Integrand size = 120, antiderivative size = 26 \[ \int \frac {4 x^2+e^{\frac {-4+x-x^2}{x}} \left (-16+4 x^2\right )+\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )}{\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )} \, dx=x-\log \left (\log ^4\left (e^{\frac {-4+x-x^2}{x}}-x\right )\right ) \] Output:
x-ln(ln(exp((-x^2+x-4)/x)-x)^4)
Time = 0.46 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {4 x^2+e^{\frac {-4+x-x^2}{x}} \left (-16+4 x^2\right )+\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )}{\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )} \, dx=x-4 \log \left (\log \left (e^{1-\frac {4}{x}-x}-x\right )\right ) \] Input:
Integrate[(4*x^2 + E^((-4 + x - x^2)/x)*(-16 + 4*x^2) + (E^((-4 + x - x^2) /x)*x^2 - x^3)*Log[E^((-4 + x - x^2)/x) - x])/((E^((-4 + x - x^2)/x)*x^2 - x^3)*Log[E^((-4 + x - x^2)/x) - x]),x]
Output:
x - 4*Log[Log[E^(1 - 4/x - x) - x]]
Time = 0.96 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {7239, 2009}
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.
\(\displaystyle \int \frac {4 x^2+e^{\frac {-x^2+x-4}{x}} \left (4 x^2-16\right )+\left (e^{\frac {-x^2+x-4}{x}} x^2-x^3\right ) \log \left (e^{\frac {-x^2+x-4}{x}}-x\right )}{\left (e^{\frac {-x^2+x-4}{x}} x^2-x^3\right ) \log \left (e^{\frac {-x^2+x-4}{x}}-x\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (1-\frac {4 \left (e^{x+\frac {4}{x}} x^2+e \left (x^2-4\right )\right )}{x^2 \left (e^{x+\frac {4}{x}} x-e\right ) \log \left (e^{-x-\frac {4}{x}+1}-x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x-4 \log \left (\log \left (e^{-x-\frac {4}{x}+1}-x\right )\right )\) |
Input:
Int[(4*x^2 + E^((-4 + x - x^2)/x)*(-16 + 4*x^2) + (E^((-4 + x - x^2)/x)*x^ 2 - x^3)*Log[E^((-4 + x - x^2)/x) - x])/((E^((-4 + x - x^2)/x)*x^2 - x^3)* Log[E^((-4 + x - x^2)/x) - x]),x]
Output:
x - 4*Log[Log[E^(1 - 4/x - x) - x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
norman | \(x -4 \ln \left (\ln \left ({\mathrm e}^{\frac {-x^{2}+x -4}{x}}-x \right )\right )\) | \(24\) |
risch | \(x -4 \ln \left (\ln \left ({\mathrm e}^{-\frac {x^{2}-x +4}{x}}-x \right )\right )\) | \(25\) |
parallelrisch | \(x -4 \ln \left (\ln \left ({\mathrm e}^{-\frac {x^{2}-x +4}{x}}-x \right )\right )\) | \(25\) |
Input:
int(((x^2*exp((-x^2+x-4)/x)-x^3)*ln(exp((-x^2+x-4)/x)-x)+(4*x^2-16)*exp((- x^2+x-4)/x)+4*x^2)/(x^2*exp((-x^2+x-4)/x)-x^3)/ln(exp((-x^2+x-4)/x)-x),x,m ethod=_RETURNVERBOSE)
Output:
x-4*ln(ln(exp((-x^2+x-4)/x)-x))
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {4 x^2+e^{\frac {-4+x-x^2}{x}} \left (-16+4 x^2\right )+\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )}{\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )} \, dx=x - 4 \, \log \left (\log \left (-x + e^{\left (-\frac {x^{2} - x + 4}{x}\right )}\right )\right ) \] Input:
integrate(((x^2*exp((-x^2+x-4)/x)-x^3)*log(exp((-x^2+x-4)/x)-x)+(4*x^2-16) *exp((-x^2+x-4)/x)+4*x^2)/(x^2*exp((-x^2+x-4)/x)-x^3)/log(exp((-x^2+x-4)/x )-x),x, algorithm="fricas")
Output:
x - 4*log(log(-x + e^(-(x^2 - x + 4)/x)))
Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {4 x^2+e^{\frac {-4+x-x^2}{x}} \left (-16+4 x^2\right )+\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )}{\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )} \, dx=x - 4 \log {\left (\log {\left (- x + e^{\frac {- x^{2} + x - 4}{x}} \right )} \right )} \] Input:
integrate(((x**2*exp((-x**2+x-4)/x)-x**3)*ln(exp((-x**2+x-4)/x)-x)+(4*x**2 -16)*exp((-x**2+x-4)/x)+4*x**2)/(x**2*exp((-x**2+x-4)/x)-x**3)/ln(exp((-x* *2+x-4)/x)-x),x)
Output:
x - 4*log(log(-x + exp((-x**2 + x - 4)/x)))
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {4 x^2+e^{\frac {-4+x-x^2}{x}} \left (-16+4 x^2\right )+\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )}{\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )} \, dx=x - 4 \, \log \left (-\frac {x^{2} - x \log \left (-x e^{\left (x + \frac {4}{x}\right )} + e\right ) + 4}{x}\right ) \] Input:
integrate(((x^2*exp((-x^2+x-4)/x)-x^3)*log(exp((-x^2+x-4)/x)-x)+(4*x^2-16) *exp((-x^2+x-4)/x)+4*x^2)/(x^2*exp((-x^2+x-4)/x)-x^3)/log(exp((-x^2+x-4)/x )-x),x, algorithm="maxima")
Output:
x - 4*log(-(x^2 - x*log(-x*e^(x + 4/x) + e) + 4)/x)
Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {4 x^2+e^{\frac {-4+x-x^2}{x}} \left (-16+4 x^2\right )+\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )}{\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )} \, dx=x - 4 \, \log \left (\log \left (-x + e^{\left (-\frac {x^{2} - x + 4}{x}\right )}\right )\right ) \] Input:
integrate(((x^2*exp((-x^2+x-4)/x)-x^3)*log(exp((-x^2+x-4)/x)-x)+(4*x^2-16) *exp((-x^2+x-4)/x)+4*x^2)/(x^2*exp((-x^2+x-4)/x)-x^3)/log(exp((-x^2+x-4)/x )-x),x, algorithm="giac")
Output:
x - 4*log(log(-x + e^(-(x^2 - x + 4)/x)))
Time = 3.49 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {4 x^2+e^{\frac {-4+x-x^2}{x}} \left (-16+4 x^2\right )+\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )}{\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )} \, dx=x-4\,\ln \left (\ln \left ({\mathrm {e}}^{-x}\,\mathrm {e}\,{\mathrm {e}}^{-\frac {4}{x}}-x\right )\right ) \] Input:
int(-(exp(-(x^2 - x + 4)/x)*(4*x^2 - 16) - log(exp(-(x^2 - x + 4)/x) - x)* (x^3 - x^2*exp(-(x^2 - x + 4)/x)) + 4*x^2)/(log(exp(-(x^2 - x + 4)/x) - x) *(x^3 - x^2*exp(-(x^2 - x + 4)/x))),x)
Output:
x - 4*log(log(exp(-x)*exp(1)*exp(-4/x) - x))
\[ \int \frac {4 x^2+e^{\frac {-4+x-x^2}{x}} \left (-16+4 x^2\right )+\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )}{\left (e^{\frac {-4+x-x^2}{x}} x^2-x^3\right ) \log \left (e^{\frac {-4+x-x^2}{x}}-x\right )} \, dx=\int \frac {\left (x^{2} {\mathrm e}^{\frac {-x^{2}+x -4}{x}}-x^{3}\right ) \mathrm {log}\left ({\mathrm e}^{\frac {-x^{2}+x -4}{x}}-x \right )+\left (4 x^{2}-16\right ) {\mathrm e}^{\frac {-x^{2}+x -4}{x}}+4 x^{2}}{\left (x^{2} {\mathrm e}^{\frac {-x^{2}+x -4}{x}}-x^{3}\right ) \mathrm {log}\left ({\mathrm e}^{\frac {-x^{2}+x -4}{x}}-x \right )}d x \] Input:
int(((x^2*exp((-x^2+x-4)/x)-x^3)*log(exp((-x^2+x-4)/x)-x)+(4*x^2-16)*exp(( -x^2+x-4)/x)+4*x^2)/(x^2*exp((-x^2+x-4)/x)-x^3)/log(exp((-x^2+x-4)/x)-x),x )
Output:
int(((x^2*exp((-x^2+x-4)/x)-x^3)*log(exp((-x^2+x-4)/x)-x)+(4*x^2-16)*exp(( -x^2+x-4)/x)+4*x^2)/(x^2*exp((-x^2+x-4)/x)-x^3)/log(exp((-x^2+x-4)/x)-x),x )