Integrand size = 93, antiderivative size = 27 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=x+\frac {-\frac {2}{5}-e^{5 x}}{1+e^{-4/x} x} \]
Timed out. \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\text {\$Aborted} \]
Integrate[(5*x + (5*x^3)/E^(8/x) + (8 + 2*x + 10*x^2)/E^(4/x) + E^(5*x)*(- 25*x + (20 + 5*x - 25*x^2)/E^(4/x)))/(5*x + (10*x^2)/E^(4/x) + (5*x^3)/E^( 8/x)),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.
\(\displaystyle \int \frac {5 e^{-8/x} x^3+e^{-4/x} \left (10 x^2+2 x+8\right )+e^{5 x} \left (e^{-4/x} \left (-25 x^2+5 x+20\right )-25 x\right )+5 x}{5 e^{-8/x} x^3+10 e^{-4/x} x^2+5 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{8/x} \left (5 e^{-8/x} x^3+e^{-4/x} \left (10 x^2+2 x+8\right )+e^{5 x} \left (e^{-4/x} \left (-25 x^2+5 x+20\right )-25 x\right )+5 x\right )}{5 x \left (x+e^{4/x}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {e^{8/x} \left (5 e^{-8/x} x^3+5 x+2 e^{-4/x} \left (5 x^2+x+4\right )-5 e^{5 x} \left (5 x-e^{-4/x} \left (-5 x^2+x+4\right )\right )\right )}{x \left (x+e^{4/x}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (\frac {5 x^3+10 e^{4/x} x^2+2 e^{4/x} x+5 e^{8/x} x+8 e^{4/x}}{x \left (x+e^{4/x}\right )^2}-\frac {5 e^{5 x+\frac {4}{x}} \left (5 x^2+5 e^{4/x} x-x-4\right )}{x \left (x+e^{4/x}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-8 \int \frac {1}{\left (x+e^{4/x}\right )^2}dx+5 \int \frac {e^{5 x+\frac {4}{x}}}{\left (x+e^{4/x}\right )^2}dx+20 \int \frac {e^{5 x+\frac {4}{x}}}{x \left (x+e^{4/x}\right )^2}dx-2 \int \frac {x}{\left (x+e^{4/x}\right )^2}dx+2 \int \frac {1}{x+e^{4/x}}dx-25 \int \frac {e^{5 x+\frac {4}{x}}}{x+e^{4/x}}dx+8 \int \frac {1}{x \left (x+e^{4/x}\right )}dx+5 x\right )\) |
Int[(5*x + (5*x^3)/E^(8/x) + (8 + 2*x + 10*x^2)/E^(4/x) + E^(5*x)*(-25*x + (20 + 5*x - 25*x^2)/E^(4/x)))/(5*x + (10*x^2)/E^(4/x) + (5*x^3)/E^(8/x)), x]
3.16.50.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
risch | \(x -\frac {5 \,{\mathrm e}^{5 x}+2}{5 \left (x \,{\mathrm e}^{-\frac {4}{x}}+1\right )}\) | \(25\) |
norman | \(\frac {x +x^{2} {\mathrm e}^{-\frac {4}{x}}-{\mathrm e}^{5 x}-\frac {2}{5}}{x \,{\mathrm e}^{-\frac {4}{x}}+1}\) | \(33\) |
parallelrisch | \(\frac {5 x^{2} {\mathrm e}^{-\frac {4}{x}}+2 x \,{\mathrm e}^{-\frac {4}{x}}+5 x -5 \,{\mathrm e}^{5 x}}{5 x \,{\mathrm e}^{-\frac {4}{x}}+5}\) | \(45\) |
parts | \(\frac {x +x^{2} {\mathrm e}^{-\frac {4}{x}}-\frac {2}{5}}{x \,{\mathrm e}^{-\frac {4}{x}}+1}-\frac {{\mathrm e}^{5 x}}{x \,{\mathrm e}^{-\frac {4}{x}}+1}\) | \(46\) |
int((((-25*x^2+5*x+20)*exp(-4/x)-25*x)*exp(5*x)+5*x^3*exp(-4/x)^2+(10*x^2+ 2*x+8)*exp(-4/x)+5*x)/(5*x^3*exp(-4/x)^2+10*x^2*exp(-4/x)+5*x),x,method=_R ETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\frac {5 \, x^{2} e^{\left (-\frac {4}{x}\right )} + 5 \, x - 5 \, e^{\left (5 \, x\right )} - 2}{5 \, {\left (x e^{\left (-\frac {4}{x}\right )} + 1\right )}} \]
integrate((((-25*x^2+5*x+20)*exp(-4/x)-25*x)*exp(5*x)+5*x^3*exp(-4/x)^2+(1 0*x^2+2*x+8)*exp(-4/x)+5*x)/(5*x^3*exp(-4/x)^2+10*x^2*exp(-4/x)+5*x),x, al gorithm=\
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=x - \frac {2}{5 x e^{- \frac {4}{x}} + 5} - \frac {e^{5 x}}{x e^{- \frac {4}{x}} + 1} \]
integrate((((-25*x**2+5*x+20)*exp(-4/x)-25*x)*exp(5*x)+5*x**3*exp(-4/x)**2 +(10*x**2+2*x+8)*exp(-4/x)+5*x)/(5*x**3*exp(-4/x)**2+10*x**2*exp(-4/x)+5*x ),x)
Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\frac {5 \, x^{2} + 5 \, {\left (x - e^{\left (5 \, x\right )}\right )} e^{\frac {4}{x}} + 2 \, x}{5 \, {\left (x + e^{\frac {4}{x}}\right )}} \]
integrate((((-25*x^2+5*x+20)*exp(-4/x)-25*x)*exp(5*x)+5*x^3*exp(-4/x)^2+(1 0*x^2+2*x+8)*exp(-4/x)+5*x)/(5*x^3*exp(-4/x)^2+10*x^2*exp(-4/x)+5*x),x, al gorithm=\
Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\frac {5 \, x^{2} + 5 \, x e^{\frac {4}{x}} + 2 \, x - 5 \, e^{\left (5 \, x + \frac {4}{x}\right )}}{5 \, {\left (x + e^{\frac {4}{x}}\right )}} \]
integrate((((-25*x^2+5*x+20)*exp(-4/x)-25*x)*exp(5*x)+5*x^3*exp(-4/x)^2+(1 0*x^2+2*x+8)*exp(-4/x)+5*x)/(5*x^3*exp(-4/x)^2+10*x^2*exp(-4/x)+5*x),x, al gorithm=\
Time = 12.76 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=x-\frac {\frac {2\,{\mathrm {e}}^{4/x}}{5}+{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{4/x}}{x+{\mathrm {e}}^{4/x}} \]