Optimal. Leaf size=27 \[ \frac {1}{5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)} \]
________________________________________________________________________________________
Rubi [F] time = 26.50, 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+e^{-2+x+\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left ((2-x) \log (2)+e^{5/x} \left (5 x-x^3\right ) \log (2)\right )}{25 x^3+10 x^4+x^5+e^{\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left (10 x^3+2 x^4\right ) \log (2)+e^{\frac {2 e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} x^3 \log ^2(2)} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x^3+e^{-2+x+\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left ((2-x) \log (2)+e^{5/x} \left (5 x-x^3\right ) \log (2)\right )}{x^3 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx\\ &=\int \left (-\frac {1}{\left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2}-\frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+x} \left (-2+x-5 e^{5/x} x+e^{5/x} x^3\right ) \log (2)}{x^3 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2}\right ) \, dx\\ &=-\left (\log (2) \int \frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+x} \left (-2+x-5 e^{5/x} x+e^{5/x} x^3\right )}{x^3 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx\right )-\int \frac {1}{\left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx\\ &=-\left (\log (2) \int \left (\frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+x} (-2+x)}{x^3 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2}+\frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+\frac {5}{x}+x} \left (-5+x^2\right )}{x^2 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2}\right ) \, dx\right )-\int \frac {1}{\left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx\\ &=-\left (\log (2) \int \frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+x} (-2+x)}{x^3 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx\right )-\log (2) \int \frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+\frac {5}{x}+x} \left (-5+x^2\right )}{x^2 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx-\int \frac {1}{\left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx\\ &=-\left (\log (2) \int \left (-\frac {2 e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+x}}{x^3 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2}+\frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+x}}{x^2 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2}\right ) \, dx\right )-\log (2) \int \left (\frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+\frac {5}{x}+x}}{\left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2}-\frac {5 e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+\frac {5}{x}+x}}{x^2 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2}\right ) \, dx-\int \frac {1}{\left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx\\ &=-\left (\log (2) \int \frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+\frac {5}{x}+x}}{\left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx\right )-\log (2) \int \frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+x}}{x^2 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx+(2 \log (2)) \int \frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+x}}{x^3 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx+(5 \log (2)) \int \frac {e^{-2+e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )+\frac {5}{x}+x}}{x^2 \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx-\int \frac {1}{\left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )^2} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [B] time = 0.58, size = 137, normalized size = 5.07 \begin {gather*} \frac {-e^4 x^3 \log (2)+e^{2+x} (5+x) (x \log (2)-\log (4))+e^{2+\frac {5}{x}+x} x (5+x) \left (x^2 \log (2)-\log (32)\right )}{e^2 \left (-e^2 x^3+e^x \left (-10+3 x+x^2\right )+e^{\frac {5}{x}+x} x \left (-25-5 x+5 x^2+x^3\right )\right ) \log (2) \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.53, size = 50, normalized size = 1.85 \begin {gather*} \frac {e^{\left (x - 2\right )}}{{\left (x + 5\right )} e^{\left (x - 2\right )} + e^{\left (\frac {x^{3} - 2 \, x^{2} + {\left (x^{2} e^{\frac {5}{x}} + 1\right )} e^{\left (x - 2\right )}}{x^{2}}\right )} \log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [B] time = 7.41, size = 564, normalized size = 20.89 \begin {gather*} \frac {x^{4} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} - x^{3} e^{2} + 5 \, x^{3} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + x^{2} e^{x} - 5 \, x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + 3 \, x e^{x} - 25 \, x e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} - 10 \, e^{x}}{x^{5} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + x^{4} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \relax (2) - x^{4} e^{2} + 10 \, x^{4} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + 5 \, x^{3} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \relax (2) - x^{3} e^{\left (\frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \relax (2) - 5 \, x^{3} e^{2} + x^{3} e^{x} + 20 \, x^{3} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + x^{2} e^{\left (x + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}}\right )} \log \relax (2) - 5 \, x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \relax (2) + 8 \, x^{2} e^{x} - 50 \, x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + 3 \, x e^{\left (x + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}}\right )} \log \relax (2) - 25 \, x e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \relax (2) + 5 \, x e^{x} - 125 \, x e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} - 10 \, e^{\left (x + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}}\right )} \log \relax (2) - 50 \, e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.21, size = 30, normalized size = 1.11
method | result | size |
risch | \(\frac {1}{\ln \relax (2) {\mathrm e}^{\frac {\left (x^{2} {\mathrm e}^{\frac {5}{x}}+1\right ) {\mathrm e}^{x -2}}{x^{2}}}+x +5}\) | \(30\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.94, size = 27, normalized size = 1.00 \begin {gather*} \frac {1}{e^{\left (\frac {e^{\left (x - 2\right )}}{x^{2}} + e^{\left (x + \frac {5}{x} - 2\right )}\right )} \log \relax (2) + x + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 0.71, size = 93, normalized size = 3.44 \begin {gather*} \frac {x\,\left (\ln \relax (2)-{\mathrm {e}}^{5/x}\,\ln \left (32\right )\right )-\ln \relax (4)+x^3\,{\mathrm {e}}^{5/x}\,\ln \relax (2)}{{\ln \relax (2)}^2\,\left ({\mathrm {e}}^{{\mathrm {e}}^{-2}\,{\mathrm {e}}^{5/x}\,{\mathrm {e}}^x+\frac {{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{x^2}}+\frac {x+5}{\ln \relax (2)}\right )\,\left (x-5\,x\,{\mathrm {e}}^{5/x}+x^3\,{\mathrm {e}}^{5/x}-2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.72, size = 27, normalized size = 1.00 \begin {gather*} \frac {1}{x + e^{\frac {\left (x^{2} e^{\frac {5}{x}} + 1\right ) e^{x - 2}}{x^{2}}} \log {\relax (2 )} + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________