Integrand size = 109, antiderivative size = 25 \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \]
Time = 5.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \]
Integrate[((E^x/(2 + E^x + x))^((2*E^(-3 + x^2))/x)*(E^(-3 + x^2)*(2*x + 2 *x^2) + E^(-3 + x^2)*(-4 - 2*x + 8*x^2 + 4*x^3 + E^x*(-2 + 4*x^2))*Log[E^x /(2 + E^x + x)]))/(2*x^2 + E^x*x^2 + x^3),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.
\(\displaystyle \int \frac {\left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \left (e^{x^2-3} \left (2 x^2+2 x\right )+e^{x^2-3} \left (4 x^3+8 x^2+e^x \left (4 x^2-2\right )-2 x-4\right ) \log \left (\frac {e^x}{x+e^x+2}\right )\right )}{x^3+e^x x^2+2 x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x+e^x+2}+\frac {2 e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x \left (x+e^x+2\right )}+\frac {8 e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x+e^x+2}+\frac {4 e^{x^2+x-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x+e^x+2}+\frac {4 e^{x^2-3} x \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x+e^x+2}-\frac {2 e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x \left (x+e^x+2\right )}-\frac {4 e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x^2 \left (x+e^x+2\right )}-\frac {2 e^{x^2+x-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x^2 \left (x+e^x+2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 8 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x+e^x+2}dx+2 \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x+e^x+2}dx+4 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2+x-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x+e^x+2}dx-4 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x^2 \left (x+e^x+2\right )}dx-2 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2+x-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x^2 \left (x+e^x+2\right )}dx-2 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x \left (x+e^x+2\right )}dx+2 \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x \left (x+e^x+2\right )}dx+4 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2-3} x \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x+e^x+2}dx-4 \int \frac {\int e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx-4 \int \frac {x \int e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx-8 \int \frac {\int e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx-8 \int \frac {x \int e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx+2 \int \frac {\int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x^2}dx}{x+e^x+2}dx+2 \int \frac {x \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x^2}dx}{x+e^x+2}dx+4 \int \frac {\int \frac {e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x^2}dx}{x+e^x+2}dx+4 \int \frac {x \int \frac {e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x^2}dx}{x+e^x+2}dx+2 \int \frac {\int \frac {e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x}dx}{x+e^x+2}dx+2 \int \frac {x \int \frac {e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x}dx}{x+e^x+2}dx-4 \int \frac {\int e^{x^2-x-3} x \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx-4 \int \frac {x \int e^{x^2-x-3} x \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx\) |
Int[((E^x/(2 + E^x + x))^((2*E^(-3 + x^2))/x)*(E^(-3 + x^2)*(2*x + 2*x^2) + E^(-3 + x^2)*(-4 - 2*x + 8*x^2 + 4*x^3 + E^x*(-2 + 4*x^2))*Log[E^x/(2 + E^x + x)]))/(2*x^2 + E^x*x^2 + x^3),x]
3.18.54.3.1 Defintions of rubi rules used
Time = 34.56 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {2 \,{\mathrm e}^{x^{2}-3} \ln \left (\frac {{\mathrm e}^{x}}{{\mathrm e}^{x}+2+x}\right )}{x}}\) | \(25\) |
risch | \(\left ({\mathrm e}^{x}+2+x \right )^{-\frac {2 \,{\mathrm e}^{x^{2}-3}}{x}} \left ({\mathrm e}^{x}\right )^{\frac {2 \,{\mathrm e}^{x^{2}-3}}{x}} {\mathrm e}^{-\frac {i {\mathrm e}^{x^{2}-3} \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}+2+x}\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}+2+x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}+2+x}\right )+\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}+2+x}\right )\right )}{x}}\) | \(112\) |
int((((4*x^2-2)*exp(x)+4*x^3+8*x^2-2*x-4)*exp(x^2-3)*ln(exp(x)/(exp(x)+2+x ))+(2*x^2+2*x)*exp(x^2-3))*exp(exp(x^2-3)*ln(exp(x)/(exp(x)+2+x))/x)^2/(ex p(x)*x^2+x^3+2*x^2),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=\left (\frac {e^{x}}{x + e^{x} + 2}\right )^{\frac {2 \, e^{\left (x^{2} - 3\right )}}{x}} \]
integrate((((4*x^2-2)*exp(x)+4*x^3+8*x^2-2*x-4)*exp(x^2-3)*log(exp(x)/(exp (x)+2+x))+(2*x^2+2*x)*exp(x^2-3))*exp(exp(x^2-3)*log(exp(x)/(exp(x)+2+x))/ x)^2/(exp(x)*x^2+x^3+2*x^2),x, algorithm=\
Timed out. \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=\text {Timed out} \]
integrate((((4*x**2-2)*exp(x)+4*x**3+8*x**2-2*x-4)*exp(x**2-3)*ln(exp(x)/( exp(x)+2+x))+(2*x**2+2*x)*exp(x**2-3))*exp(exp(x**2-3)*ln(exp(x)/(exp(x)+2 +x))/x)**2/(exp(x)*x**2+x**3+2*x**2),x)
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=e^{\left (-\frac {2 \, e^{\left (x^{2} - 3\right )} \log \left (x + e^{x} + 2\right )}{x} + 2 \, e^{\left (x^{2} - 3\right )}\right )} \]
integrate((((4*x^2-2)*exp(x)+4*x^3+8*x^2-2*x-4)*exp(x^2-3)*log(exp(x)/(exp (x)+2+x))+(2*x^2+2*x)*exp(x^2-3))*exp(exp(x^2-3)*log(exp(x)/(exp(x)+2+x))/ x)^2/(exp(x)*x^2+x^3+2*x^2),x, algorithm=\
\[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=\int { \frac {2 \, {\left ({\left (2 \, x^{3} + 4 \, x^{2} + {\left (2 \, x^{2} - 1\right )} e^{x} - x - 2\right )} e^{\left (x^{2} - 3\right )} \log \left (\frac {e^{x}}{x + e^{x} + 2}\right ) + {\left (x^{2} + x\right )} e^{\left (x^{2} - 3\right )}\right )} \left (\frac {e^{x}}{x + e^{x} + 2}\right )^{\frac {2 \, e^{\left (x^{2} - 3\right )}}{x}}}{x^{3} + x^{2} e^{x} + 2 \, x^{2}} \,d x } \]
integrate((((4*x^2-2)*exp(x)+4*x^3+8*x^2-2*x-4)*exp(x^2-3)*log(exp(x)/(exp (x)+2+x))+(2*x^2+2*x)*exp(x^2-3))*exp(exp(x^2-3)*log(exp(x)/(exp(x)+2+x))/ x)^2/(exp(x)*x^2+x^3+2*x^2),x, algorithm=\
integrate(2*((2*x^3 + 4*x^2 + (2*x^2 - 1)*e^x - x - 2)*e^(x^2 - 3)*log(e^x /(x + e^x + 2)) + (x^2 + x)*e^(x^2 - 3))*(e^x/(x + e^x + 2))^(2*e^(x^2 - 3 )/x)/(x^3 + x^2*e^x + 2*x^2), x)
Time = 11.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx={\mathrm {e}}^{2\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-3}}\,{\left (\frac {1}{x+{\mathrm {e}}^x+2}\right )}^{\frac {2\,{\mathrm {e}}^{x^2-3}}{x}} \]