Integrand size = 117, antiderivative size = 34 \[ \int \frac {2 e^{\frac {-1+x^2}{x^2}} x^2+2 x^5+e^{2 x} \left (2 e^{\frac {-1+x^2}{x^2}}+2 x^3\right )+e^x \left (-4 e^{\frac {-1+x^2}{x^2}} x-4 x^4\right )+2 \left (-2 x^3+e^x \left (x^3+x^4\right )\right ) \log (\log (5))}{e^{2 x} x^3-2 e^x x^4+x^5} \, dx=e^{\frac {-\frac {1}{x}+x}{x}}+(2+x) \left (2+\frac {2 \log (\log (5))}{-e^x+x}\right ) \]
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^{\frac {-1+x^2}{x^2}} x^2+2 x^5+e^{2 x} \left (2 e^{\frac {-1+x^2}{x^2}}+2 x^3\right )+e^x \left (-4 e^{\frac {-1+x^2}{x^2}} x-4 x^4\right )+2 \left (-2 x^3+e^x \left (x^3+x^4\right )\right ) \log (\log (5))}{e^{2 x} x^3-2 e^x x^4+x^5} \, dx=2 \left (\frac {1}{2} e^{1-\frac {1}{x^2}}+x-\frac {(2+x) \log (\log (5))}{e^x-x}\right ) \]
Integrate[(2*E^((-1 + x^2)/x^2)*x^2 + 2*x^5 + E^(2*x)*(2*E^((-1 + x^2)/x^2 ) + 2*x^3) + E^x*(-4*E^((-1 + x^2)/x^2)*x - 4*x^4) + 2*(-2*x^3 + E^x*(x^3 + x^4))*Log[Log[5]])/(E^(2*x)*x^3 - 2*E^x*x^4 + x^5),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.
\(\displaystyle \int \frac {2 x^5+2 e^{\frac {x^2-1}{x^2}} x^2+2 \left (e^x \left (x^4+x^3\right )-2 x^3\right ) \log (\log (5))+e^x \left (-4 x^4-4 e^{\frac {x^2-1}{x^2}} x\right )+e^{2 x} \left (2 x^3+2 e^{\frac {x^2-1}{x^2}}\right )}{x^5-2 e^x x^4+e^{2 x} x^3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 x^5+2 e^{\frac {x^2-1}{x^2}} x^2+2 \left (e^x \left (x^4+x^3\right )-2 x^3\right ) \log (\log (5))+e^x \left (-4 x^4-4 e^{\frac {x^2-1}{x^2}} x\right )+e^{2 x} \left (2 x^3+2 e^{\frac {x^2-1}{x^2}}\right )}{\left (e^x-x\right )^2 x^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (x^2+x-2\right ) \log (\log (5))}{\left (e^x-x\right )^2}+\frac {2 e^{-\frac {1}{x^2}} \left (e^{\frac {1}{x^2}} x^3+e\right )}{x^3}+\frac {2 (x+1) \log (\log (5))}{e^x-x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \log (\log (5)) \int \frac {x^2}{\left (e^x-x\right )^2}dx-4 \log (\log (5)) \int \frac {1}{\left (e^x-x\right )^2}dx+2 \log (\log (5)) \int \frac {1}{e^x-x}dx+2 \log (\log (5)) \int \frac {x}{\left (e^x-x\right )^2}dx+2 \log (\log (5)) \int \frac {x}{e^x-x}dx+e^{1-\frac {1}{x^2}}+2 x\) |
Int[(2*E^((-1 + x^2)/x^2)*x^2 + 2*x^5 + E^(2*x)*(2*E^((-1 + x^2)/x^2) + 2* x^3) + E^x*(-4*E^((-1 + x^2)/x^2)*x - 4*x^4) + 2*(-2*x^3 + E^x*(x^3 + x^4) )*Log[Log[5]])/(E^(2*x)*x^3 - 2*E^x*x^4 + x^5),x]
3.19.32.3.1 Defintions of rubi rules used
Time = 0.94 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94
method | result | size |
risch | \(2 x +\frac {2 \left (2+x \right ) \ln \left (\ln \left (5\right )\right )}{x -{\mathrm e}^{x}}+{\mathrm e}^{\frac {\left (-1+x \right ) \left (1+x \right )}{x^{2}}}\) | \(32\) |
parallelrisch | \(\frac {2 x \ln \left (\ln \left (5\right )\right )+2 x^{2}-2 \,{\mathrm e}^{x} x +x \,{\mathrm e}^{\frac {x^{2}-1}{x^{2}}}-{\mathrm e}^{x} {\mathrm e}^{\frac {x^{2}-1}{x^{2}}}+4 \ln \left (\ln \left (5\right )\right )}{x -{\mathrm e}^{x}}\) | \(58\) |
int((2*((x^4+x^3)*exp(x)-2*x^3)*ln(ln(5))+(2*exp((x^2-1)/x^2)+2*x^3)*exp(x )^2+(-4*x*exp((x^2-1)/x^2)-4*x^4)*exp(x)+2*x^2*exp((x^2-1)/x^2)+2*x^5)/(ex p(x)^2*x^3-2*exp(x)*x^4+x^5),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \[ \int \frac {2 e^{\frac {-1+x^2}{x^2}} x^2+2 x^5+e^{2 x} \left (2 e^{\frac {-1+x^2}{x^2}}+2 x^3\right )+e^x \left (-4 e^{\frac {-1+x^2}{x^2}} x-4 x^4\right )+2 \left (-2 x^3+e^x \left (x^3+x^4\right )\right ) \log (\log (5))}{e^{2 x} x^3-2 e^x x^4+x^5} \, dx=\frac {2 \, x^{2} - {\left (2 \, x + e^{\left (\frac {x^{2} - 1}{x^{2}}\right )}\right )} e^{x} + x e^{\left (\frac {x^{2} - 1}{x^{2}}\right )} + 2 \, {\left (x + 2\right )} \log \left (\log \left (5\right )\right )}{x - e^{x}} \]
integrate((2*((x^4+x^3)*exp(x)-2*x^3)*log(log(5))+(2*exp((x^2-1)/x^2)+2*x^ 3)*exp(x)^2+(-4*x*exp((x^2-1)/x^2)-4*x^4)*exp(x)+2*x^2*exp((x^2-1)/x^2)+2* x^5)/(exp(x)^2*x^3-2*exp(x)*x^4+x^5),x, algorithm=\
(2*x^2 - (2*x + e^((x^2 - 1)/x^2))*e^x + x*e^((x^2 - 1)/x^2) + 2*(x + 2)*l og(log(5)))/(x - e^x)
Time = 0.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^{\frac {-1+x^2}{x^2}} x^2+2 x^5+e^{2 x} \left (2 e^{\frac {-1+x^2}{x^2}}+2 x^3\right )+e^x \left (-4 e^{\frac {-1+x^2}{x^2}} x-4 x^4\right )+2 \left (-2 x^3+e^x \left (x^3+x^4\right )\right ) \log (\log (5))}{e^{2 x} x^3-2 e^x x^4+x^5} \, dx=2 x + e^{\frac {x^{2} - 1}{x^{2}}} + \frac {- 2 x \log {\left (\log {\left (5 \right )} \right )} - 4 \log {\left (\log {\left (5 \right )} \right )}}{- x + e^{x}} \]
integrate((2*((x**4+x**3)*exp(x)-2*x**3)*ln(ln(5))+(2*exp((x**2-1)/x**2)+2 *x**3)*exp(x)**2+(-4*x*exp((x**2-1)/x**2)-4*x**4)*exp(x)+2*x**2*exp((x**2- 1)/x**2)+2*x**5)/(exp(x)**2*x**3-2*exp(x)*x**4+x**5),x)
Time = 0.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.50 \[ \int \frac {2 e^{\frac {-1+x^2}{x^2}} x^2+2 x^5+e^{2 x} \left (2 e^{\frac {-1+x^2}{x^2}}+2 x^3\right )+e^x \left (-4 e^{\frac {-1+x^2}{x^2}} x-4 x^4\right )+2 \left (-2 x^3+e^x \left (x^3+x^4\right )\right ) \log (\log (5))}{e^{2 x} x^3-2 e^x x^4+x^5} \, dx=\frac {{\left (x e + 2 \, {\left (x^{2} - x e^{x} + x \log \left (\log \left (5\right )\right ) + 2 \, \log \left (\log \left (5\right )\right )\right )} e^{\left (\frac {1}{x^{2}}\right )} - e^{\left (x + 1\right )}\right )} e^{\left (-\frac {1}{x^{2}}\right )}}{x - e^{x}} \]
integrate((2*((x^4+x^3)*exp(x)-2*x^3)*log(log(5))+(2*exp((x^2-1)/x^2)+2*x^ 3)*exp(x)^2+(-4*x*exp((x^2-1)/x^2)-4*x^4)*exp(x)+2*x^2*exp((x^2-1)/x^2)+2* x^5)/(exp(x)^2*x^3-2*exp(x)*x^4+x^5),x, algorithm=\
(x*e + 2*(x^2 - x*e^x + x*log(log(5)) + 2*log(log(5)))*e^(x^(-2)) - e^(x + 1))*e^(-1/x^2)/(x - e^x)
Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {2 e^{\frac {-1+x^2}{x^2}} x^2+2 x^5+e^{2 x} \left (2 e^{\frac {-1+x^2}{x^2}}+2 x^3\right )+e^x \left (-4 e^{\frac {-1+x^2}{x^2}} x-4 x^4\right )+2 \left (-2 x^3+e^x \left (x^3+x^4\right )\right ) \log (\log (5))}{e^{2 x} x^3-2 e^x x^4+x^5} \, dx=\frac {2 \, {\left (x^{2} - x e^{x} + x \log \left (\log \left (5\right )\right ) + 2 \, \log \left (\log \left (5\right )\right )\right )}}{x - e^{x}} + e^{\left (-\frac {1}{x^{2}} + 1\right )} \]
integrate((2*((x^4+x^3)*exp(x)-2*x^3)*log(log(5))+(2*exp((x^2-1)/x^2)+2*x^ 3)*exp(x)^2+(-4*x*exp((x^2-1)/x^2)-4*x^4)*exp(x)+2*x^2*exp((x^2-1)/x^2)+2* x^5)/(exp(x)^2*x^3-2*exp(x)*x^4+x^5),x, algorithm=\
Time = 10.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.32 \[ \int \frac {2 e^{\frac {-1+x^2}{x^2}} x^2+2 x^5+e^{2 x} \left (2 e^{\frac {-1+x^2}{x^2}}+2 x^3\right )+e^x \left (-4 e^{\frac {-1+x^2}{x^2}} x-4 x^4\right )+2 \left (-2 x^3+e^x \left (x^3+x^4\right )\right ) \log (\log (5))}{e^{2 x} x^3-2 e^x x^4+x^5} \, dx=2\,x+{\mathrm {e}}^{1-\frac {1}{x^2}}+\frac {2\,\left (\ln \left (\ln \left (5\right )\right )\,x^2+\ln \left (\ln \left (5\right )\right )\,x-2\,\ln \left (\ln \left (5\right )\right )\right )}{\left (x-{\mathrm {e}}^x\right )\,\left (x-1\right )} \]
int((2*x^2*exp((x^2 - 1)/x^2) + 2*log(log(5))*(exp(x)*(x^3 + x^4) - 2*x^3) + exp(2*x)*(2*exp((x^2 - 1)/x^2) + 2*x^3) - exp(x)*(4*x*exp((x^2 - 1)/x^2 ) + 4*x^4) + 2*x^5)/(x^3*exp(2*x) - 2*x^4*exp(x) + x^5),x)