\(\int \frac {2 x+(-6+3 x) \log (\frac {e (2-x)}{x})}{(-2 x^2+x^3) \log (\frac {e (2-x)}{x})} \, dx\) [3716]

Optimal result

Integrand size = 46, antiderivative size = 18 \[ \int \frac {2 x+(-6+3 x) \log \left (\frac {e (2-x)}{x}\right )}{\left (-2 x^2+x^3\right ) \log \left (\frac {e (2-x)}{x}\right )} \, dx=-\frac {3}{x}+\log \left (\log \left (-e+\frac {2 e}{x}\right )\right ) \]

[Out]

ln(ln(2*exp(1)/x-exp(1)))-3/x

Rubi [F]

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

[In]

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

[Out]

-3/x + Defer[Int][1/((-2 + x)*Log[-E + (2*E)/x]), x] - Defer[Int][1/(x*Log[-E + (2*E)/x]), x]

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {2 x+(-6+3 x) \log \left (\frac {e (2-x)}{x}\right )}{\left (-2 x^2+x^3\right ) \log \left (\frac {e (2-x)}{x}\right )} \, dx=-\frac {3}{x}+\log \left (\log \left (-\frac {e (-2+x)}{x}\right )\right ) \]

[In]

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

[Out]

-3/x + Log[Log[-((E*(-2 + x))/x)]]

Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11

method	result	size
norman	\(-\frac {3}{x}+\ln \left (\ln \left (\frac {\left (2-x \right ) {\mathrm e}}{x}\right )\right )\)	\(20\)
risch	\(-\frac {3}{x}+\ln \left (\ln \left (\frac {\left (2-x \right ) {\mathrm e}}{x}\right )\right )\)	\(20\)
parts	\(-\frac {3}{x}+\ln \left (\ln \left (\frac {\left (2-x \right ) {\mathrm e}}{x}\right )\right )\)	\(20\)
parallelrisch	\(-\frac {3-\ln \left (\ln \left (\frac {\left (2-x \right ) {\mathrm e}}{x}\right )\right ) x}{x}\)	\(24\)
derivativedivides	\(-\frac {{\mathrm e} \left (6 \,{\mathrm e}^{-2} \left (\frac {2 \,{\mathrm e}}{x}-{\mathrm e}\right )-4 \,{\mathrm e}^{-1} \ln \left (\ln \left (\frac {2 \,{\mathrm e}}{x}-{\mathrm e}\right )\right )\right )}{4}\)	\(44\)
default	\(-\frac {{\mathrm e} \left (6 \,{\mathrm e}^{-2} \left (\frac {2 \,{\mathrm e}}{x}-{\mathrm e}\right )-4 \,{\mathrm e}^{-1} \ln \left (\ln \left (\frac {2 \,{\mathrm e}}{x}-{\mathrm e}\right )\right )\right )}{4}\)	\(44\)

[In]

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

[Out]

-3/x+ln(ln((2-x)*exp(1)/x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {2 x+(-6+3 x) \log \left (\frac {e (2-x)}{x}\right )}{\left (-2 x^2+x^3\right ) \log \left (\frac {e (2-x)}{x}\right )} \, dx=\frac {x \log \left (\log \left (-\frac {{\left (x - 2\right )} e}{x}\right )\right ) - 3}{x} \]

[In]

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

[Out]

(x*log(log(-(x - 2)*e/x)) - 3)/x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {2 x+(-6+3 x) \log \left (\frac {e (2-x)}{x}\right )}{\left (-2 x^2+x^3\right ) \log \left (\frac {e (2-x)}{x}\right )} \, dx=\log {\left (\log {\left (\frac {e \left (2 - x\right )}{x} \right )} \right )} - \frac {3}{x} \]

[In]

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

[Out]

log(log(E*(2 - x)/x)) - 3/x

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (20) = 40\).

Time = 0.24 (sec) , antiderivative size = 103, normalized size of antiderivative = 5.72 \[ \int \frac {2 x+(-6+3 x) \log \left (\frac {e (2-x)}{x}\right )}{\left (-2 x^2+x^3\right ) \log \left (\frac {e (2-x)}{x}\right )} \, dx=\frac {3}{2} \, {\left (e^{\left (-1\right )} \log \left (\frac {2 \, e}{x} - e\right ) \log \left (-\log \left (x\right ) + \log \left (-x + 2\right ) + 1\right ) + {\left ({\left (\log \left (x\right ) - \log \left (-x + 2\right ) - 1\right )} \log \left (-\log \left (x\right ) + \log \left (-x + 2\right ) + 1\right ) - \log \left (x\right ) + \log \left (-x + 2\right )\right )} e^{\left (-1\right )}\right )} e - \frac {3}{x} - \frac {3}{2} \, \log \left (x - 2\right ) + \frac {3}{2} \, \log \left (x\right ) + \log \left (-\log \left (x\right ) + \log \left (-x + 2\right ) + 1\right ) \]

[In]

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

[Out]

3/2*(e^(-1)*log(2*e/x - e)*log(-log(x) + log(-x + 2) + 1) + ((log(x) - log(-x + 2) - 1)*log(-log(x) + log(-x +
2) + 1) - log(x) + log(-x + 2))*e^(-1))*e - 3/x - 3/2*log(x - 2) + 3/2*log(x) + log(-log(x) + log(-x + 2) + 1
)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.17 \[ \int \frac {2 x+(-6+3 x) \log \left (\frac {e (2-x)}{x}\right )}{\left (-2 x^2+x^3\right ) \log \left (\frac {e (2-x)}{x}\right )} \, dx=\frac {1}{2} \, {\left (2 \, e \log \left (\log \left (-\frac {x e - 2 \, e}{x}\right )\right ) + \frac {3 \, {\left (x e - 2 \, e\right )}}{x}\right )} e^{\left (-1\right )} \]

[In]

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

[Out]

1/2*(2*e*log(log(-(x*e - 2*e)/x)) + 3*(x*e - 2*e)/x)*e^(-1)

Mupad [B] (verification not implemented)

Time = 9.34 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {2 x+(-6+3 x) \log \left (\frac {e (2-x)}{x}\right )}{\left (-2 x^2+x^3\right ) \log \left (\frac {e (2-x)}{x}\right )} \, dx=\ln \left (\ln \left (-\frac {\mathrm {e}\,\left (x-2\right )}{x}\right )\right )-\frac {3}{x} \]

[In]

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

[Out]

log(log(-(exp(1)*(x - 2))/x)) - 3/x