Optimal. Leaf size=32 \[ x (x-\log (x)) \left (-1-e^x-e^x \log \left (\frac {x^2}{5 \log (x)}\right )\right ) \]
________________________________________________________________________________________
Rubi [B] time = 1.46, antiderivative size = 70, normalized size of antiderivative = 2.19, number of steps used = 4, number of rules used = 3, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6742, 2295, 2288} \begin {gather*} -x^2-\frac {e^x \left (-x \log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )+x^2 \log (x)+x^2 \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-x \log ^2(x)\right )}{\log (x)}+x \log (x) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 2288
Rule 2295
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-2 x+\log (x)+\frac {e^x \left (x-4 x \log (x)-x^2 \log (x)+3 \log ^2(x)+x \log ^2(x)+\log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-2 x \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-x^2 \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )+\log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )+x \log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )\right )}{\log (x)}\right ) \, dx\\ &=x-x^2+\int \log (x) \, dx+\int \frac {e^x \left (x-4 x \log (x)-x^2 \log (x)+3 \log ^2(x)+x \log ^2(x)+\log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-2 x \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-x^2 \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )+\log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )+x \log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )\right )}{\log (x)} \, dx\\ &=-x^2+x \log (x)-\frac {e^x \left (x^2 \log (x)-x \log ^2(x)+x^2 \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-x \log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )\right )}{\log (x)}\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.22, size = 30, normalized size = 0.94 \begin {gather*} -x (x-\log (x)) \left (1+e^x+e^x \log \left (\frac {x^2}{5 \log (x)}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.54, size = 48, normalized size = 1.50 \begin {gather*} -x^{2} e^{x} - x^{2} + {\left (x e^{x} + x\right )} \log \relax (x) - {\left (x^{2} e^{x} - x e^{x} \log \relax (x)\right )} \log \left (\frac {x^{2}}{5 \, \log \relax (x)}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [B] time = 0.17, size = 64, normalized size = 2.00 \begin {gather*} -2 \, x^{2} e^{x} \log \relax (x) + 2 \, x e^{x} \log \relax (x)^{2} + x^{2} e^{x} \log \left (5 \, \log \relax (x)\right ) - x e^{x} \log \relax (x) \log \left (5 \, \log \relax (x)\right ) - x^{2} e^{x} + x e^{x} \log \relax (x) - x^{2} + x \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [C] time = 0.23, size = 437, normalized size = 13.66
method | result | size |
risch | \(x^{2} \ln \relax (5) {\mathrm e}^{x}+2 x \,{\mathrm e}^{x} \ln \relax (x )^{2}+x \,{\mathrm e}^{x} \ln \relax (x )-2 x^{2} {\mathrm e}^{x} \ln \relax (x )-x^{2}-{\mathrm e}^{x} x^{2}+x \ln \relax (x )-x \ln \relax (5) {\mathrm e}^{x} \ln \relax (x )-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}^{x} \ln \relax (x )}{2}+\frac {i \pi x \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x} \ln \relax (x )}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}^{x}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}^{x}}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}}{2}+\left ({\mathrm e}^{x} x^{2}-x \,{\mathrm e}^{x} \ln \relax (x )\right ) \ln \left (\ln \relax (x )\right )+\frac {i \pi x \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}^{x} \ln \relax (x )}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{3} {\mathrm e}^{x} \ln \relax (x )}{2}+i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x} \ln \relax (x )+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}}{2}-\frac {i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x} \ln \relax (x )}{2}-\frac {i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x} \ln \relax (x )}{2}-i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{3} {\mathrm e}^{x}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}}{2}\) | \(437\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 0.48, size = 81, normalized size = 2.53 \begin {gather*} {\left (x^{2} - x \log \relax (x)\right )} e^{x} \log \left (\log \relax (x)\right ) - x^{2} + {\left (x^{2} \log \relax (5) + 2 \, x \log \relax (x)^{2} - {\left (2 \, x^{2} + x {\left (\log \relax (5) - 1\right )}\right )} \log \relax (x) + 2 \, x - 2\right )} e^{x} - {\left (x^{2} - 2 \, x + 2\right )} e^{x} - 4 \, {\left (x - 1\right )} e^{x} + x \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 5.97, size = 57, normalized size = 1.78 \begin {gather*} \ln \left (\frac {x^2}{5\,\ln \relax (x)}\right )\,\left (\frac {{\mathrm {e}}^x\,\left (x-x^3\right )}{x}-{\mathrm {e}}^x+x\,{\mathrm {e}}^x\,\ln \relax (x)\right )-x^2\,{\mathrm {e}}^x+\ln \relax (x)\,\left (x+x\,{\mathrm {e}}^x\right )-x^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 6.62, size = 48, normalized size = 1.50 \begin {gather*} - x^{2} + x \log {\relax (x )} + \left (- x^{2} \log {\left (\frac {x^{2}}{5 \log {\relax (x )}} \right )} - x^{2} + x \log {\relax (x )} \log {\left (\frac {x^{2}}{5 \log {\relax (x )}} \right )} + x \log {\relax (x )}\right ) e^{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________