Integrand size = 77, antiderivative size = 25 \[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=x+e^{\frac {e^{-4 x^2} x}{\log (x)}} (4-5 (7+x)) \]
Time = 5.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=e^{\frac {e^{-4 x^2} x}{\log (x)}} (-31-5 x)+x \]
Integrate[(E^(4*x^2)*Log[x]^2 + E^(x/(E^(4*x^2)*Log[x]))*(31 + 5*x + (-31 - 5*x + 248*x^2 + 40*x^3)*Log[x] - 5*E^(4*x^2)*Log[x]^2))/(E^(4*x^2)*Log[x ]^2),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (-5 e^{4 x^2} \log ^2(x)+\left (40 x^3+248 x^2-5 x-31\right ) \log (x)+5 x+31\right )\right )}{\log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(5 x+31) e^{\frac {e^{-4 x^2} x}{\log (x)}-4 x^2} \left (8 x^2 \log (x)-\log (x)+1\right )}{\log ^2(x)}-5 e^{\frac {e^{-4 x^2} x}{\log (x)}}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 31 \int \frac {e^{\frac {e^{-4 x^2} x}{\log (x)}-4 x^2}}{\log ^2(x)}dx+5 \int \frac {e^{\frac {e^{-4 x^2} x}{\log (x)}-4 x^2} x}{\log ^2(x)}dx-5 \int e^{\frac {e^{-4 x^2} x}{\log (x)}}dx-31 \int \frac {e^{\frac {e^{-4 x^2} x}{\log (x)}-4 x^2}}{\log (x)}dx-5 \int \frac {e^{\frac {e^{-4 x^2} x}{\log (x)}-4 x^2} x}{\log (x)}dx+248 \int \frac {e^{\frac {e^{-4 x^2} x}{\log (x)}-4 x^2} x^2}{\log (x)}dx+40 \int \frac {e^{\frac {e^{-4 x^2} x}{\log (x)}-4 x^2} x^3}{\log (x)}dx+x\) |
Int[(E^(4*x^2)*Log[x]^2 + E^(x/(E^(4*x^2)*Log[x]))*(31 + 5*x + (-31 - 5*x + 248*x^2 + 40*x^3)*Log[x] - 5*E^(4*x^2)*Log[x]^2))/(E^(4*x^2)*Log[x]^2),x ]
3.18.85.3.1 Defintions of rubi rules used
Time = 0.53 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
risch | \(x +{\mathrm e}^{\frac {x \,{\mathrm e}^{-4 x^{2}}}{\ln \left (x \right )}} \left (-31-5 x \right )\) | \(22\) |
parallelrisch | \(-5 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-4 x^{2}}}{\ln \left (x \right )}} x -31 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-4 x^{2}}}{\ln \left (x \right )}}+x\) | \(38\) |
int(((-5*exp(4*x^2)*ln(x)^2+(40*x^3+248*x^2-5*x-31)*ln(x)+5*x+31)*exp(x/ex p(4*x^2)/ln(x))+exp(4*x^2)*ln(x)^2)/exp(4*x^2)/ln(x)^2,x,method=_RETURNVER BOSE)
Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=-{\left (5 \, x + 31\right )} e^{\left (\frac {x e^{\left (-4 \, x^{2}\right )}}{\log \left (x\right )}\right )} + x \]
integrate(((-5*exp(4*x^2)*log(x)^2+(40*x^3+248*x^2-5*x-31)*log(x)+5*x+31)* exp(x/exp(4*x^2)/log(x))+exp(4*x^2)*log(x)^2)/exp(4*x^2)/log(x)^2,x, algor ithm=\
Time = 57.98 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=x + \left (- 5 x - 31\right ) e^{\frac {x e^{- 4 x^{2}}}{\log {\left (x \right )}}} \]
integrate(((-5*exp(4*x**2)*ln(x)**2+(40*x**3+248*x**2-5*x-31)*ln(x)+5*x+31 )*exp(x/exp(4*x**2)/ln(x))+exp(4*x**2)*ln(x)**2)/exp(4*x**2)/ln(x)**2,x)
\[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=\int { \frac {{\left (e^{\left (4 \, x^{2}\right )} \log \left (x\right )^{2} - {\left (5 \, e^{\left (4 \, x^{2}\right )} \log \left (x\right )^{2} - {\left (40 \, x^{3} + 248 \, x^{2} - 5 \, x - 31\right )} \log \left (x\right ) - 5 \, x - 31\right )} e^{\left (\frac {x e^{\left (-4 \, x^{2}\right )}}{\log \left (x\right )}\right )}\right )} e^{\left (-4 \, x^{2}\right )}}{\log \left (x\right )^{2}} \,d x } \]
integrate(((-5*exp(4*x^2)*log(x)^2+(40*x^3+248*x^2-5*x-31)*log(x)+5*x+31)* exp(x/exp(4*x^2)/log(x))+exp(4*x^2)*log(x)^2)/exp(4*x^2)/log(x)^2,x, algor ithm=\
x - integrate((5*e^(4*x^2)*log(x)^2 - (40*x^3 + 248*x^2 - 5*x - 31)*log(x) - 5*x - 31)*e^(-4*x^2 + x*e^(-4*x^2)/log(x))/log(x)^2, x)
\[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=\int { \frac {{\left (e^{\left (4 \, x^{2}\right )} \log \left (x\right )^{2} - {\left (5 \, e^{\left (4 \, x^{2}\right )} \log \left (x\right )^{2} - {\left (40 \, x^{3} + 248 \, x^{2} - 5 \, x - 31\right )} \log \left (x\right ) - 5 \, x - 31\right )} e^{\left (\frac {x e^{\left (-4 \, x^{2}\right )}}{\log \left (x\right )}\right )}\right )} e^{\left (-4 \, x^{2}\right )}}{\log \left (x\right )^{2}} \,d x } \]
integrate(((-5*exp(4*x^2)*log(x)^2+(40*x^3+248*x^2-5*x-31)*log(x)+5*x+31)* exp(x/exp(4*x^2)/log(x))+exp(4*x^2)*log(x)^2)/exp(4*x^2)/log(x)^2,x, algor ithm=\
integrate((e^(4*x^2)*log(x)^2 - (5*e^(4*x^2)*log(x)^2 - (40*x^3 + 248*x^2 - 5*x - 31)*log(x) - 5*x - 31)*e^(x*e^(-4*x^2)/log(x)))*e^(-4*x^2)/log(x)^ 2, x)
Timed out. \[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=\int \frac {{\mathrm {e}}^{-4\,x^2}\,\left ({\mathrm {e}}^{4\,x^2}\,{\ln \left (x\right )}^2+{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-4\,x^2}}{\ln \left (x\right )}}\,\left (-5\,{\mathrm {e}}^{4\,x^2}\,{\ln \left (x\right )}^2+\left (40\,x^3+248\,x^2-5\,x-31\right )\,\ln \left (x\right )+5\,x+31\right )\right )}{{\ln \left (x\right )}^2} \,d x \]
int((exp(-4*x^2)*(exp(4*x^2)*log(x)^2 + exp((x*exp(-4*x^2))/log(x))*(5*x - 5*exp(4*x^2)*log(x)^2 - log(x)*(5*x - 248*x^2 - 40*x^3 + 31) + 31)))/log( x)^2,x)