Integrand size = 111, antiderivative size = 29 \[ \int \frac {4 x-x^2+e^x \left (-4 x+x^2\right )+e^{e^x} \left (-x^2+e^{2 x} \left (-4 x+x^2\right )+e^x \left (x+4 x^2-x^3\right )\right )+\left (-2 e^x+2 x\right ) \log \left (\frac {x}{5}\right )+\left (e^x x-x^2\right ) \log \left (-e^x+x\right )}{e^x x-x^2} \, dx=-\log ^2\left (\frac {x}{5}\right )+(-4+x) \left (e^{e^x}+\log \left (-e^x+x\right )\right ) \]
Time = 0.51 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {4 x-x^2+e^x \left (-4 x+x^2\right )+e^{e^x} \left (-x^2+e^{2 x} \left (-4 x+x^2\right )+e^x \left (x+4 x^2-x^3\right )\right )+\left (-2 e^x+2 x\right ) \log \left (\frac {x}{5}\right )+\left (e^x x-x^2\right ) \log \left (-e^x+x\right )}{e^x x-x^2} \, dx=e^{e^x} (-4+x)-4 \log \left (e^x-x\right )-\log ^2\left (\frac {x}{5}\right )+x \log \left (-e^x+x\right ) \]
Integrate[(4*x - x^2 + E^x*(-4*x + x^2) + E^E^x*(-x^2 + E^(2*x)*(-4*x + x^ 2) + E^x*(x + 4*x^2 - x^3)) + (-2*E^x + 2*x)*Log[x/5] + (E^x*x - x^2)*Log[ -E^x + x])/(E^x*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.
\(\displaystyle \int \frac {-x^2+e^x \left (x^2-4 x\right )+\left (e^x x-x^2\right ) \log \left (x-e^x\right )+e^{e^x} \left (-x^2+e^{2 x} \left (x^2-4 x\right )+e^x \left (-x^3+4 x^2+x\right )\right )+4 x+\left (2 x-2 e^x\right ) \log \left (\frac {x}{5}\right )}{e^x x-x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2-5 x+4}{e^x-x}+\frac {x^2+e^{e^x} x-4 x+x \log \left (x-e^x\right )-2 \log \left (\frac {x}{5}\right )}{x}+e^{x+e^x} (x-4)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \frac {1}{e^x-x}dx+\int e^{x+e^x} xdx-4 \int \frac {x}{e^x-x}dx+\operatorname {ExpIntegralEi}\left (e^x\right )-4 e^{e^x}-4 x-\log ^2\left (\frac {x}{5}\right )+x \log \left (x-e^x\right )\) |
Int[(4*x - x^2 + E^x*(-4*x + x^2) + E^E^x*(-x^2 + E^(2*x)*(-4*x + x^2) + E ^x*(x + 4*x^2 - x^3)) + (-2*E^x + 2*x)*Log[x/5] + (E^x*x - x^2)*Log[-E^x + x])/(E^x*x - x^2),x]
3.26.19.3.1 Defintions of rubi rules used
Time = 1.82 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31
method | result | size |
risch | \(\ln \left (x -{\mathrm e}^{x}\right ) x +x \,{\mathrm e}^{{\mathrm e}^{x}}-\ln \left (\frac {x}{5}\right )^{2}-4 \ln \left ({\mathrm e}^{x}-x \right )-4 \,{\mathrm e}^{{\mathrm e}^{x}}\) | \(38\) |
parallelrisch | \(x \,{\mathrm e}^{{\mathrm e}^{x}}+\ln \left (x -{\mathrm e}^{x}\right ) x -\ln \left (\frac {x}{5}\right )^{2}-4 \,{\mathrm e}^{{\mathrm e}^{x}}-4 \ln \left (x -{\mathrm e}^{x}\right )\) | \(38\) |
int(((exp(x)*x-x^2)*ln(x-exp(x))+((x^2-4*x)*exp(x)^2+(-x^3+4*x^2+x)*exp(x) -x^2)*exp(exp(x))+(-2*exp(x)+2*x)*ln(1/5*x)+(x^2-4*x)*exp(x)-x^2+4*x)/(exp (x)*x-x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {4 x-x^2+e^x \left (-4 x+x^2\right )+e^{e^x} \left (-x^2+e^{2 x} \left (-4 x+x^2\right )+e^x \left (x+4 x^2-x^3\right )\right )+\left (-2 e^x+2 x\right ) \log \left (\frac {x}{5}\right )+\left (e^x x-x^2\right ) \log \left (-e^x+x\right )}{e^x x-x^2} \, dx={\left (x - 4\right )} e^{\left (e^{x}\right )} - \log \left (\frac {1}{5} \, x\right )^{2} + {\left (x - 4\right )} \log \left (x - e^{x}\right ) \]
integrate(((exp(x)*x-x^2)*log(x-exp(x))+((x^2-4*x)*exp(x)^2+(-x^3+4*x^2+x) *exp(x)-x^2)*exp(exp(x))+(-2*exp(x)+2*x)*log(1/5*x)+(x^2-4*x)*exp(x)-x^2+4 *x)/(exp(x)*x-x^2),x, algorithm=\
Exception generated. \[ \int \frac {4 x-x^2+e^x \left (-4 x+x^2\right )+e^{e^x} \left (-x^2+e^{2 x} \left (-4 x+x^2\right )+e^x \left (x+4 x^2-x^3\right )\right )+\left (-2 e^x+2 x\right ) \log \left (\frac {x}{5}\right )+\left (e^x x-x^2\right ) \log \left (-e^x+x\right )}{e^x x-x^2} \, dx=\text {Exception raised: TypeError} \]
integrate(((exp(x)*x-x**2)*ln(x-exp(x))+((x**2-4*x)*exp(x)**2+(-x**3+4*x** 2+x)*exp(x)-x**2)*exp(exp(x))+(-2*exp(x)+2*x)*ln(1/5*x)+(x**2-4*x)*exp(x)- x**2+4*x)/(exp(x)*x-x**2),x)
Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {4 x-x^2+e^x \left (-4 x+x^2\right )+e^{e^x} \left (-x^2+e^{2 x} \left (-4 x+x^2\right )+e^x \left (x+4 x^2-x^3\right )\right )+\left (-2 e^x+2 x\right ) \log \left (\frac {x}{5}\right )+\left (e^x x-x^2\right ) \log \left (-e^x+x\right )}{e^x x-x^2} \, dx={\left (x - 4\right )} e^{\left (e^{x}\right )} + {\left (x - 4\right )} \log \left (x - e^{x}\right ) + 2 \, \log \left (5\right ) \log \left (x\right ) - \log \left (x\right )^{2} \]
integrate(((exp(x)*x-x^2)*log(x-exp(x))+((x^2-4*x)*exp(x)^2+(-x^3+4*x^2+x) *exp(x)-x^2)*exp(exp(x))+(-2*exp(x)+2*x)*log(1/5*x)+(x^2-4*x)*exp(x)-x^2+4 *x)/(exp(x)*x-x^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (24) = 48\).
Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.00 \[ \int \frac {4 x-x^2+e^x \left (-4 x+x^2\right )+e^{e^x} \left (-x^2+e^{2 x} \left (-4 x+x^2\right )+e^x \left (x+4 x^2-x^3\right )\right )+\left (-2 e^x+2 x\right ) \log \left (\frac {x}{5}\right )+\left (e^x x-x^2\right ) \log \left (-e^x+x\right )}{e^x x-x^2} \, dx={\left (x e^{x} \log \left (x - e^{x}\right ) + 2 \, e^{x} \log \left (5\right ) \log \left (x\right ) - e^{x} \log \left (x\right )^{2} + x e^{\left (x + e^{x}\right )} - 4 \, e^{x} \log \left (x - e^{x}\right ) - 4 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \]
integrate(((exp(x)*x-x^2)*log(x-exp(x))+((x^2-4*x)*exp(x)^2+(-x^3+4*x^2+x) *exp(x)-x^2)*exp(exp(x))+(-2*exp(x)+2*x)*log(1/5*x)+(x^2-4*x)*exp(x)-x^2+4 *x)/(exp(x)*x-x^2),x, algorithm=\
(x*e^x*log(x - e^x) + 2*e^x*log(5)*log(x) - e^x*log(x)^2 + x*e^(x + e^x) - 4*e^x*log(x - e^x) - 4*e^(x + e^x))*e^(-x)
Timed out. \[ \int \frac {4 x-x^2+e^x \left (-4 x+x^2\right )+e^{e^x} \left (-x^2+e^{2 x} \left (-4 x+x^2\right )+e^x \left (x+4 x^2-x^3\right )\right )+\left (-2 e^x+2 x\right ) \log \left (\frac {x}{5}\right )+\left (e^x x-x^2\right ) \log \left (-e^x+x\right )}{e^x x-x^2} \, dx=\int \frac {4\,x+\ln \left (\frac {x}{5}\right )\,\left (2\,x-2\,{\mathrm {e}}^x\right )+\ln \left (x-{\mathrm {e}}^x\right )\,\left (x\,{\mathrm {e}}^x-x^2\right )-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^{2\,x}\,\left (4\,x-x^2\right )-{\mathrm {e}}^x\,\left (-x^3+4\,x^2+x\right )+x^2\right )-{\mathrm {e}}^x\,\left (4\,x-x^2\right )-x^2}{x\,{\mathrm {e}}^x-x^2} \,d x \]
int((4*x + log(x/5)*(2*x - 2*exp(x)) + log(x - exp(x))*(x*exp(x) - x^2) - exp(exp(x))*(exp(2*x)*(4*x - x^2) - exp(x)*(x + 4*x^2 - x^3) + x^2) - exp( x)*(4*x - x^2) - x^2)/(x*exp(x) - x^2),x)