Integrand size = 111, antiderivative size = 22 \[ \int \frac {1+x+e^x \left (1+2 x-x^2\right )+e^{x^2} \left (-2 x^2+e^x \left (x-2 x^2\right )\right )-e^x x \log (x)}{x+e^{2 x^2} x-2 x^2+x^3+e^{x^2} \left (2 x-2 x^2\right )+\left (-2 x-2 e^{x^2} x+2 x^2\right ) \log (x)+x \log ^2(x)} \, dx=\frac {1+e^x}{1+e^{x^2}-x-\log (x)} \]
Time = 0.64 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1+x+e^x \left (1+2 x-x^2\right )+e^{x^2} \left (-2 x^2+e^x \left (x-2 x^2\right )\right )-e^x x \log (x)}{x+e^{2 x^2} x-2 x^2+x^3+e^{x^2} \left (2 x-2 x^2\right )+\left (-2 x-2 e^{x^2} x+2 x^2\right ) \log (x)+x \log ^2(x)} \, dx=\frac {1+e^x}{1+e^{x^2}-x-\log (x)} \]
Integrate[(1 + x + E^x*(1 + 2*x - x^2) + E^x^2*(-2*x^2 + E^x*(x - 2*x^2)) - E^x*x*Log[x])/(x + E^(2*x^2)*x - 2*x^2 + x^3 + E^x^2*(2*x - 2*x^2) + (-2 *x - 2*E^x^2*x + 2*x^2)*Log[x] + x*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.
\(\displaystyle \int \frac {e^x \left (-x^2+2 x+1\right )+e^{x^2} \left (e^x \left (x-2 x^2\right )-2 x^2\right )+x-e^x x \log (x)+1}{x^3-2 x^2+e^{2 x^2} x+e^{x^2} \left (2 x-2 x^2\right )+\left (2 x^2-2 e^{x^2} x-2 x\right ) \log (x)+x+x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x \left (-x^2+2 x+1\right )+e^{x^2} \left (e^x \left (x-2 x^2\right )-2 x^2\right )+x-e^x x \log (x)+1}{x \left (e^{x^2}-x-\log (x)+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 e^x x+2 x-e^x}{-e^{x^2}+x+\log (x)-1}-\frac {\left (e^x+1\right ) \left (2 x^3-2 x^2+2 x^2 \log (x)-x-1\right )}{x \left (-e^{x^2}+x+\log (x)-1\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{\left (-x+e^{x^2}-\log (x)+1\right )^2}dx+\int \frac {e^x}{\left (-x+e^{x^2}-\log (x)+1\right )^2}dx+\int \frac {e^x}{-x+e^{x^2}-\log (x)+1}dx+\int \frac {1}{x \left (x-e^{x^2}+\log (x)-1\right )^2}dx+\int \frac {e^x}{x \left (x-e^{x^2}+\log (x)-1\right )^2}dx+2 \int \frac {x}{\left (x-e^{x^2}+\log (x)-1\right )^2}dx+2 \int \frac {e^x x}{\left (x-e^{x^2}+\log (x)-1\right )^2}dx-2 \int \frac {x^2}{\left (x-e^{x^2}+\log (x)-1\right )^2}dx-2 \int \frac {e^x x^2}{\left (x-e^{x^2}+\log (x)-1\right )^2}dx-2 \int \frac {x \log (x)}{\left (x-e^{x^2}+\log (x)-1\right )^2}dx-2 \int \frac {e^x x \log (x)}{\left (x-e^{x^2}+\log (x)-1\right )^2}dx+2 \int \frac {x}{x-e^{x^2}+\log (x)-1}dx+2 \int \frac {e^x x}{x-e^{x^2}+\log (x)-1}dx\) |
Int[(1 + x + E^x*(1 + 2*x - x^2) + E^x^2*(-2*x^2 + E^x*(x - 2*x^2)) - E^x* x*Log[x])/(x + E^(2*x^2)*x - 2*x^2 + x^3 + E^x^2*(2*x - 2*x^2) + (-2*x - 2 *E^x^2*x + 2*x^2)*Log[x] + x*Log[x]^2),x]
3.27.25.3.1 Defintions of rubi rules used
Time = 0.38 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {{\mathrm e}^{x}+1}{x -1+\ln \left (x \right )-{\mathrm e}^{x^{2}}}\) | \(20\) |
parallelrisch | \(\frac {-{\mathrm e}^{x}-1}{x -1+\ln \left (x \right )-{\mathrm e}^{x^{2}}}\) | \(21\) |
int((-x*exp(x)*ln(x)+((-2*x^2+x)*exp(x)-2*x^2)*exp(x^2)+(-x^2+2*x+1)*exp(x )+x+1)/(x*ln(x)^2+(-2*exp(x^2)*x+2*x^2-2*x)*ln(x)+x*exp(x^2)^2+(-2*x^2+2*x )*exp(x^2)+x^3-2*x^2+x),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1+x+e^x \left (1+2 x-x^2\right )+e^{x^2} \left (-2 x^2+e^x \left (x-2 x^2\right )\right )-e^x x \log (x)}{x+e^{2 x^2} x-2 x^2+x^3+e^{x^2} \left (2 x-2 x^2\right )+\left (-2 x-2 e^{x^2} x+2 x^2\right ) \log (x)+x \log ^2(x)} \, dx=-\frac {e^{x} + 1}{x - e^{\left (x^{2}\right )} + \log \left (x\right ) - 1} \]
integrate((-x*exp(x)*log(x)+((-2*x^2+x)*exp(x)-2*x^2)*exp(x^2)+(-x^2+2*x+1 )*exp(x)+x+1)/(x*log(x)^2+(-2*exp(x^2)*x+2*x^2-2*x)*log(x)+x*exp(x^2)^2+(- 2*x^2+2*x)*exp(x^2)+x^3-2*x^2+x),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {1+x+e^x \left (1+2 x-x^2\right )+e^{x^2} \left (-2 x^2+e^x \left (x-2 x^2\right )\right )-e^x x \log (x)}{x+e^{2 x^2} x-2 x^2+x^3+e^{x^2} \left (2 x-2 x^2\right )+\left (-2 x-2 e^{x^2} x+2 x^2\right ) \log (x)+x \log ^2(x)} \, dx=\frac {e^{x} + 1}{- x + e^{x^{2}} - \log {\left (x \right )} + 1} \]
integrate((-x*exp(x)*ln(x)+((-2*x**2+x)*exp(x)-2*x**2)*exp(x**2)+(-x**2+2* x+1)*exp(x)+x+1)/(x*ln(x)**2+(-2*exp(x**2)*x+2*x**2-2*x)*ln(x)+x*exp(x**2) **2+(-2*x**2+2*x)*exp(x**2)+x**3-2*x**2+x),x)
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1+x+e^x \left (1+2 x-x^2\right )+e^{x^2} \left (-2 x^2+e^x \left (x-2 x^2\right )\right )-e^x x \log (x)}{x+e^{2 x^2} x-2 x^2+x^3+e^{x^2} \left (2 x-2 x^2\right )+\left (-2 x-2 e^{x^2} x+2 x^2\right ) \log (x)+x \log ^2(x)} \, dx=-\frac {e^{x} + 1}{x - e^{\left (x^{2}\right )} + \log \left (x\right ) - 1} \]
integrate((-x*exp(x)*log(x)+((-2*x^2+x)*exp(x)-2*x^2)*exp(x^2)+(-x^2+2*x+1 )*exp(x)+x+1)/(x*log(x)^2+(-2*exp(x^2)*x+2*x^2-2*x)*log(x)+x*exp(x^2)^2+(- 2*x^2+2*x)*exp(x^2)+x^3-2*x^2+x),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1+x+e^x \left (1+2 x-x^2\right )+e^{x^2} \left (-2 x^2+e^x \left (x-2 x^2\right )\right )-e^x x \log (x)}{x+e^{2 x^2} x-2 x^2+x^3+e^{x^2} \left (2 x-2 x^2\right )+\left (-2 x-2 e^{x^2} x+2 x^2\right ) \log (x)+x \log ^2(x)} \, dx=-\frac {e^{x} + 1}{x - e^{\left (x^{2}\right )} + \log \left (x\right ) - 1} \]
integrate((-x*exp(x)*log(x)+((-2*x^2+x)*exp(x)-2*x^2)*exp(x^2)+(-x^2+2*x+1 )*exp(x)+x+1)/(x*log(x)^2+(-2*exp(x^2)*x+2*x^2-2*x)*log(x)+x*exp(x^2)^2+(- 2*x^2+2*x)*exp(x^2)+x^3-2*x^2+x),x, algorithm=\
Time = 12.94 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.86 \[ \int \frac {1+x+e^x \left (1+2 x-x^2\right )+e^{x^2} \left (-2 x^2+e^x \left (x-2 x^2\right )\right )-e^x x \log (x)}{x+e^{2 x^2} x-2 x^2+x^3+e^{x^2} \left (2 x-2 x^2\right )+\left (-2 x-2 e^{x^2} x+2 x^2\right ) \log (x)+x \log ^2(x)} \, dx=-\frac {x\,\left ({\mathrm {e}}^x+1\right )+x^2\,\left ({\mathrm {e}}^x+1\right )+x^3\,\left (2\,{\mathrm {e}}^x-\ln \left (x\right )\,\left (2\,{\mathrm {e}}^x+2\right )+2\right )-x^4\,\left (2\,{\mathrm {e}}^x+2\right )}{\left (x-{\mathrm {e}}^{x^2}+\ln \left (x\right )-1\right )\,\left (x-2\,x^3\,\ln \left (x\right )+x^2+2\,x^3-2\,x^4\right )} \]
int((x + exp(x^2)*(exp(x)*(x - 2*x^2) - 2*x^2) + exp(x)*(2*x - x^2 + 1) - x*exp(x)*log(x) + 1)/(x + exp(x^2)*(2*x - 2*x^2) + x*log(x)^2 + x*exp(2*x^ 2) - log(x)*(2*x + 2*x*exp(x^2) - 2*x^2) - 2*x^2 + x^3),x)