Optimal. Leaf size=26 \[ \frac {-1+2 x}{\log \left (\left (2-x+\frac {x}{e^5}\right )^2-\log (2)\right )} \]
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Rubi [F] time = 7.18, 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 {-2 x+4 x^2+e^5 \left (-4+12 x-8 x^2\right )+e^{10} \left (4-10 x+4 x^2\right )+\left (-2 x^2+e^{10} \left (-8+8 x-2 x^2\right )+e^5 \left (-8 x+4 x^2\right )+2 e^{10} \log (2)\right ) \log \left (\frac {x^2+e^5 \left (4 x-2 x^2\right )+e^{10} \left (4-4 x+x^2\right )-e^{10} \log (2)}{e^{10}}\right )}{\left (-x^2+e^{10} \left (-4+4 x-x^2\right )+e^5 \left (-4 x+2 x^2\right )+e^{10} \log (2)\right ) \log ^2\left (\frac {x^2+e^5 \left (4 x-2 x^2\right )+e^{10} \left (4-4 x+x^2\right )-e^{10} \log (2)}{e^{10}}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x-4 x^2-e^5 \left (-4+12 x-8 x^2\right )-e^{10} \left (4-10 x+4 x^2\right )-\left (-2 x^2+e^{10} \left (-8+8 x-2 x^2\right )+e^5 \left (-8 x+4 x^2\right )+2 e^{10} \log (2)\right ) \log \left (\frac {x^2+e^5 \left (4 x-2 x^2\right )+e^{10} \left (4-4 x+x^2\right )-e^{10} \log (2)}{e^{10}}\right )}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (\frac {4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))}{e^{10}}\right )} \, dx\\ &=\int \left (\frac {4 x^2}{\left (-4 e^5 \left (1-e^5\right ) x-\left (1-e^5\right )^2 x^2-e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {4 e^5 (1-2 x) (1-x)}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {2 e^{10} (1-2 x) (-2+x)}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {2 x}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {2}{\log \left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}\right ) \, dx\\ &=2 \int \frac {x}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx+2 \int \frac {1}{\log \left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx+4 \int \frac {x^2}{\left (-4 e^5 \left (1-e^5\right ) x-\left (1-e^5\right )^2 x^2-e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx+\left (4 e^5\right ) \int \frac {(1-2 x) (1-x)}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx+\left (2 e^{10}\right ) \int \frac {(1-2 x) (-2+x)}{\left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx\\ &=2 \int \left (\frac {1+\frac {2}{\sqrt {\log (2)}}}{\left (4 e^5 \left (1-e^5\right )+2 \left (1-e^5\right )^2 x-2 e^5 \left (-1+e^5\right ) \sqrt {\log (2)}\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {1-\frac {2}{\sqrt {\log (2)}}}{\left (4 e^5 \left (1-e^5\right )+2 \left (1-e^5\right )^2 x+2 e^5 \left (-1+e^5\right ) \sqrt {\log (2)}\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}\right ) \, dx+2 \int \frac {1}{\log \left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \, dx+4 \int \left (-\frac {1}{\left (1-e^5\right )^2 \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {e^5 \left (4 \left (1-e^5\right ) x+e^5 (4-\log (2))\right )}{\left (1-e^5\right )^2 \left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}\right ) \, dx+\left (4 e^5\right ) \int \left (\frac {2}{\left (1-e^5\right )^2 \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {1-2 e^5-\left (3+2 e^5-5 e^{10}\right ) x-e^{10} (7-\log (4))}{\left (1-e^5\right )^2 \left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}\right ) \, dx+\left (2 e^{10}\right ) \int \left (-\frac {2}{\left (1-e^5\right )^2 \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}+\frac {-2+4 e^5+\left (5-2 e^5-3 e^{10}\right ) x+e^{10} (6-\log (4))}{\left (1-e^5\right )^2 \left (4 e^5 \left (1-e^5\right ) x+\left (1-e^5\right )^2 x^2+e^{10} (4-\log (2))\right ) \log ^2\left (4-4 \left (1-\frac {1}{e^5}\right ) x+\frac {\left (1-e^5\right )^2 x^2}{e^{10}}-\log (2)\right )}\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 38, normalized size = 1.46 \begin {gather*} \frac {-1+2 x}{\log \left (4+\left (-4+\frac {4}{e^5}\right ) x+\frac {\left (-1+e^5\right )^2 x^2}{e^{10}}-\log (2)\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.13, size = 44, normalized size = 1.69 \begin {gather*} \frac {2 \, x - 1}{\log \left ({\left (x^{2} + {\left (x^{2} - 4 \, x + 4\right )} e^{10} - 2 \, {\left (x^{2} - 2 \, x\right )} e^{5} - e^{10} \log \relax (2)\right )} e^{\left (-10\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.80, size = 51, normalized size = 1.96 \begin {gather*} \frac {2 \, x - 1}{\log \left (x^{2} e^{15} - 2 \, x^{2} e^{10} + x^{2} e^{5} - 4 \, x e^{15} + 4 \, x e^{10} - e^{15} \log \relax (2) + 4 \, e^{15}\right ) - 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.46, size = 46, normalized size = 1.77
method | result | size |
risch | \(\frac {2 x -1}{\ln \left (\left (-{\mathrm e}^{10} \ln \relax (2)+\left (x^{2}-4 x +4\right ) {\mathrm e}^{10}+\left (-2 x^{2}+4 x \right ) {\mathrm e}^{5}+x^{2}\right ) {\mathrm e}^{-10}\right )}\) | \(46\) |
norman | \(\frac {2 x -1}{\ln \left (\left (-{\mathrm e}^{10} \ln \relax (2)+\left (x^{2}-4 x +4\right ) {\mathrm e}^{10}+\left (-2 x^{2}+4 x \right ) {\mathrm e}^{5}+x^{2}\right ) {\mathrm e}^{-10}\right )}\) | \(52\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.76, size = 44, normalized size = 1.69 \begin {gather*} \frac {2 \, x - 1}{\log \left (x^{2} {\left (e^{10} - 2 \, e^{5} + 1\right )} - 4 \, x {\left (e^{10} - e^{5}\right )} - e^{10} \log \relax (2) + 4 \, e^{10}\right ) - 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.35, size = 101, normalized size = 3.88 \begin {gather*} \frac {2\,x\,{\mathrm {e}}^5-{\mathrm {e}}^5-2\,x+2\,{\mathrm {e}}^5\,\ln \left ({\mathrm {e}}^{-10}\,\left ({\mathrm {e}}^5\,\left (4\,x-2\,x^2\right )-{\mathrm {e}}^{10}\,\ln \relax (2)+{\mathrm {e}}^{10}\,\left (x^2-4\,x+4\right )+x^2\right )\right )+1}{\ln \left ({\mathrm {e}}^{-10}\,\left ({\mathrm {e}}^5\,\left (4\,x-2\,x^2\right )-{\mathrm {e}}^{10}\,\ln \relax (2)+{\mathrm {e}}^{10}\,\left (x^2-4\,x+4\right )+x^2\right )\right )\,\left ({\mathrm {e}}^5-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.40, size = 42, normalized size = 1.62 \begin {gather*} \frac {2 x - 1}{\log {\left (\frac {x^{2} + \left (- 2 x^{2} + 4 x\right ) e^{5} + \left (x^{2} - 4 x + 4\right ) e^{10} - e^{10} \log {\relax (2 )}}{e^{10}} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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