Integrand size = 85, antiderivative size = 31 \[ \int \frac {12+22 x^2-2 x^3+e^{e^{-4+x}} \left (-x^2+2 x^3+e^{-4+x} \left (-x^3+x^4\right )\right )}{-12 x-9 x^2+22 x^3-x^4+e^{e^{-4+x}} \left (-x^3+x^4\right )} \, dx=\log (1-x)+\log \left (6+3 \left (5+\frac {4}{x}\right )-x+e^{e^{-4+x}} x\right ) \] Output:
ln(1-x)+ln(21+12/x-x+x*exp(exp(-4+x)))
Time = 0.50 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {12+22 x^2-2 x^3+e^{e^{-4+x}} \left (-x^2+2 x^3+e^{-4+x} \left (-x^3+x^4\right )\right )}{-12 x-9 x^2+22 x^3-x^4+e^{e^{-4+x}} \left (-x^3+x^4\right )} \, dx=\frac {e^4 \log (1-x)-e^4 \log (x)+e^4 \log \left (12+21 x-x^2+e^{e^{-4+x}} x^2\right )}{e^4} \] Input:
Integrate[(12 + 22*x^2 - 2*x^3 + E^E^(-4 + x)*(-x^2 + 2*x^3 + E^(-4 + x)*( -x^3 + x^4)))/(-12*x - 9*x^2 + 22*x^3 - x^4 + E^E^(-4 + x)*(-x^3 + x^4)),x ]
Output:
(E^4*Log[1 - x] - E^4*Log[x] + E^4*Log[12 + 21*x - x^2 + E^E^(-4 + x)*x^2] )/E^4
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.
\(\displaystyle \int \frac {-2 x^3+22 x^2+e^{e^{x-4}} \left (2 x^3-x^2+e^{x-4} \left (x^4-x^3\right )\right )+12}{-x^4+22 x^3-9 x^2+e^{e^{x-4}} \left (x^4-x^3\right )-12 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 x^3-22 x^2-e^{e^{x-4}} \left (2 x^3-x^2+e^{x-4} \left (x^4-x^3\right )\right )-12}{(1-x) x \left (e^{e^{x-4}} x^2-x^2+21 x+12\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x+e^{x-4}-4} x^2}{e^{e^{x-4}} x^2-x^2+21 x+12}+\frac {2 e^{e^{x-4}} x^2}{(x-1) \left (e^{e^{x-4}} x^2-x^2+21 x+12\right )}-\frac {2 x^2}{(x-1) \left (e^{e^{x-4}} x^2-x^2+21 x+12\right )}-\frac {e^{e^{x-4}} x}{(x-1) \left (e^{e^{x-4}} x^2-x^2+21 x+12\right )}+\frac {22 x}{(x-1) \left (e^{e^{x-4}} x^2-x^2+21 x+12\right )}+\frac {12}{(x-1) \left (e^{e^{x-4}} x^2-x^2+21 x+12\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 20 \int \frac {1}{e^{e^{x-4}} x^2-x^2+21 x+12}dx+\int \frac {e^{e^{x-4}}}{e^{e^{x-4}} x^2-x^2+21 x+12}dx+32 \int \frac {1}{(x-1) \left (e^{e^{x-4}} x^2-x^2+21 x+12\right )}dx+\int \frac {e^{e^{x-4}}}{(x-1) \left (e^{e^{x-4}} x^2-x^2+21 x+12\right )}dx-12 \int \frac {1}{x \left (e^{e^{x-4}} x^2-x^2+21 x+12\right )}dx-2 \int \frac {x}{e^{e^{x-4}} x^2-x^2+21 x+12}dx+2 \int \frac {e^{e^{x-4}} x}{e^{e^{x-4}} x^2-x^2+21 x+12}dx+\int \frac {e^{x+e^{x-4}-4} x^2}{e^{e^{x-4}} x^2-x^2+21 x+12}dx\) |
Input:
Int[(12 + 22*x^2 - 2*x^3 + E^E^(-4 + x)*(-x^2 + 2*x^3 + E^(-4 + x)*(-x^3 + x^4)))/(-12*x - 9*x^2 + 22*x^3 - x^4 + E^E^(-4 + x)*(-x^3 + x^4)),x]
Output:
$Aborted
Time = 0.75 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97
method | result | size |
norman | \(-\ln \left (x \right )+\ln \left (-1+x \right )+\ln \left ({\mathrm e}^{{\mathrm e}^{x -4}} x^{2}-x^{2}+21 x +12\right )\) | \(30\) |
risch | \(\ln \left (x^{2}-x \right )+\ln \left ({\mathrm e}^{{\mathrm e}^{x -4}}-\frac {x^{2}-21 x -12}{x^{2}}\right )\) | \(30\) |
parallelrisch | \(-\ln \left (x \right )+\ln \left (-1+x \right )+\ln \left ({\mathrm e}^{{\mathrm e}^{x -4}} x^{2}-x^{2}+21 x +12\right )\) | \(30\) |
Input:
int((((x^4-x^3)*exp(x-4)+2*x^3-x^2)*exp(exp(x-4))-2*x^3+22*x^2+12)/((x^4-x ^3)*exp(exp(x-4))-x^4+22*x^3-9*x^2-12*x),x,method=_RETURNVERBOSE)
Output:
-ln(x)+ln(-1+x)+ln(exp(exp(x-4))*x^2-x^2+21*x+12)
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {12+22 x^2-2 x^3+e^{e^{-4+x}} \left (-x^2+2 x^3+e^{-4+x} \left (-x^3+x^4\right )\right )}{-12 x-9 x^2+22 x^3-x^4+e^{e^{-4+x}} \left (-x^3+x^4\right )} \, dx=\log \left (x^{2} - x\right ) + \log \left (\frac {x^{2} e^{\left (e^{\left (x - 4\right )}\right )} - x^{2} + 21 \, x + 12}{x^{2}}\right ) \] Input:
integrate((((x^4-x^3)*exp(-4+x)+2*x^3-x^2)*exp(exp(-4+x))-2*x^3+22*x^2+12) /((x^4-x^3)*exp(exp(-4+x))-x^4+22*x^3-9*x^2-12*x),x, algorithm="fricas")
Output:
log(x^2 - x) + log((x^2*e^(e^(x - 4)) - x^2 + 21*x + 12)/x^2)
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {12+22 x^2-2 x^3+e^{e^{-4+x}} \left (-x^2+2 x^3+e^{-4+x} \left (-x^3+x^4\right )\right )}{-12 x-9 x^2+22 x^3-x^4+e^{e^{-4+x}} \left (-x^3+x^4\right )} \, dx=\log {\left (x^{2} - x \right )} + \log {\left (e^{e^{x - 4}} + \frac {- x^{2} + 21 x + 12}{x^{2}} \right )} \] Input:
integrate((((x**4-x**3)*exp(-4+x)+2*x**3-x**2)*exp(exp(-4+x))-2*x**3+22*x* *2+12)/((x**4-x**3)*exp(exp(-4+x))-x**4+22*x**3-9*x**2-12*x),x)
Output:
log(x**2 - x) + log(exp(exp(x - 4)) + (-x**2 + 21*x + 12)/x**2)
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {12+22 x^2-2 x^3+e^{e^{-4+x}} \left (-x^2+2 x^3+e^{-4+x} \left (-x^3+x^4\right )\right )}{-12 x-9 x^2+22 x^3-x^4+e^{e^{-4+x}} \left (-x^3+x^4\right )} \, dx=\log \left (x - 1\right ) + \log \left (x\right ) + \log \left (\frac {x^{2} e^{\left (e^{\left (x - 4\right )}\right )} - x^{2} + 21 \, x + 12}{x^{2}}\right ) \] Input:
integrate((((x^4-x^3)*exp(-4+x)+2*x^3-x^2)*exp(exp(-4+x))-2*x^3+22*x^2+12) /((x^4-x^3)*exp(exp(-4+x))-x^4+22*x^3-9*x^2-12*x),x, algorithm="maxima")
Output:
log(x - 1) + log(x) + log((x^2*e^(e^(x - 4)) - x^2 + 21*x + 12)/x^2)
\[ \int \frac {12+22 x^2-2 x^3+e^{e^{-4+x}} \left (-x^2+2 x^3+e^{-4+x} \left (-x^3+x^4\right )\right )}{-12 x-9 x^2+22 x^3-x^4+e^{e^{-4+x}} \left (-x^3+x^4\right )} \, dx=\int { \frac {2 \, x^{3} - 22 \, x^{2} - {\left (2 \, x^{3} - x^{2} + {\left (x^{4} - x^{3}\right )} e^{\left (x - 4\right )}\right )} e^{\left (e^{\left (x - 4\right )}\right )} - 12}{x^{4} - 22 \, x^{3} + 9 \, x^{2} - {\left (x^{4} - x^{3}\right )} e^{\left (e^{\left (x - 4\right )}\right )} + 12 \, x} \,d x } \] Input:
integrate((((x^4-x^3)*exp(-4+x)+2*x^3-x^2)*exp(exp(-4+x))-2*x^3+22*x^2+12) /((x^4-x^3)*exp(exp(-4+x))-x^4+22*x^3-9*x^2-12*x),x, algorithm="giac")
Output:
integrate((2*x^3 - 22*x^2 - (2*x^3 - x^2 + (x^4 - x^3)*e^(x - 4))*e^(e^(x - 4)) - 12)/(x^4 - 22*x^3 + 9*x^2 - (x^4 - x^3)*e^(e^(x - 4)) + 12*x), x)
Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {12+22 x^2-2 x^3+e^{e^{-4+x}} \left (-x^2+2 x^3+e^{-4+x} \left (-x^3+x^4\right )\right )}{-12 x-9 x^2+22 x^3-x^4+e^{e^{-4+x}} \left (-x^3+x^4\right )} \, dx=\ln \left (x\,\left (x-1\right )\right )+\ln \left (\frac {21\,x-x^2+x^2\,{\mathrm {e}}^{{\mathrm {e}}^{x-4}}+12}{x^2}\right ) \] Input:
int((exp(exp(x - 4))*(exp(x - 4)*(x^3 - x^4) + x^2 - 2*x^3) - 22*x^2 + 2*x ^3 - 12)/(12*x + exp(exp(x - 4))*(x^3 - x^4) + 9*x^2 - 22*x^3 + x^4),x)
Output:
log(x*(x - 1)) + log((21*x - x^2 + x^2*exp(exp(x - 4)) + 12)/x^2)
Time = 0.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {12+22 x^2-2 x^3+e^{e^{-4+x}} \left (-x^2+2 x^3+e^{-4+x} \left (-x^3+x^4\right )\right )}{-12 x-9 x^2+22 x^3-x^4+e^{e^{-4+x}} \left (-x^3+x^4\right )} \, dx=\mathrm {log}\left (e^{\frac {e^{x}}{e^{4}}} x^{2}-x^{2}+21 x +12\right )+\mathrm {log}\left (x -1\right )-\mathrm {log}\left (x \right ) \] Input:
int((((x^4-x^3)*exp(-4+x)+2*x^3-x^2)*exp(exp(-4+x))-2*x^3+22*x^2+12)/((x^4 -x^3)*exp(exp(-4+x))-x^4+22*x^3-9*x^2-12*x),x)
Output:
log(e**(e**x/e**4)*x**2 - x**2 + 21*x + 12) + log(x - 1) - log(x)