Integrand size = 75, antiderivative size = 30 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=\frac {\left (-e^2+x\right ) \left (4 e^{\frac {1}{(-2-x) x}}+3 x\right )}{x} \] Output:
(x-exp(1)^2)/x*(3*x+4*exp(1/x/(-2-x)))
Time = 5.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=3 x+\frac {4 e^{-\frac {1}{2 x+x^2}} \left (-e^2+x\right )}{x} \] Input:
Integrate[(12*x^3 + 12*x^4 + 3*x^5 + (8*x + 8*x^2 + E^2*(-8 + 8*x + 16*x^2 + 4*x^3))/E^(2*x + x^2)^(-1))/(4*x^3 + 4*x^4 + x^5),x]
Output:
3*x + (4*(-E^2 + x))/(E^(2*x + x^2)^(-1)*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.
\(\displaystyle \int \frac {3 x^5+12 x^4+12 x^3+e^{-\frac {1}{x^2+2 x}} \left (8 x^2+e^2 \left (4 x^3+16 x^2+8 x-8\right )+8 x\right )}{x^5+4 x^4+4 x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {3 x^5+12 x^4+12 x^3+e^{-\frac {1}{x^2+2 x}} \left (8 x^2+e^2 \left (4 x^3+16 x^2+8 x-8\right )+8 x\right )}{x^3 \left (x^2+4 x+4\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {3 x^5+12 x^4+12 x^3+e^{-\frac {1}{x^2+2 x}} \left (8 x^2+e^2 \left (4 x^3+16 x^2+8 x-8\right )+8 x\right )}{x^3 (x+2)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 e^{-\frac {1}{x (x+2)}} \left (e^2 x^3+2 \left (1+2 e^2\right ) x^2+2 \left (1+e^2\right ) x-2 e^2\right )}{x^3 (x+2)^2}+3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {e^{2-\frac {1}{x (x+2)}}}{x^3}dx+2 \left (1+2 e^2\right ) \int \frac {e^{-\frac {1}{x (x+2)}}}{x^2}dx+\frac {1}{2} \int \frac {e^{2-\frac {1}{x (x+2)}}}{x}dx-\left (2+e^2\right ) \int \frac {e^{-\frac {1}{x (x+2)}}}{(x+2)^2}dx-\frac {1}{2} \int \frac {e^{2-\frac {1}{x (x+2)}}}{x+2}dx+3 x\) |
Input:
Int[(12*x^3 + 12*x^4 + 3*x^5 + (8*x + 8*x^2 + E^2*(-8 + 8*x + 16*x^2 + 4*x ^3))/E^(2*x + x^2)^(-1))/(4*x^3 + 4*x^4 + x^5),x]
Output:
$Aborted
Time = 1.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90
method | result | size |
risch | \(3 x -\frac {4 \left ({\mathrm e}^{2}-x \right ) {\mathrm e}^{-\frac {1}{x \left (2+x \right )}}}{x}\) | \(27\) |
parallelrisch | \(-\frac {4 \,{\mathrm e}^{2} {\mathrm e}^{-\frac {1}{x \left (2+x \right )}}-3 x^{2}-4 \,{\mathrm e}^{-\frac {1}{x \left (2+x \right )}} x +24 x}{x}\) | \(46\) |
parts | \(3 x +\frac {\left (-4 \,{\mathrm e}^{2}+8\right ) x^{2} {\mathrm e}^{-\frac {1}{x^{2}+2 x}}+4 x^{3} {\mathrm e}^{-\frac {1}{x^{2}+2 x}}-8 \,{\mathrm e}^{2} {\mathrm e}^{-\frac {1}{x^{2}+2 x}} x}{x^{2} \left (2+x \right )}\) | \(75\) |
norman | \(\frac {-12 x^{2}+\left (-4 \,{\mathrm e}^{2}+8\right ) x^{2} {\mathrm e}^{-\frac {1}{x^{2}+2 x}}+3 x^{4}+4 x^{3} {\mathrm e}^{-\frac {1}{x^{2}+2 x}}-8 \,{\mathrm e}^{2} {\mathrm e}^{-\frac {1}{x^{2}+2 x}} x}{x^{2} \left (2+x \right )}\) | \(81\) |
Input:
int((((4*x^3+16*x^2+8*x-8)*exp(1)^2+8*x^2+8*x)*exp(-1/(x^2+2*x))+3*x^5+12* x^4+12*x^3)/(x^5+4*x^4+4*x^3),x,method=_RETURNVERBOSE)
Output:
3*x-4*(exp(2)-x)/x*exp(-1/x/(2+x))
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=\frac {3 \, x^{2} + 4 \, {\left (x - e^{2}\right )} e^{\left (-\frac {1}{x^{2} + 2 \, x}\right )}}{x} \] Input:
integrate((((4*x^3+16*x^2+8*x-8)*exp(1)^2+8*x^2+8*x)*exp(-1/(x^2+2*x))+3*x ^5+12*x^4+12*x^3)/(x^5+4*x^4+4*x^3),x, algorithm="fricas")
Output:
(3*x^2 + 4*(x - e^2)*e^(-1/(x^2 + 2*x)))/x
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=3 x + \frac {\left (4 x - 4 e^{2}\right ) e^{- \frac {1}{x^{2} + 2 x}}}{x} \] Input:
integrate((((4*x**3+16*x**2+8*x-8)*exp(1)**2+8*x**2+8*x)*exp(-1/(x**2+2*x) )+3*x**5+12*x**4+12*x**3)/(x**5+4*x**4+4*x**3),x)
Output:
3*x + (4*x - 4*exp(2))*exp(-1/(x**2 + 2*x))/x
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=3 \, x + \frac {4 \, {\left (x - e^{2}\right )} e^{\left (\frac {1}{2 \, {\left (x + 2\right )}} - \frac {1}{2 \, x}\right )}}{x} \] Input:
integrate((((4*x^3+16*x^2+8*x-8)*exp(1)^2+8*x^2+8*x)*exp(-1/(x^2+2*x))+3*x ^5+12*x^4+12*x^3)/(x^5+4*x^4+4*x^3),x, algorithm="maxima")
Output:
3*x + 4*(x - e^2)*e^(1/2/(x + 2) - 1/2/x)/x
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=\frac {3 \, x^{2} - 4 \, e^{\left (\frac {2 \, x^{2} + 4 \, x - 1}{x^{2} + 2 \, x}\right )}}{x} + 4 \, e^{\left (-\frac {1}{x^{2} + 2 \, x}\right )} \] Input:
integrate((((4*x^3+16*x^2+8*x-8)*exp(1)^2+8*x^2+8*x)*exp(-1/(x^2+2*x))+3*x ^5+12*x^4+12*x^3)/(x^5+4*x^4+4*x^3),x, algorithm="giac")
Output:
(3*x^2 - 4*e^((2*x^2 + 4*x - 1)/(x^2 + 2*x)))/x + 4*e^(-1/(x^2 + 2*x))
Time = 2.97 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=3\,x+4\,{\mathrm {e}}^{-\frac {1}{x^2+2\,x}}-\frac {4\,{\mathrm {e}}^{-\frac {1}{x^2+2\,x}}\,{\mathrm {e}}^2}{x} \] Input:
int((exp(-1/(2*x + x^2))*(8*x + exp(2)*(8*x + 16*x^2 + 4*x^3 - 8) + 8*x^2) + 12*x^3 + 12*x^4 + 3*x^5)/(4*x^3 + 4*x^4 + x^5),x)
Output:
3*x + 4*exp(-1/(2*x + x^2)) - (4*exp(-1/(2*x + x^2))*exp(2))/x
Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=\frac {3 e^{\frac {1}{x^{2}+2 x}} x^{2}-4 e^{2}+4 x}{e^{\frac {1}{x^{2}+2 x}} x} \] Input:
int((((4*x^3+16*x^2+8*x-8)*exp(1)^2+8*x^2+8*x)*exp(-1/(x^2+2*x))+3*x^5+12* x^4+12*x^3)/(x^5+4*x^4+4*x^3),x)
Output:
(3*e**(1/(x**2 + 2*x))*x**2 - 4*e**2 + 4*x)/(e**(1/(x**2 + 2*x))*x)