\(\int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} (2 x+4 x^2-8 x \log (x)+2 \log ^2(x))}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx\) [7794]

Optimal result

Integrand size = 76, antiderivative size = 24 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{2+\frac {-1+x+\frac {\log (x)}{2 x-\log (x)}}{x}} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=\int \frac {\exp \left (\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}\right ) \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx \]

[In]

Int[(E^((2*x - 6*x^2 + (-2 + 3*x)*Log[x])/(-2*x^2 + x*Log[x]))*(2*x + 4*x^2 - 8*x*Log[x] + 2*Log[x]^2))/(4*x^4
- 4*x^3*Log[x] + x^2*Log[x]^2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 3.61 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{3-\frac {2}{x}-\frac {2}{-2 x+\log (x)}} \]

[In]

Integrate[(E^((2*x - 6*x^2 + (-2 + 3*x)*Log[x])/(-2*x^2 + x*Log[x]))*(2*x + 4*x^2 - 8*x*Log[x] + 2*Log[x]^2))/
(4*x^4 - 4*x^3*Log[x] + x^2*Log[x]^2),x]

[Out]

E^(3 - 2/x - 2/(-2*x + Log[x]))

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\left (-2+3 x \right ) \ln \left (x \right )-6 x^{2}+2 x}{x \left (\ln \left (x \right )-2 x \right )}}\)	\(31\)
risch	\({\mathrm e}^{\frac {3 x \ln \left (x \right )-6 x^{2}-2 \ln \left (x \right )+2 x}{x \left (\ln \left (x \right )-2 x \right )}}\)	\(32\)

[In]

int((2*ln(x)^2-8*x*ln(x)+4*x^2+2*x)*exp(((-2+3*x)*ln(x)-6*x^2+2*x)/(x*ln(x)-2*x^2))/(x^2*ln(x)^2-4*x^3*ln(x)+4
*x^4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{\left (\frac {6 \, x^{2} - {\left (3 \, x - 2\right )} \log \left (x\right ) - 2 \, x}{2 \, x^{2} - x \log \left (x\right )}\right )} \]

[In]

integrate((2*log(x)^2-8*x*log(x)+4*x^2+2*x)*exp(((-2+3*x)*log(x)-6*x^2+2*x)/(x*log(x)-2*x^2))/(x^2*log(x)^2-4*
x^3*log(x)+4*x^4),x, algorithm="fricas")

[Out]

e^((6*x^2 - (3*x - 2)*log(x) - 2*x)/(2*x^2 - x*log(x)))

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{\frac {- 6 x^{2} + 2 x + \left (3 x - 2\right ) \log {\left (x \right )}}{- 2 x^{2} + x \log {\left (x \right )}}} \]

[In]

integrate((2*ln(x)**2-8*x*ln(x)+4*x**2+2*x)*exp(((-2+3*x)*ln(x)-6*x**2+2*x)/(x*ln(x)-2*x**2))/(x**2*ln(x)**2-4
*x**3*ln(x)+4*x**4),x)

[Out]

exp((-6*x**2 + 2*x + (3*x - 2)*log(x))/(-2*x**2 + x*log(x)))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{\left (\frac {2}{2 \, x - \log \left (x\right )} - \frac {2}{x} + 3\right )} \]

[In]

integrate((2*log(x)^2-8*x*log(x)+4*x^2+2*x)*exp(((-2+3*x)*log(x)-6*x^2+2*x)/(x*log(x)-2*x^2))/(x^2*log(x)^2-4*
x^3*log(x)+4*x^4),x, algorithm="maxima")

[Out]

e^(2/(2*x - log(x)) - 2/x + 3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (23) = 46\).

Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.96 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{\left (\frac {6 \, x^{2}}{2 \, x^{2} - x \log \left (x\right )} - \frac {3 \, x \log \left (x\right )}{2 \, x^{2} - x \log \left (x\right )} - \frac {2 \, x}{2 \, x^{2} - x \log \left (x\right )} + \frac {2 \, \log \left (x\right )}{2 \, x^{2} - x \log \left (x\right )}\right )} \]

[In]

integrate((2*log(x)^2-8*x*log(x)+4*x^2+2*x)*exp(((-2+3*x)*log(x)-6*x^2+2*x)/(x*log(x)-2*x^2))/(x^2*log(x)^2-4*
x^3*log(x)+4*x^4),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.96 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.29 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=x^{\frac {3\,x-2}{x\,\ln \left (x\right )-2\,x^2}}\,{\mathrm {e}}^{-\frac {6\,x^2}{x\,\ln \left (x\right )-2\,x^2}}\,{\mathrm {e}}^{\frac {2\,x}{x\,\ln \left (x\right )-2\,x^2}} \]

[In]

int((exp((2*x + log(x)*(3*x - 2) - 6*x^2)/(x*log(x) - 2*x^2))*(2*x + 2*log(x)^2 - 8*x*log(x) + 4*x^2))/(x^2*lo
g(x)^2 - 4*x^3*log(x) + 4*x^4),x)

[Out]

x^((3*x - 2)/(x*log(x) - 2*x^2))*exp(-(6*x^2)/(x*log(x) - 2*x^2))*exp((2*x)/(x*log(x) - 2*x^2))