\(\int \frac {-4 x+x^3+e^x (4-x^2)+(-4+x^2) \log (\frac {1}{x})+(2 x-2 x^2+2 e^x x^2) \log (e^x-x-\log (\frac {1}{x}))}{-e^x x^2+x^3+x^2 \log (\frac {1}{x})} \, dx\) [1819]

Optimal result

Integrand size = 83, antiderivative size = 29 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=-2-\frac {-4+x}{x}+x-\log ^2\left (e^x-x-\log \left (\frac {1}{x}\right )\right ) \]

[Out]

-2-(x-4)/x+x-ln(-ln(1/x)+exp(x)-x)^2

Rubi [F]

\[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=\int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx \]

[In]

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]

[Out]

4/x + x - 2*Defer[Int][Log[E^x - x - Log[x^(-1)]], x] + 2*Defer[Int][Log[E^x - x - Log[x^(-1)]]/(E^x - x - Log
[x^(-1)]), x] + 2*Defer[Int][Log[E^x - x - Log[x^(-1)]]/(x*(-E^x + x + Log[x^(-1)])), x] + 2*Defer[Int][(x*Log
[E^x - x - Log[x^(-1)]])/(-E^x + x + Log[x^(-1)]), x] + 2*Defer[Int][(Log[x^(-1)]*Log[E^x - x - Log[x^(-1)]])/
(-E^x + x + Log[x^(-1)]), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 \left (1-x+x^2+x \log \left (\frac {1}{x}\right )\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{x \left (-e^x+x+\log \left (\frac {1}{x}\right )\right )}+\frac {-4+x^2-2 x^2 \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{x^2}\right ) \, dx \\ & = 2 \int \frac {\left (1-x+x^2+x \log \left (\frac {1}{x}\right )\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{x \left (-e^x+x+\log \left (\frac {1}{x}\right )\right )} \, dx+\int \frac {-4+x^2-2 x^2 \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{x^2} \, dx \\ & = 2 \int \left (\frac {\log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{e^x-x-\log \left (\frac {1}{x}\right )}+\frac {\log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{x \left (-e^x+x+\log \left (\frac {1}{x}\right )\right )}+\frac {x \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x+x+\log \left (\frac {1}{x}\right )}+\frac {\log \left (\frac {1}{x}\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x+x+\log \left (\frac {1}{x}\right )}\right ) \, dx+\int \left (\frac {-4+x^2}{x^2}-2 \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )\right ) \, dx \\ & = -\left (2 \int \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right ) \, dx\right )+2 \int \frac {\log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{e^x-x-\log \left (\frac {1}{x}\right )} \, dx+2 \int \frac {\log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{x \left (-e^x+x+\log \left (\frac {1}{x}\right )\right )} \, dx+2 \int \frac {x \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x+x+\log \left (\frac {1}{x}\right )} \, dx+2 \int \frac {\log \left (\frac {1}{x}\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x+x+\log \left (\frac {1}{x}\right )} \, dx+\int \frac {-4+x^2}{x^2} \, dx \\ & = -\left (2 \int \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right ) \, dx\right )+2 \int \frac {\log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{e^x-x-\log \left (\frac {1}{x}\right )} \, dx+2 \int \frac {\log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{x \left (-e^x+x+\log \left (\frac {1}{x}\right )\right )} \, dx+2 \int \frac {x \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x+x+\log \left (\frac {1}{x}\right )} \, dx+2 \int \frac {\log \left (\frac {1}{x}\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x+x+\log \left (\frac {1}{x}\right )} \, dx+\int \left (1-\frac {4}{x^2}\right ) \, dx \\ & = \frac {4}{x}+x-2 \int \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right ) \, dx+2 \int \frac {\log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{e^x-x-\log \left (\frac {1}{x}\right )} \, dx+2 \int \frac {\log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{x \left (-e^x+x+\log \left (\frac {1}{x}\right )\right )} \, dx+2 \int \frac {x \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x+x+\log \left (\frac {1}{x}\right )} \, dx+2 \int \frac {\log \left (\frac {1}{x}\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x+x+\log \left (\frac {1}{x}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=\frac {4}{x}+x-\log ^2\left (e^x-x-\log \left (\frac {1}{x}\right )\right ) \]

[In]

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]

[Out]

4/x + x - Log[E^x - x - Log[x^(-1)]]^2

Maple [A] (verified)

Time = 0.61 (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	\(-\frac {-8+2 \ln \left (-\ln \left (\frac {1}{x}\right )+{\mathrm e}^{x}-x \right )^{2} x +2 x \ln \left (x \right )-2 x^{2}+2 x \ln \left (\frac {1}{x}\right )}{2 x}\)	\(43\)

[In]

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)

[Out]

-ln(ln(x)+exp(x)-x)^2+1/x*(x^2+4)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=-\frac {x \log \left (-x + e^{x} - \log \left (\frac {1}{x}\right )\right )^{2} - x^{2} - 4}{x} \]

[In]

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="fricas")

[Out]

-(x*log(-x + e^x - log(1/x))^2 - x^2 - 4)/x

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.59 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=x - \log {\left (- x + e^{x} - \log {\left (\frac {1}{x} \right )} \right )}^{2} + \frac {4}{x} \]

[In]

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)

[Out]

x - log(-x + exp(x) - log(1/x))**2 + 4/x

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=-\frac {x \log \left (-x + e^{x} + \log \left (x\right )\right )^{2} - x^{2} - 4}{x} \]

[In]

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="maxima")

[Out]

-(x*log(-x + e^x + log(x))^2 - x^2 - 4)/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=-\frac {x \log \left (-x + e^{x} + \log \left (x\right )\right )^{2} - x^{2} - 4}{x} \]

[In]

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="giac")

[Out]

-(x*log(-x + e^x + log(x))^2 - x^2 - 4)/x

Mupad [B] (verification not implemented)

Time = 8.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx=x-{\ln \left ({\mathrm {e}}^x-\ln \left (\frac {1}{x}\right )-x\right )}^2+\frac {4}{x} \]

[In]

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)

[Out]

x - log(exp(x) - log(1/x) - x)^2 + 4/x