Optimal. Leaf size=30 \[ e^{\frac {x+\frac {1}{x \log (4)}+\frac {2}{x-\log (2 x)}}{5+x}} \]
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Rubi [F] time = 20.91, 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 {\exp \left (\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{\left (-5 x^2-x^3\right ) \log (4)+\left (5 x+x^2\right ) \log (4) \log (2 x)}\right ) \left (-5 x^2-2 x^3+\left (10 x-8 x^2-4 x^3+5 x^4\right ) \log (4)+\left (10 x+4 x^2+\left (2 x^2-10 x^3\right ) \log (4)\right ) \log (2 x)+\left (-5-2 x+5 x^2 \log (4)\right ) \log ^2(2 x)\right )}{\left (25 x^4+10 x^5+x^6\right ) \log (4)+\left (-50 x^3-20 x^4-2 x^5\right ) \log (4) \log (2 x)+\left (25 x^2+10 x^3+x^4\right ) \log (4) \log ^2(2 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) \left (-5 x^2-2 x^3+\left (10 x-8 x^2-4 x^3+5 x^4\right ) \log (4)+\left (10 x+4 x^2+\left (2 x^2-10 x^3\right ) \log (4)\right ) \log (2 x)+\left (-5-2 x+5 x^2 \log (4)\right ) \log ^2(2 x)\right )}{x^2 (5+x)^2 \log (4) (x-\log (2 x))^2} \, dx\\ &=\frac {\int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) \left (-5 x^2-2 x^3+\left (10 x-8 x^2-4 x^3+5 x^4\right ) \log (4)+\left (10 x+4 x^2+\left (2 x^2-10 x^3\right ) \log (4)\right ) \log (2 x)+\left (-5-2 x+5 x^2 \log (4)\right ) \log ^2(2 x)\right )}{x^2 (5+x)^2 (x-\log (2 x))^2} \, dx}{\log (4)}\\ &=\frac {\int \left (\frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) \left (-5-2 x+5 x^2 \log (4)\right )}{x^2 (5+x)^2}-\frac {2 \exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) (-1+x) \log (4)}{x (5+x) (x-\log (2 x))^2}-\frac {2 \exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) \log (4)}{(5+x)^2 (x-\log (2 x))}\right ) \, dx}{\log (4)}\\ &=-\left (2 \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) (-1+x)}{x (5+x) (x-\log (2 x))^2} \, dx\right )-2 \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{(5+x)^2 (x-\log (2 x))} \, dx+\frac {\int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) \left (-5-2 x+5 x^2 \log (4)\right )}{x^2 (5+x)^2} \, dx}{\log (4)}\\ &=-\left (2 \int \left (-\frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{5 x (x-\log (2 x))^2}+\frac {6 \exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{5 (5+x) (x-\log (2 x))^2}\right ) \, dx\right )-2 \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{(5+x)^2 (x-\log (2 x))} \, dx+\frac {\int \left (-\frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{5 x^2}+\frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right ) (1+25 \log (4))}{5 (5+x)^2}\right ) \, dx}{\log (4)}\\ &=\frac {2}{5} \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{x (x-\log (2 x))^2} \, dx-2 \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{(5+x)^2 (x-\log (2 x))} \, dx-\frac {12}{5} \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{(5+x) (x-\log (2 x))^2} \, dx-\frac {\int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{x^2} \, dx}{5 \log (4)}+\frac {(1+25 \log (4)) \int \frac {\exp \left (-\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{x (5+x) \log (4) (x-\log (2 x))}\right )}{(5+x)^2} \, dx}{5 \log (4)}\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 16.97, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {-x+\left (-2 x-x^3\right ) \log (4)+\left (1+x^2 \log (4)\right ) \log (2 x)}{\left (-5 x^2-x^3\right ) \log (4)+\left (5 x+x^2\right ) \log (4) \log (2 x)}} \left (-5 x^2-2 x^3+\left (10 x-8 x^2-4 x^3+5 x^4\right ) \log (4)+\left (10 x+4 x^2+\left (2 x^2-10 x^3\right ) \log (4)\right ) \log (2 x)+\left (-5-2 x+5 x^2 \log (4)\right ) \log ^2(2 x)\right )}{\left (25 x^4+10 x^5+x^6\right ) \log (4)+\left (-50 x^3-20 x^4-2 x^5\right ) \log (4) \log (2 x)+\left (25 x^2+10 x^3+x^4\right ) \log (4) \log ^2(2 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 2.37, size = 61, normalized size = 2.03 \begin {gather*} e^{\left (-\frac {2 \, {\left (x^{3} + 2 \, x\right )} \log \relax (2) - {\left (2 \, x^{2} \log \relax (2) + 1\right )} \log \left (2 \, x\right ) + x}{2 \, {\left ({\left (x^{2} + 5 \, x\right )} \log \relax (2) \log \left (2 \, x\right ) - {\left (x^{3} + 5 \, x^{2}\right )} \log \relax (2)\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 [A] time = 0.07, size = 57, normalized size = 1.90
method | result | size |
risch | \({\mathrm e}^{\frac {2 \ln \relax (2) \ln \left (2 x \right ) x^{2}-2 x^{3} \ln \relax (2)-4 x \ln \relax (2)+\ln \left (2 x \right )-x}{2 \ln \relax (2) x \left (5+x \right ) \left (\ln \left (2 x \right )-x \right )}}\) | \(57\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.79, size = 197, normalized size = 6.57 \begin {gather*} e^{\left (-\frac {5 \, \log \relax (2)}{x {\left (\log \relax (2) + 5\right )} + {\left (x + 5\right )} \log \relax (x) + 5 \, \log \relax (2) + 25} - \frac {\log \relax (x)}{10 \, {\left ({\left (\log \relax (2)^{2} + 5 \, \log \relax (2)\right )} x + 5 \, \log \relax (2)^{2} + {\left (x \log \relax (2) + 5 \, \log \relax (2)\right )} \log \relax (x) + 25 \, \log \relax (2)\right )}} - \frac {5 \, \log \relax (x)}{x {\left (\log \relax (2) + 5\right )} + {\left (x + 5\right )} \log \relax (x) + 5 \, \log \relax (2) + 25} - \frac {1}{2 \, {\left ({\left (\log \relax (2)^{2} + 5 \, \log \relax (2)\right )} x + 5 \, \log \relax (2)^{2} + {\left (x \log \relax (2) + 5 \, \log \relax (2)\right )} \log \relax (x) + 25 \, \log \relax (2)\right )}} + \frac {2}{x {\left (\log \relax (2) + 5\right )} - \log \relax (2)^{2} + {\left (x - 2 \, \log \relax (2) - 5\right )} \log \relax (x) - \log \relax (x)^{2} - 5 \, \log \relax (2)} - \frac {271}{10 \, {\left (x {\left (\log \relax (2) + 5\right )} + {\left (x + 5\right )} \log \relax (x) + 5 \, \log \relax (2) + 25\right )}} + \frac {1}{10 \, x \log \relax (2)} + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.71, size = 272, normalized size = 9.07 \begin {gather*} 2^{\frac {1}{2\,x^2\,{\ln \relax (2)}^2+10\,x\,{\ln \relax (2)}^2-10\,x^2\,\ln \relax (2)-2\,x^3\,\ln \relax (2)+10\,x\,\ln \relax (2)\,\ln \relax (x)+2\,x^2\,\ln \relax (2)\,\ln \relax (x)}-\frac {x^2+2}{x\,{\ln \relax (2)}^2-5\,x\,\ln \relax (2)-x^2\,\ln \relax (2)+5\,\ln \relax (2)\,\ln \relax (x)+5\,{\ln \relax (2)}^2+x\,\ln \relax (2)\,\ln \relax (x)}}\,x^{\frac {2\,\ln \relax (2)\,x^2+1}{2\,\left (x^2\,{\ln \relax (2)}^2+5\,x\,{\ln \relax (2)}^2-5\,x^2\,\ln \relax (2)-x^3\,\ln \relax (2)+5\,x\,\ln \relax (2)\,\ln \relax (x)+x^2\,\ln \relax (2)\,\ln \relax (x)\right )}}\,{\mathrm {e}}^{-\frac {x}{2\,x^2\,{\ln \relax (2)}^2+10\,x\,{\ln \relax (2)}^2-10\,x^2\,\ln \relax (2)-2\,x^3\,\ln \relax (2)+10\,x\,\ln \relax (2)\,\ln \relax (x)+2\,x^2\,\ln \relax (2)\,\ln \relax (x)}}\,{\mathrm {e}}^{\frac {2\,x^2\,{\ln \relax (2)}^2}{2\,x^2\,{\ln \relax (2)}^2+10\,x\,{\ln \relax (2)}^2-10\,x^2\,\ln \relax (2)-2\,x^3\,\ln \relax (2)+10\,x\,\ln \relax (2)\,\ln \relax (x)+2\,x^2\,\ln \relax (2)\,\ln \relax (x)}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 2.17, size = 63, normalized size = 2.10 \begin {gather*} e^{\frac {- x + \left (- 2 x^{3} - 4 x\right ) \log {\relax (2 )} + \left (2 x^{2} \log {\relax (2 )} + 1\right ) \log {\left (2 x \right )}}{\left (2 x^{2} + 10 x\right ) \log {\relax (2 )} \log {\left (2 x \right )} + \left (- 2 x^{3} - 10 x^{2}\right ) \log {\relax (2 )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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