Integrand size = 85, antiderivative size = 31 \[ \int \frac {e^{\frac {4+x-x^2}{x}} (-4-x)-x^3}{\left (e^{\frac {4+x-x^2}{x}} x^2-x^3\right ) \log \left (\frac {e^{e^3+x} \left (-e^{\frac {4+x-x^2}{x}}+x\right )}{x}\right )} \, dx=\log \left (\log \left (\frac {e^{e^3+x} \left (-e^{\frac {4+x-x^2}{x}}+x\right )}{x}\right )\right ) \]
Time = 0.39 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {4+x-x^2}{x}} (-4-x)-x^3}{\left (e^{\frac {4+x-x^2}{x}} x^2-x^3\right ) \log \left (\frac {e^{e^3+x} \left (-e^{\frac {4+x-x^2}{x}}+x\right )}{x}\right )} \, dx=\log \left (\log \left (e^{e^3} \left (e^x-\frac {e^{1+\frac {4}{x}}}{x}\right )\right )\right ) \]
Integrate[(E^((4 + x - x^2)/x)*(-4 - x) - x^3)/((E^((4 + x - x^2)/x)*x^2 - x^3)*Log[(E^(E^3 + x)*(-E^((4 + x - x^2)/x) + x))/x]),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.
\(\displaystyle \int \frac {e^{\frac {-x^2+x+4}{x}} (-x-4)-x^3}{\left (e^{\frac {-x^2+x+4}{x}} x^2-x^3\right ) \log \left (\frac {e^{x+e^3} \left (x-e^{\frac {-x^2+x+4}{x}}\right )}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {1}{\log \left (e^{e^3} \left (e^x-\frac {e^{\frac {4}{x}+1}}{x}\right )\right )}-\frac {e^{\frac {4}{x}+1} \left (x^2+x+4\right )}{x^2 \left (e^{\frac {4}{x}+1}-e^x x\right ) \log \left (e^{e^3} \left (e^x-\frac {e^{\frac {4}{x}+1}}{x}\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \frac {e^{1+\frac {4}{x}}}{x^2 \left (e^x x-e^{1+\frac {4}{x}}\right ) \log \left (e^{e^3} \left (e^x-\frac {e^{1+\frac {4}{x}}}{x}\right )\right )}dx+\int \frac {1}{\log \left (e^{e^3} \left (e^x-\frac {e^{1+\frac {4}{x}}}{x}\right )\right )}dx-\int \frac {e^{1+\frac {4}{x}}}{\left (e^{1+\frac {4}{x}}-e^x x\right ) \log \left (e^{e^3} \left (e^x-\frac {e^{1+\frac {4}{x}}}{x}\right )\right )}dx+\int \frac {e^{1+\frac {4}{x}}}{x \left (e^x x-e^{1+\frac {4}{x}}\right ) \log \left (e^{e^3} \left (e^x-\frac {e^{1+\frac {4}{x}}}{x}\right )\right )}dx\) |
Int[(E^((4 + x - x^2)/x)*(-4 - x) - x^3)/((E^((4 + x - x^2)/x)*x^2 - x^3)* Log[(E^(E^3 + x)*(-E^((4 + x - x^2)/x) + x))/x]),x]
3.18.43.3.1 Defintions of rubi rules used
Time = 0.58 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(\ln \left (\ln \left (\frac {\left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right ) {\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{3}}}{x}\right )\right )\) | \(30\) |
risch | \(\ln \left (\ln \left ({\mathrm e}^{x}\right )-\frac {i \left (-\pi \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right )\right )^{2}+\pi \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )+\pi \operatorname {csgn}\left (i {\mathrm e}^{x} \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right )\right )^{3}-\pi \operatorname {csgn}\left (i {\mathrm e}^{x} \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{x} \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right )\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right ) {\mathrm e}^{x}}{x}\right )}^{2}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{x} \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right ) {\mathrm e}^{x}}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )+\pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right ) {\mathrm e}^{x}}{x}\right )}^{3}-\pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right ) {\mathrm e}^{x}}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 i {\mathrm e}^{3}-2 i \ln \left (x \right )+2 i \ln \left (-{\mathrm e}^{-\frac {x^{2}-x -4}{x}}+x \right )\right )}{2}\right )\) | \(396\) |
int(((-4-x)*exp((-x^2+x+4)/x)-x^3)/(x^2*exp((-x^2+x+4)/x)-x^3)/ln((-exp((- x^2+x+4)/x)+x)*exp(x)*exp(exp(3))/x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {4+x-x^2}{x}} (-4-x)-x^3}{\left (e^{\frac {4+x-x^2}{x}} x^2-x^3\right ) \log \left (\frac {e^{e^3+x} \left (-e^{\frac {4+x-x^2}{x}}+x\right )}{x}\right )} \, dx=\log \left (\log \left (\frac {{\left (x - e^{\left (-\frac {x^{2} - x - 4}{x}\right )}\right )} e^{\left (x + e^{3}\right )}}{x}\right )\right ) \]
integrate(((-4-x)*exp((-x^2+x+4)/x)-x^3)/(x^2*exp((-x^2+x+4)/x)-x^3)/log(( -exp((-x^2+x+4)/x)+x)*exp(x)*exp(exp(3))/x),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {4+x-x^2}{x}} (-4-x)-x^3}{\left (e^{\frac {4+x-x^2}{x}} x^2-x^3\right ) \log \left (\frac {e^{e^3+x} \left (-e^{\frac {4+x-x^2}{x}}+x\right )}{x}\right )} \, dx=\log {\left (\log {\left (\frac {\left (x - e^{\frac {- x^{2} + x + 4}{x}}\right ) e^{x} e^{e^{3}}}{x} \right )} \right )} \]
integrate(((-4-x)*exp((-x**2+x+4)/x)-x**3)/(x**2*exp((-x**2+x+4)/x)-x**3)/ ln((-exp((-x**2+x+4)/x)+x)*exp(x)*exp(exp(3))/x),x)
Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {4+x-x^2}{x}} (-4-x)-x^3}{\left (e^{\frac {4+x-x^2}{x}} x^2-x^3\right ) \log \left (\frac {e^{e^3+x} \left (-e^{\frac {4+x-x^2}{x}}+x\right )}{x}\right )} \, dx=\log \left (e^{3} + \log \left (x e^{x} - e^{\left (\frac {4}{x} + 1\right )}\right ) - \log \left (x\right )\right ) \]
integrate(((-4-x)*exp((-x^2+x+4)/x)-x^3)/(x^2*exp((-x^2+x+4)/x)-x^3)/log(( -exp((-x^2+x+4)/x)+x)*exp(x)*exp(exp(3))/x),x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {4+x-x^2}{x}} (-4-x)-x^3}{\left (e^{\frac {4+x-x^2}{x}} x^2-x^3\right ) \log \left (\frac {e^{e^3+x} \left (-e^{\frac {4+x-x^2}{x}}+x\right )}{x}\right )} \, dx=\log \left (\log \left (\frac {x e^{\left (x + e^{3}\right )} - e^{\left (\frac {x e^{3} + x + 4}{x}\right )}}{x}\right )\right ) \]
integrate(((-4-x)*exp((-x^2+x+4)/x)-x^3)/(x^2*exp((-x^2+x+4)/x)-x^3)/log(( -exp((-x^2+x+4)/x)+x)*exp(x)*exp(exp(3))/x),x, algorithm=\
Time = 10.82 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {4+x-x^2}{x}} (-4-x)-x^3}{\left (e^{\frac {4+x-x^2}{x}} x^2-x^3\right ) \log \left (\frac {e^{e^3+x} \left (-e^{\frac {4+x-x^2}{x}}+x\right )}{x}\right )} \, dx=\ln \left (\ln \left ({\mathrm {e}}^{{\mathrm {e}}^3}\,{\mathrm {e}}^x-\frac {\mathrm {e}\,{\mathrm {e}}^{4/x}\,{\mathrm {e}}^{{\mathrm {e}}^3}}{x}\right )\right ) \]