+e^5 \left (2+e^5\right )\right ) x^3-e^x \left (1+e^5 \left (3+3 e^5+e^{10}\right )\right ) x^3\right )}{\left (1+e^5\right )^2 \left (e^{2 x}+e^2 x^2+2 e^{1+x} \left (1+e^5\right ) x^2+e^{2 x} \left (1+e^5 \left (2+e^5\right )\right ) x^2\right )}+\frac {\left (4+6 e^5+3 e^{10}\right ) x \left (-e-2 e^x \left (1+e^5\right )+e x+e^x \left (1+e^5\right ) x+e \left (1+e^5 \left (2+e^5\right )\right ) x^3+e^x \left (1+e^5 \left (3+3 e^5+e^{10}\right )\right ) x^3\right )}{\left (1+e^5\right )^2 \left (1+\left (1+e^5\right )^2 x^2\right ) \left (e^{2 x}+e^2 x^2+2 e^{1+x} \left (1+e^5\right ) x^2+e^{2 x} \left (1+e^5 \left (2+e^5\right )\right ) x^2\right )}\right ) \, dx\\ &=2 \int \left (\frac {\left (1+e^5\right )^2 x \left (-3+x^2\right )}{1+\left (1+e^5\right )^2 x^2}+x \log \left (1+e^{-2 x} \left (e+e^x+e^{5+x}\right )^2 x^2\right )\right ) \, dx+\frac {(2 e) \int \frac {x \left (e+2 e^x \left (1+e^5\right )-e x-e^x \left (1+e^5\right ) x-e \left (1+e^5 \left (2+e^5\right )\right ) x^3-e^x \left (1+e^5 \left (3+3 e^5+e^{10}\right )\right ) x^3\right )}{e^{2 x}+e^2 x^2+2 e^{1+x} \left (1+e^5\right ) x^2+e^{2 x} \left (1+e^5 \left (2+e^5\right )\right ) x^2} \, dx}{\left (1+e^5\right )^2}+\frac {\left (2 e \left (4+6 e^5+3 e^{10}\right )\right ) \int \frac {x \left (-e-2 e^x \left (1+e^5\right )+e x+e^x \left (1+e^5\right ) x+e \left (1+e^5 \left (2+e^5\right )\right ) x^3+e^x \left (1+e^5 \left (3+3 e^5+e^{10}\right )\right ) x^3\right )}{\left (1+\left (1+e^5\right )^2 x^2\right ) \left (e^{2 x}+e^2 x^2+2 e^{1+x} \left (1+e^5\right ) x^2+e^{2 x} \left (1+e^5 \left (2+e^5\right )\right ) x^2\right )} \, dx}{\left (1+e^5\right )^2}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [B] time = 0.26, size = 94, normalized size = 3.48 \begin {gather*} 2 \left (-\frac {3}{2} \log \left (1+x^2+2 e^5 x^2+e^{10} x^2+e^{2-2 x} x^2+2 e^{1-x} x^2+2 e^{6-x} x^2\right )+\frac {1}{2} x^2 \log \left (1+e^{-2 x} \left (e+e^x+e^{5+x}\right )^2 x^2\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 1.08, size = 53, normalized size = 1.96 \begin {gather*} {\left (x^{2} - 3\right )} \log \left (x^{2} e^{10} + 2 \, x^{2} e^{5} + x^{2} e^{\left (-2 \, x + 2\right )} + x^{2} + 2 \, {\left (x^{2} e^{5} + x^{2}\right )} e^{\left (-x + 1\right )} + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.31, size = 133, normalized size = 4.93 \begin {gather*} -2 \, x^{3} + x^{2} \log \left (x^{2} e^{2} + x^{2} e^{\left (2 \, x\right )} + x^{2} e^{\left (2 \, x + 10\right )} + 2 \, x^{2} e^{\left (2 \, x + 5\right )} + 2 \, x^{2} e^{\left (x + 6\right )} + 2 \, x^{2} e^{\left (x + 1\right )} + e^{\left (2 \, x\right )}\right ) + 6 \, x - 3 \, \log \left (x^{2} e^{2} + x^{2} e^{\left (2 \, x\right )} + x^{2} e^{\left (2 \, x + 10\right )} + 2 \, x^{2} e^{\left (2 \, x + 5\right )} + 2 \, x^{2} e^{\left (x + 6\right )} + 2 \, x^{2} e^{\left (x + 1\right )} + e^{\left (2 \, x\right )}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.07, size = 105, normalized size = 3.89
method | result | size |
risch | \(x^{2} \ln \left (x^{2} {\mathrm e}^{-2 x +2}+\left (2 x^{2} {\mathrm e}^{5}+2 x^{2}\right ) {\mathrm e}^{1-x}+x^{2} {\mathrm e}^{10}+2 x^{2} {\mathrm e}^{5}+x^{2}+1\right )-6 \ln \relax (x )+6-3 \ln \left ({\mathrm e}^{-2 x +2}+\left (2 \,{\mathrm e}^{5}+2\right ) {\mathrm e}^{1-x}+\frac {2 x^{2} {\mathrm e}^{5}+x^{2} {\mathrm e}^{10}+x^{2}+1}{x^{2}}\right )\) | \(105\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.46, size = 127, normalized size = 4.70 \begin {gather*} -2 \, x^{3} + x^{2} \log \left (2 \, x^{2} {\left (e^{6} + e\right )} e^{x} + x^{2} e^{2} + {\left (x^{2} {\left (e^{10} + 2 \, e^{5} + 1\right )} + 1\right )} e^{\left (2 \, x\right )}\right ) + 6 \, x - 3 \, \log \left (x^{2} {\left (e^{10} + 2 \, e^{5} + 1\right )} + 1\right ) - 3 \, \log \left (\frac {2 \, x^{2} {\left (e^{6} + e\right )} e^{x} + x^{2} e^{2} + {\left (x^{2} {\left (e^{10} + 2 \, e^{5} + 1\right )} + 1\right )} e^{\left (2 \, x\right )}}{x^{2} {\left (e^{10} + 2 \, e^{5} + 1\right )} + 1}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.88, size = 119, normalized size = 4.41 \begin {gather*} x^2\,\ln \left (2\,x^2\,{\mathrm {e}}^5+x^2\,{\mathrm {e}}^{10}+x^2+2\,x^2\,{\mathrm {e}}^{-x}\,\mathrm {e}+x^2\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^2+2\,x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^6+1\right )-3\,\ln \left (2\,x^2\,{\mathrm {e}}^5+x^2\,{\mathrm {e}}^{10}+x^2+2\,x^2\,{\mathrm {e}}^{-x}\,\mathrm {e}+x^2\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^2+2\,x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^6+1\right )-6\,\ln \relax (x)-3\,\ln \left (\frac {1}{x^2}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 1.07, size = 104, normalized size = 3.85 \begin {gather*} x^{2} \log {\left (x^{2} e^{2 - 2 x} + x^{2} + 2 x^{2} e^{5} + x^{2} e^{10} + \left (2 x^{2} + 2 x^{2} e^{5}\right ) e^{1 - x} + 1 \right )} - 6 \log {\relax (x )} - 3 \log {\left (\left (2 + 2 e^{5}\right ) e^{1 - x} + e^{2 - 2 x} + \frac {x^{2} + 2 x^{2} e^{5} + x^{2} e^{10} + 1}{x^{2}} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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