Integrand size = 127, antiderivative size = 24 \[ \int \frac {e^{x^2} \left (x+e^{-2+2 x} x+x^2+\left (x+2 x^3+2 x^4+e^{-2+2 x} \left (2 x+2 x^3\right )\right ) \log (x)+\left (1+e^{-2+2 x}+x+\left (-1-x+2 x^2+2 x^3+e^{-2+2 x} \left (-1+2 x^2\right )\right ) \log (x)\right ) \log \left (1+e^{-2+2 x}+x\right )\right )}{x^2+e^{-2+2 x} x^2+x^3} \, dx=\frac {e^{x^2} \log (x) \left (x+\log \left (1+e^{-2+2 x}+x\right )\right )}{x} \]
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x^2} \left (x+e^{-2+2 x} x+x^2+\left (x+2 x^3+2 x^4+e^{-2+2 x} \left (2 x+2 x^3\right )\right ) \log (x)+\left (1+e^{-2+2 x}+x+\left (-1-x+2 x^2+2 x^3+e^{-2+2 x} \left (-1+2 x^2\right )\right ) \log (x)\right ) \log \left (1+e^{-2+2 x}+x\right )\right )}{x^2+e^{-2+2 x} x^2+x^3} \, dx=\frac {e^{x^2} \log (x) \left (x+\log \left (1+e^{-2+2 x}+x\right )\right )}{x} \]
Integrate[(E^x^2*(x + E^(-2 + 2*x)*x + x^2 + (x + 2*x^3 + 2*x^4 + E^(-2 + 2*x)*(2*x + 2*x^3))*Log[x] + (1 + E^(-2 + 2*x) + x + (-1 - x + 2*x^2 + 2*x ^3 + E^(-2 + 2*x)*(-1 + 2*x^2))*Log[x])*Log[1 + E^(-2 + 2*x) + x]))/(x^2 + E^(-2 + 2*x)*x^2 + x^3),x]
Time = 10.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {e^{x^2} \left (x^2+\left (2 x^4+2 x^3+e^{2 x-2} \left (2 x^3+2 x\right )+x\right ) \log (x)+\left (\left (2 x^3+2 x^2+e^{2 x-2} \left (2 x^2-1\right )-x-1\right ) \log (x)+x+e^{2 x-2}+1\right ) \log \left (x+e^{2 x-2}+1\right )+e^{2 x-2} x+x\right )}{x^3+e^{2 x-2} x^2+x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x^2} \left (2 x^3 \log (x)+2 x^2 \log (x) \log \left (x+e^{2 x-2}+1\right )+x+2 x \log (x)-\log (x) \log \left (x+e^{2 x-2}+1\right )+\log \left (x+e^{2 x-2}+1\right )\right )}{x^2}-\frac {e^{x^2+2} (2 x+1) \log (x)}{x \left (e^2 x+e^{2 x}+e^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^{x^2} \log (x)+\frac {e^{x^2} \log \left (x+e^{2 x-2}+1\right ) \log (x)}{x}\) |
Int[(E^x^2*(x + E^(-2 + 2*x)*x + x^2 + (x + 2*x^3 + 2*x^4 + E^(-2 + 2*x)*( 2*x + 2*x^3))*Log[x] + (1 + E^(-2 + 2*x) + x + (-1 - x + 2*x^2 + 2*x^3 + E ^(-2 + 2*x)*(-1 + 2*x^2))*Log[x])*Log[1 + E^(-2 + 2*x) + x]))/(x^2 + E^(-2 + 2*x)*x^2 + x^3),x]
3.10.95.3.1 Defintions of rubi rules used
Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
\[\frac {\left (\ln \left ({\mathrm e}^{-2+2 x}+x +1\right )+x \right ) {\mathrm e}^{x^{2}} \ln \left (x \right )}{x}\]
int(((((2*x^2-1)*exp(-1+x)^2+2*x^3+2*x^2-x-1)*ln(x)+exp(-1+x)^2+x+1)*ln(ex p(-1+x)^2+x+1)+((2*x^3+2*x)*exp(-1+x)^2+2*x^4+2*x^3+x)*ln(x)+x*exp(-1+x)^2 +x^2+x)*exp(ln(ln(x))+x^2)/(x^2*exp(-1+x)^2+x^3+x^2)/ln(x),x)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x^2} \left (x+e^{-2+2 x} x+x^2+\left (x+2 x^3+2 x^4+e^{-2+2 x} \left (2 x+2 x^3\right )\right ) \log (x)+\left (1+e^{-2+2 x}+x+\left (-1-x+2 x^2+2 x^3+e^{-2+2 x} \left (-1+2 x^2\right )\right ) \log (x)\right ) \log \left (1+e^{-2+2 x}+x\right )\right )}{x^2+e^{-2+2 x} x^2+x^3} \, dx=\frac {{\left (x + \log \left (x + e^{\left (2 \, x - 2\right )} + 1\right )\right )} e^{\left (x^{2} + \log \left (\log \left (x\right )\right )\right )}}{x} \]
integrate(((((2*x^2-1)*exp(-1+x)^2+2*x^3+2*x^2-x-1)*log(x)+exp(-1+x)^2+x+1 )*log(exp(-1+x)^2+x+1)+((2*x^3+2*x)*exp(-1+x)^2+2*x^4+2*x^3+x)*log(x)+x*ex p(-1+x)^2+x^2+x)*exp(log(log(x))+x^2)/(x^2*exp(-1+x)^2+x^3+x^2)/log(x),x, algorithm=\
Timed out. \[ \int \frac {e^{x^2} \left (x+e^{-2+2 x} x+x^2+\left (x+2 x^3+2 x^4+e^{-2+2 x} \left (2 x+2 x^3\right )\right ) \log (x)+\left (1+e^{-2+2 x}+x+\left (-1-x+2 x^2+2 x^3+e^{-2+2 x} \left (-1+2 x^2\right )\right ) \log (x)\right ) \log \left (1+e^{-2+2 x}+x\right )\right )}{x^2+e^{-2+2 x} x^2+x^3} \, dx=\text {Timed out} \]
integrate(((((2*x**2-1)*exp(-1+x)**2+2*x**3+2*x**2-x-1)*ln(x)+exp(-1+x)**2 +x+1)*ln(exp(-1+x)**2+x+1)+((2*x**3+2*x)*exp(-1+x)**2+2*x**4+2*x**3+x)*ln( x)+x*exp(-1+x)**2+x**2+x)*exp(ln(ln(x))+x**2)/(x**2*exp(-1+x)**2+x**3+x**2 )/ln(x),x)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{x^2} \left (x+e^{-2+2 x} x+x^2+\left (x+2 x^3+2 x^4+e^{-2+2 x} \left (2 x+2 x^3\right )\right ) \log (x)+\left (1+e^{-2+2 x}+x+\left (-1-x+2 x^2+2 x^3+e^{-2+2 x} \left (-1+2 x^2\right )\right ) \log (x)\right ) \log \left (1+e^{-2+2 x}+x\right )\right )}{x^2+e^{-2+2 x} x^2+x^3} \, dx=\frac {{\left (x - 2\right )} e^{\left (x^{2}\right )} \log \left (x\right ) + e^{\left (x^{2}\right )} \log \left (x e^{2} + e^{2} + e^{\left (2 \, x\right )}\right ) \log \left (x\right )}{x} \]
integrate(((((2*x^2-1)*exp(-1+x)^2+2*x^3+2*x^2-x-1)*log(x)+exp(-1+x)^2+x+1 )*log(exp(-1+x)^2+x+1)+((2*x^3+2*x)*exp(-1+x)^2+2*x^4+2*x^3+x)*log(x)+x*ex p(-1+x)^2+x^2+x)*exp(log(log(x))+x^2)/(x^2*exp(-1+x)^2+x^3+x^2)/log(x),x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {e^{x^2} \left (x+e^{-2+2 x} x+x^2+\left (x+2 x^3+2 x^4+e^{-2+2 x} \left (2 x+2 x^3\right )\right ) \log (x)+\left (1+e^{-2+2 x}+x+\left (-1-x+2 x^2+2 x^3+e^{-2+2 x} \left (-1+2 x^2\right )\right ) \log (x)\right ) \log \left (1+e^{-2+2 x}+x\right )\right )}{x^2+e^{-2+2 x} x^2+x^3} \, dx=\frac {x e^{\left (x^{2}\right )} \log \left (x\right ) + e^{\left (x^{2}\right )} \log \left (x e^{2} + e^{2} + e^{\left (2 \, x\right )}\right ) \log \left (x\right ) - 2 \, e^{\left (x^{2}\right )} \log \left (x\right )}{x} \]
integrate(((((2*x^2-1)*exp(-1+x)^2+2*x^3+2*x^2-x-1)*log(x)+exp(-1+x)^2+x+1 )*log(exp(-1+x)^2+x+1)+((2*x^3+2*x)*exp(-1+x)^2+2*x^4+2*x^3+x)*log(x)+x*ex p(-1+x)^2+x^2+x)*exp(log(log(x))+x^2)/(x^2*exp(-1+x)^2+x^3+x^2)/log(x),x, algorithm=\
Timed out. \[ \int \frac {e^{x^2} \left (x+e^{-2+2 x} x+x^2+\left (x+2 x^3+2 x^4+e^{-2+2 x} \left (2 x+2 x^3\right )\right ) \log (x)+\left (1+e^{-2+2 x}+x+\left (-1-x+2 x^2+2 x^3+e^{-2+2 x} \left (-1+2 x^2\right )\right ) \log (x)\right ) \log \left (1+e^{-2+2 x}+x\right )\right )}{x^2+e^{-2+2 x} x^2+x^3} \, dx=\int \frac {{\mathrm {e}}^{\ln \left (\ln \left (x\right )\right )+x^2}\,\left (x+\ln \left (x+{\mathrm {e}}^{2\,x-2}+1\right )\,\left (x+{\mathrm {e}}^{2\,x-2}+\ln \left (x\right )\,\left ({\mathrm {e}}^{2\,x-2}\,\left (2\,x^2-1\right )-x+2\,x^2+2\,x^3-1\right )+1\right )+x\,{\mathrm {e}}^{2\,x-2}+x^2+\ln \left (x\right )\,\left (x+{\mathrm {e}}^{2\,x-2}\,\left (2\,x^3+2\,x\right )+2\,x^3+2\,x^4\right )\right )}{\ln \left (x\right )\,\left (x^2\,{\mathrm {e}}^{2\,x-2}+x^2+x^3\right )} \,d x \]
int((exp(log(log(x)) + x^2)*(x + log(x + exp(2*x - 2) + 1)*(x + exp(2*x - 2) + log(x)*(exp(2*x - 2)*(2*x^2 - 1) - x + 2*x^2 + 2*x^3 - 1) + 1) + x*ex p(2*x - 2) + x^2 + log(x)*(x + exp(2*x - 2)*(2*x + 2*x^3) + 2*x^3 + 2*x^4) ))/(log(x)*(x^2*exp(2*x - 2) + x^2 + x^3)),x)