Integrand size = 83, antiderivative size = 29 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=-2-\frac {-4+x}{x}+x-\log ^2\left (e^x-x-\log \left (\frac {1}{x}\right )\right ) \] Output:
-2-(-4+x)/x+x-ln(-ln(1/x)+exp(x)-x)^2
Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=\frac {4}{x}+x-\log ^2\left (e^x-x-\log \left (\frac {1}{x}\right )\right ) \] Input:
Integrate[(-4*x + x^3 + E^x*(4 - x^2) + (-4 + x^2)*Log[x^(-1)] + (2*x - 2* x^2 + 2*E^x*x^2)*Log[E^x - x - Log[x^(-1)]])/(-(E^x*x^2) + x^3 + x^2*Log[x ^(-1)]),x]
Output:
4/x + x - Log[E^x - x - Log[x^(-1)]]^2
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.
\(\displaystyle \int \frac {x^3+e^x \left (4-x^2\right )+\left (x^2-4\right ) \log \left (\frac {1}{x}\right )+\left (2 e^x x^2-2 x^2+2 x\right ) \log \left (-x+e^x-\log \left (\frac {1}{x}\right )\right )-4 x}{x^3-e^x x^2+x^2 \log \left (\frac {1}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (x^2-x+x \log \left (\frac {1}{x}\right )+1\right ) \log \left (-x+e^x-\log \left (\frac {1}{x}\right )\right )}{x \left (x-e^x+\log \left (\frac {1}{x}\right )\right )}+\frac {x^2-2 x^2 \log \left (-x+e^x-\log \left (\frac {1}{x}\right )\right )-4}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \log \left (-x+e^x-\log \left (\frac {1}{x}\right )\right )dx+2 \int \frac {\log \left (-x+e^x-\log \left (\frac {1}{x}\right )\right )}{-x+e^x-\log \left (\frac {1}{x}\right )}dx+2 \int \frac {\log \left (-x+e^x-\log \left (\frac {1}{x}\right )\right )}{x \left (x-e^x+\log \left (\frac {1}{x}\right )\right )}dx+2 \int \frac {x \log \left (-x+e^x-\log \left (\frac {1}{x}\right )\right )}{x-e^x+\log \left (\frac {1}{x}\right )}dx+2 \int \frac {\log \left (\frac {1}{x}\right ) \log \left (-x+e^x-\log \left (\frac {1}{x}\right )\right )}{x-e^x+\log \left (\frac {1}{x}\right )}dx+x+\frac {4}{x}\) |
Input:
Int[(-4*x + x^3 + E^x*(4 - x^2) + (-4 + x^2)*Log[x^(-1)] + (2*x - 2*x^2 + 2*E^x*x^2)*Log[E^x - x - Log[x^(-1)]])/(-(E^x*x^2) + x^3 + x^2*Log[x^(-1)] ),x]
Output:
$Aborted
Time = 0.77 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-\ln \left (\ln \left (x \right )+{\mathrm e}^{x}-x \right )^{2}+\frac {x^{2}+4}{x}\) | \(24\) |
parallelrisch | \(-\frac {-8+2 \ln \left (-\ln \left (\frac {1}{x}\right )+{\mathrm e}^{x}-x \right )^{2} x +2 x \ln \left (x \right )-2 x^{2}+2 x \ln \left (\frac {1}{x}\right )}{2 x}\) | \(43\) |
Input:
int(((2*exp(x)*x^2-2*x^2+2*x)*ln(-ln(1/x)+exp(x)-x)+(x^2-4)*ln(1/x)+(-x^2+ 4)*exp(x)+x^3-4*x)/(x^2*ln(1/x)-exp(x)*x^2+x^3),x,method=_RETURNVERBOSE)
Output:
-ln(ln(x)+exp(x)-x)^2+1/x*(x^2+4)
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=-\frac {x \log \left (-x + e^{x} - \log \left (\frac {1}{x}\right )\right )^{2} - x^{2} - 4}{x} \] Input:
integrate(((2*exp(x)*x^2-2*x^2+2*x)*log(-log(1/x)+exp(x)-x)+(x^2-4)*log(1/ x)+(-x^2+4)*exp(x)+x^3-4*x)/(x^2*log(1/x)-exp(x)*x^2+x^3),x, algorithm="fr icas")
Output:
-(x*log(-x + e^x - log(1/x))^2 - x^2 - 4)/x
Time = 0.40 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.59 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=x - \log {\left (- x + e^{x} - \log {\left (\frac {1}{x} \right )} \right )}^{2} + \frac {4}{x} \] Input:
integrate(((2*exp(x)*x**2-2*x**2+2*x)*ln(-ln(1/x)+exp(x)-x)+(x**2-4)*ln(1/ x)+(-x**2+4)*exp(x)+x**3-4*x)/(x**2*ln(1/x)-exp(x)*x**2+x**3),x)
Output:
x - log(-x + exp(x) - log(1/x))**2 + 4/x
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=-\frac {x \log \left (-x + e^{x} + \log \left (x\right )\right )^{2} - x^{2} - 4}{x} \] Input:
integrate(((2*exp(x)*x^2-2*x^2+2*x)*log(-log(1/x)+exp(x)-x)+(x^2-4)*log(1/ x)+(-x^2+4)*exp(x)+x^3-4*x)/(x^2*log(1/x)-exp(x)*x^2+x^3),x, algorithm="ma xima")
Output:
-(x*log(-x + e^x + log(x))^2 - x^2 - 4)/x
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=-\frac {x \log \left (-x + e^{x} + \log \left (x\right )\right )^{2} - x^{2} - 4}{x} \] Input:
integrate(((2*exp(x)*x^2-2*x^2+2*x)*log(-log(1/x)+exp(x)-x)+(x^2-4)*log(1/ x)+(-x^2+4)*exp(x)+x^3-4*x)/(x^2*log(1/x)-exp(x)*x^2+x^3),x, algorithm="gi ac")
Output:
-(x*log(-x + e^x + log(x))^2 - x^2 - 4)/x
Time = 1.73 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=x-{\ln \left ({\mathrm {e}}^x-\ln \left (\frac {1}{x}\right )-x\right )}^2+\frac {4}{x} \] Input:
int((log(exp(x) - log(1/x) - x)*(2*x + 2*x^2*exp(x) - 2*x^2) - 4*x - exp(x )*(x^2 - 4) + x^3 + log(1/x)*(x^2 - 4))/(x^2*log(1/x) - x^2*exp(x) + x^3), x)
Output:
x - log(exp(x) - log(1/x) - x)^2 + 4/x
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=\frac {-\mathrm {log}\left (e^{x}+\mathrm {log}\left (x \right )-x \right )^{2} x +x^{2}+4}{x} \] Input:
int(((2*exp(x)*x^2-2*x^2+2*x)*log(-log(1/x)+exp(x)-x)+(x^2-4)*log(1/x)+(-x ^2+4)*exp(x)+x^3-4*x)/(x^2*log(1/x)-exp(x)*x^2+x^3),x)
Output:
( - log(e**x + log(x) - x)**2*x + x**2 + 4)/x