Integrand size = 108, antiderivative size = 21 \[ \int \frac {-x^2+x^3+\left (2 e^x x+2 x^2\right ) \log \left (e^x+x\right )}{x^3-4 x^4+4 x^5+e^x \left (x^2-4 x^3+4 x^4\right )+\left (2 x^2-4 x^3+e^x \left (2 x-4 x^2\right )\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=3+\frac {x^2}{x-2 x^2+\log \left (e^x+x\right )} \]
Time = 0.40 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-x^2+x^3+\left (2 e^x x+2 x^2\right ) \log \left (e^x+x\right )}{x^3-4 x^4+4 x^5+e^x \left (x^2-4 x^3+4 x^4\right )+\left (2 x^2-4 x^3+e^x \left (2 x-4 x^2\right )\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=\frac {x^2}{x-2 x^2+\log \left (e^x+x\right )} \]
Integrate[(-x^2 + x^3 + (2*E^x*x + 2*x^2)*Log[E^x + x])/(x^3 - 4*x^4 + 4*x ^5 + E^x*(x^2 - 4*x^3 + 4*x^4) + (2*x^2 - 4*x^3 + E^x*(2*x - 4*x^2))*Log[E ^x + x] + (E^x + x)*Log[E^x + 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.
\(\displaystyle \int \frac {x^3-x^2+\left (2 x^2+2 e^x x\right ) \log \left (x+e^x\right )}{4 x^5-4 x^4+x^3+\left (-4 x^3+2 x^2+e^x \left (2 x-4 x^2\right )\right ) \log \left (x+e^x\right )+e^x \left (4 x^4-4 x^3+x^2\right )+\left (x+e^x\right ) \log ^2\left (x+e^x\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x \left ((x-1) x+2 \left (x+e^x\right ) \log \left (x+e^x\right )\right )}{\left (x+e^x\right ) \left (-2 x^2+x+\log \left (x+e^x\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(x-1) x^2}{\left (x+e^x\right ) \left (2 x^2-x-\log \left (x+e^x\right )\right )^2}+\frac {2 x \log \left (x+e^x\right )}{\left (2 x^2-x-\log \left (x+e^x\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {x^2}{\left (2 x^2-x-\log \left (x+e^x\right )\right )^2}dx-\int \frac {x^2}{\left (x+e^x\right ) \left (2 x^2-x-\log \left (x+e^x\right )\right )^2}dx-2 \int \frac {x}{2 x^2-x-\log \left (x+e^x\right )}dx+4 \int \frac {x^3}{\left (2 x^2-x-\log \left (x+e^x\right )\right )^2}dx+\int \frac {x^3}{\left (x+e^x\right ) \left (2 x^2-x-\log \left (x+e^x\right )\right )^2}dx\) |
Int[(-x^2 + x^3 + (2*E^x*x + 2*x^2)*Log[E^x + x])/(x^3 - 4*x^4 + 4*x^5 + E ^x*(x^2 - 4*x^3 + 4*x^4) + (2*x^2 - 4*x^3 + E^x*(2*x - 4*x^2))*Log[E^x + x ] + (E^x + x)*Log[E^x + x]^2),x]
3.20.9.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14
method | result | size |
risch | \(-\frac {x^{2}}{2 x^{2}-x -\ln \left ({\mathrm e}^{x}+x \right )}\) | \(24\) |
parallelrisch | \(-\frac {x^{2}}{2 x^{2}-x -\ln \left ({\mathrm e}^{x}+x \right )}\) | \(24\) |
int(((2*exp(x)*x+2*x^2)*ln(exp(x)+x)+x^3-x^2)/((exp(x)+x)*ln(exp(x)+x)^2+( (-4*x^2+2*x)*exp(x)-4*x^3+2*x^2)*ln(exp(x)+x)+(4*x^4-4*x^3+x^2)*exp(x)+4*x ^5-4*x^4+x^3),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {-x^2+x^3+\left (2 e^x x+2 x^2\right ) \log \left (e^x+x\right )}{x^3-4 x^4+4 x^5+e^x \left (x^2-4 x^3+4 x^4\right )+\left (2 x^2-4 x^3+e^x \left (2 x-4 x^2\right )\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=-\frac {x^{2}}{2 \, x^{2} - x - \log \left (x + e^{x}\right )} \]
integrate(((2*exp(x)*x+2*x^2)*log(exp(x)+x)+x^3-x^2)/((exp(x)+x)*log(exp(x )+x)^2+((-4*x^2+2*x)*exp(x)-4*x^3+2*x^2)*log(exp(x)+x)+(4*x^4-4*x^3+x^2)*e xp(x)+4*x^5-4*x^4+x^3),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {-x^2+x^3+\left (2 e^x x+2 x^2\right ) \log \left (e^x+x\right )}{x^3-4 x^4+4 x^5+e^x \left (x^2-4 x^3+4 x^4\right )+\left (2 x^2-4 x^3+e^x \left (2 x-4 x^2\right )\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=\frac {x^{2}}{- 2 x^{2} + x + \log {\left (x + e^{x} \right )}} \]
integrate(((2*exp(x)*x+2*x**2)*ln(exp(x)+x)+x**3-x**2)/((exp(x)+x)*ln(exp( x)+x)**2+((-4*x**2+2*x)*exp(x)-4*x**3+2*x**2)*ln(exp(x)+x)+(4*x**4-4*x**3+ x**2)*exp(x)+4*x**5-4*x**4+x**3),x)
Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {-x^2+x^3+\left (2 e^x x+2 x^2\right ) \log \left (e^x+x\right )}{x^3-4 x^4+4 x^5+e^x \left (x^2-4 x^3+4 x^4\right )+\left (2 x^2-4 x^3+e^x \left (2 x-4 x^2\right )\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=-\frac {x^{2}}{2 \, x^{2} - x - \log \left (x + e^{x}\right )} \]
integrate(((2*exp(x)*x+2*x^2)*log(exp(x)+x)+x^3-x^2)/((exp(x)+x)*log(exp(x )+x)^2+((-4*x^2+2*x)*exp(x)-4*x^3+2*x^2)*log(exp(x)+x)+(4*x^4-4*x^3+x^2)*e xp(x)+4*x^5-4*x^4+x^3),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {-x^2+x^3+\left (2 e^x x+2 x^2\right ) \log \left (e^x+x\right )}{x^3-4 x^4+4 x^5+e^x \left (x^2-4 x^3+4 x^4\right )+\left (2 x^2-4 x^3+e^x \left (2 x-4 x^2\right )\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=-\frac {x^{2}}{2 \, x^{2} - x - \log \left (x + e^{x}\right )} \]
integrate(((2*exp(x)*x+2*x^2)*log(exp(x)+x)+x^3-x^2)/((exp(x)+x)*log(exp(x )+x)^2+((-4*x^2+2*x)*exp(x)-4*x^3+2*x^2)*log(exp(x)+x)+(4*x^4-4*x^3+x^2)*e xp(x)+4*x^5-4*x^4+x^3),x, algorithm=\
Timed out. \[ \int \frac {-x^2+x^3+\left (2 e^x x+2 x^2\right ) \log \left (e^x+x\right )}{x^3-4 x^4+4 x^5+e^x \left (x^2-4 x^3+4 x^4\right )+\left (2 x^2-4 x^3+e^x \left (2 x-4 x^2\right )\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=\int \frac {\ln \left (x+{\mathrm {e}}^x\right )\,\left (2\,x\,{\mathrm {e}}^x+2\,x^2\right )-x^2+x^3}{{\ln \left (x+{\mathrm {e}}^x\right )}^2\,\left (x+{\mathrm {e}}^x\right )+x^3-4\,x^4+4\,x^5+{\mathrm {e}}^x\,\left (4\,x^4-4\,x^3+x^2\right )+\ln \left (x+{\mathrm {e}}^x\right )\,\left ({\mathrm {e}}^x\,\left (2\,x-4\,x^2\right )+2\,x^2-4\,x^3\right )} \,d x \]
int((log(x + exp(x))*(2*x*exp(x) + 2*x^2) - x^2 + x^3)/(log(x + exp(x))^2* (x + exp(x)) + x^3 - 4*x^4 + 4*x^5 + exp(x)*(x^2 - 4*x^3 + 4*x^4) + log(x + exp(x))*(exp(x)*(2*x - 4*x^2) + 2*x^2 - 4*x^3)),x)