Integrand size = 93, antiderivative size = 27 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=x+\frac {-\frac {2}{5}-e^{5 x}}{1+e^{-4/x} x} \] Output:
(-exp(5*x)-2/5)/(x*exp(-4/x)+1)+x
Timed out. \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\text {\$Aborted} \] Input:
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]
Output:
$Aborted
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 |
\(\displaystyle \frac {1}{5} \left (-8 \int \frac {1}{\left (x+e^{4/x}\right )^2}dx+5 \int \frac {e^{5 x+\frac {4}{x}}}{\left (x+e^{4/x}\right )^2}dx+20 \int \frac {e^{5 x+\frac {4}{x}}}{x \left (x+e^{4/x}\right )^2}dx-2 \int \frac {x}{\left (x+e^{4/x}\right )^2}dx+2 \int \frac {1}{x+e^{4/x}}dx-25 \int \frac {e^{5 x+\frac {4}{x}}}{x+e^{4/x}}dx+8 \int \frac {1}{x \left (x+e^{4/x}\right )}dx+5 x\right )\) |
Input:
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]
Output:
$Aborted
Time = 0.43 (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 | \(\frac {5 x^{2} {\mathrm e}^{-\frac {4}{x}}-5 \,{\mathrm e}^{5 x}+5 x -2}{5 x \,{\mathrm e}^{-\frac {4}{x}}+5}\) | \(37\) |
parts | \(\frac {x +x^{2} {\mathrm e}^{-\frac {4}{x}}-\frac {2}{5}}{x \,{\mathrm e}^{-\frac {4}{x}}+1}-\frac {{\mathrm e}^{5 x}}{x \,{\mathrm e}^{-\frac {4}{x}}+1}\) | \(46\) |
Input:
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)
Output:
x-1/5*(5*exp(5*x)+2)/(x*exp(-4/x)+1)
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\frac {5 \, x^{2} e^{\left (-\frac {4}{x}\right )} + 5 \, x - 5 \, e^{\left (5 \, x\right )} - 2}{5 \, {\left (x e^{\left (-\frac {4}{x}\right )} + 1\right )}} \] Input:
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="fricas")
Output:
1/5*(5*x^2*e^(-4/x) + 5*x - 5*e^(5*x) - 2)/(x*e^(-4/x) + 1)
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=x - \frac {2}{5 x e^{- \frac {4}{x}} + 5} - \frac {e^{5 x}}{x e^{- \frac {4}{x}} + 1} \] Input:
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)
Output:
x - 2/(5*x*exp(-4/x) + 5) - exp(5*x)/(x*exp(-4/x) + 1)
Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\frac {5 \, x^{2} + 5 \, {\left (x - e^{\left (5 \, x\right )}\right )} e^{\frac {4}{x}} + 2 \, x}{5 \, {\left (x + e^{\frac {4}{x}}\right )}} \] Input:
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="maxima")
Output:
1/5*(5*x^2 + 5*(x - e^(5*x))*e^(4/x) + 2*x)/(x + e^(4/x))
Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\frac {5 \, x^{2} + 5 \, x e^{\frac {4}{x}} + 2 \, x - 5 \, e^{\left (5 \, x + \frac {4}{x}\right )}}{5 \, {\left (x + e^{\frac {4}{x}}\right )}} \] Input:
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="giac")
Output:
1/5*(5*x^2 + 5*x*e^(4/x) + 2*x - 5*e^(5*x + 4/x))/(x + e^(4/x))
Time = 2.94 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=x-\frac {\frac {2\,{\mathrm {e}}^{4/x}}{5}+{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{4/x}}{x+{\mathrm {e}}^{4/x}} \] Input:
int((5*x - exp(5*x)*(25*x - exp(-4/x)*(5*x - 25*x^2 + 20)) + exp(-4/x)*(2* x + 10*x^2 + 8) + 5*x^3*exp(-8/x))/(5*x + 10*x^2*exp(-4/x) + 5*x^3*exp(-8/ x)),x)
Output:
x - ((2*exp(4/x))/5 + exp(5*x)*exp(4/x))/(x + exp(4/x))
\[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\int \frac {\left (\left (-25 x^{2}+5 x +20\right ) {\mathrm e}^{-\frac {4}{x}}-25 x \right ) {\mathrm e}^{5 x}+5 x^{3} \left ({\mathrm e}^{-\frac {4}{x}}\right )^{2}+\left (10 x^{2}+2 x +8\right ) {\mathrm e}^{-\frac {4}{x}}+5 x}{5 x^{3} \left ({\mathrm e}^{-\frac {4}{x}}\right )^{2}+10 x^{2} {\mathrm e}^{-\frac {4}{x}}+5 x}d x \] Input:
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)
Output:
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)