\(\int \frac {-1+x+2 x^2-6 x^3+4 x^4}{-6 x+x^2+x \log (\frac {1}{5} e^{x^2-2 x^3+x^4})+x \log (\frac {2}{x})} \, dx\) [2803]

Optimal result

Integrand size = 57, antiderivative size = 28 \[ \int \frac {-1+x+2 x^2-6 x^3+4 x^4}{-6 x+x^2+x \log \left (\frac {1}{5} e^{x^2-2 x^3+x^4}\right )+x \log \left (\frac {2}{x}\right )} \, dx=-1+\log \left (-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )\right ) \]

[Out]

-1+ln(ln(2/x)+ln(1/5*exp(x^2*(-1+x)^2))+x-6)

Rubi [F]

\[ \int \frac {-1+x+2 x^2-6 x^3+4 x^4}{-6 x+x^2+x \log \left (\frac {1}{5} e^{x^2-2 x^3+x^4}\right )+x \log \left (\frac {2}{x}\right )} \, dx=\int \frac {-1+x+2 x^2-6 x^3+4 x^4}{-6 x+x^2+x \log \left (\frac {1}{5} e^{x^2-2 x^3+x^4}\right )+x \log \left (\frac {2}{x}\right )} \, dx \]

[In]

Int[(-1 + x + 2*x^2 - 6*x^3 + 4*x^4)/(-6*x + x^2 + x*Log[E^(x^2 - 2*x^3 + x^4)/5] + x*Log[2/x]),x]

[Out]

Defer[Int][(-6 + x + Log[E^((-1 + x)^2*x^2)/5] + Log[2/x])^(-1), x] - Defer[Int][1/(x*(-6 + x + Log[E^((-1 + x
)^2*x^2)/5] + Log[2/x])), x] + 2*Defer[Int][x/(-6 + x + Log[E^((-1 + x)^2*x^2)/5] + Log[2/x]), x] - 6*Defer[In
t][x^2/(-6 + x + Log[E^((-1 + x)^2*x^2)/5] + Log[2/x]), x] + 4*Defer[Int][x^3/(-6 + x + Log[E^((-1 + x)^2*x^2)
/5] + Log[2/x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )}-\frac {1}{x \left (-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )\right )}+\frac {2 x}{-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )}-\frac {6 x^2}{-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )}+\frac {4 x^3}{-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )}\right ) \, dx \\ & = 2 \int \frac {x}{-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )} \, dx+4 \int \frac {x^3}{-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )} \, dx-6 \int \frac {x^2}{-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )} \, dx+\int \frac {1}{-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )} \, dx-\int \frac {1}{x \left (-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {-1+x+2 x^2-6 x^3+4 x^4}{-6 x+x^2+x \log \left (\frac {1}{5} e^{x^2-2 x^3+x^4}\right )+x \log \left (\frac {2}{x}\right )} \, dx=\log \left (-6+x+\log \left (\frac {1}{5} e^{(-1+x)^2 x^2}\right )+\log \left (\frac {2}{x}\right )\right ) \]

[In]

Integrate[(-1 + x + 2*x^2 - 6*x^3 + 4*x^4)/(-6*x + x^2 + x*Log[E^(x^2 - 2*x^3 + x^4)/5] + x*Log[2/x]),x]

[Out]

Log[-6 + x + Log[E^((-1 + x)^2*x^2)/5] + Log[2/x]]

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96

method	result	size
default	\(\ln \left (\ln \left (\frac {{\mathrm e}^{x^{4}-2 x^{3}+x^{2}}}{5}\right )+\ln \left (\frac {2}{x}\right )+x -6\right )\)	\(27\)
parallelrisch	\(\ln \left (\ln \left (\frac {{\mathrm e}^{x^{2} \left (x^{2}-2 x +1\right )}}{5}\right )+\ln \left (\frac {2}{x}\right )+x -6\right )\)	\(27\)
risch	\(\ln \left (\ln \left ({\mathrm e}^{x^{2} \left (-1+x \right )^{2}}\right )+\frac {i \left (2 i \ln \left (5\right )-2 i \ln \left (2\right )-2 i x +2 i \ln \left (x \right )+12 i\right )}{2}\right )\)	\(39\)

[In]

int((4*x^4-6*x^3+2*x^2+x-1)/(x*ln(1/5*exp(x^4-2*x^3+x^2))+x*ln(2/x)+x^2-6*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(1/5*exp(x^4-2*x^3+x^2))+ln(2/x)+x-6)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-1+x+2 x^2-6 x^3+4 x^4}{-6 x+x^2+x \log \left (\frac {1}{5} e^{x^2-2 x^3+x^4}\right )+x \log \left (\frac {2}{x}\right )} \, dx=\log \left (x^{4} - 2 \, x^{3} + x^{2} + x - \log \left (5\right ) + \log \left (\frac {2}{x}\right ) - 6\right ) \]

[In]

integrate((4*x^4-6*x^3+2*x^2+x-1)/(x*log(1/5*exp(x^4-2*x^3+x^2))+x*log(2/x)+x^2-6*x),x, algorithm="fricas")

[Out]

log(x^4 - 2*x^3 + x^2 + x - log(5) + log(2/x) - 6)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-1+x+2 x^2-6 x^3+4 x^4}{-6 x+x^2+x \log \left (\frac {1}{5} e^{x^2-2 x^3+x^4}\right )+x \log \left (\frac {2}{x}\right )} \, dx=\log {\left (x^{4} - 2 x^{3} + x^{2} + x + \log {\left (\frac {2}{x} \right )} - 6 - \log {\left (5 \right )} \right )} \]

[In]

integrate((4*x**4-6*x**3+2*x**2+x-1)/(x*ln(1/5*exp(x**4-2*x**3+x**2))+x*ln(2/x)+x**2-6*x),x)

[Out]

log(x**4 - 2*x**3 + x**2 + x + log(2/x) - 6 - log(5))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-1+x+2 x^2-6 x^3+4 x^4}{-6 x+x^2+x \log \left (\frac {1}{5} e^{x^2-2 x^3+x^4}\right )+x \log \left (\frac {2}{x}\right )} \, dx=\log \left (-x^{4} + 2 \, x^{3} - x^{2} - x + \log \left (5\right ) - \log \left (2\right ) + \log \left (x\right ) + 6\right ) \]

[In]

integrate((4*x^4-6*x^3+2*x^2+x-1)/(x*log(1/5*exp(x^4-2*x^3+x^2))+x*log(2/x)+x^2-6*x),x, algorithm="maxima")

[Out]

log(-x^4 + 2*x^3 - x^2 - x + log(5) - log(2) + log(x) + 6)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-1+x+2 x^2-6 x^3+4 x^4}{-6 x+x^2+x \log \left (\frac {1}{5} e^{x^2-2 x^3+x^4}\right )+x \log \left (\frac {2}{x}\right )} \, dx=\log \left (-x^{4} + 2 \, x^{3} - x^{2} - x + \log \left (5\right ) - \log \left (2\right ) + \log \left (x\right ) + 6\right ) \]

[In]

integrate((4*x^4-6*x^3+2*x^2+x-1)/(x*log(1/5*exp(x^4-2*x^3+x^2))+x*log(2/x)+x^2-6*x),x, algorithm="giac")

[Out]

log(-x^4 + 2*x^3 - x^2 - x + log(5) - log(2) + log(x) + 6)

Mupad [B] (verification not implemented)

Time = 9.59 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-1+x+2 x^2-6 x^3+4 x^4}{-6 x+x^2+x \log \left (\frac {1}{5} e^{x^2-2 x^3+x^4}\right )+x \log \left (\frac {2}{x}\right )} \, dx=\ln \left (x+\ln \left (\frac {2}{5\,x}\right )+x^2-2\,x^3+x^4-6\right ) \]

[In]

int((x + 2*x^2 - 6*x^3 + 4*x^4 - 1)/(x*log(exp(x^2 - 2*x^3 + x^4)/5) - 6*x + x*log(2/x) + x^2),x)

[Out]

log(x + log(2/(5*x)) + x^2 - 2*x^3 + x^4 - 6)