\(\int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{4 x^2-4 x^3+x^4} \, dx\) [9525]

Optimal result

Integrand size = 44, antiderivative size = 13 \[ \int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{4 x^2-4 x^3+x^4} \, dx=e^{5+\frac {1}{-2 x+x^2}} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{4 x^2-4 x^3+x^4} \, dx=\int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{4 x^2-4 x^3+x^4} \, dx \]

[In]

Int[(E^((1 - 10*x + 5*x^2)/(-2*x + x^2))*(2 - 2*x))/(4*x^2 - 4*x^3 + x^4),x]

[Out]

-1/2*Defer[Int][E^((1 - 10*x + 5*x^2)/((-2 + x)*x))/(-2 + x)^2, x] + Defer[Int][E^((1 - 10*x + 5*x^2)/((-2 + x
)*x))/x^2, x]/2

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{x^2 \left (4-4 x+x^2\right )} \, dx \\ & = \int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{(-2+x)^2 x^2} \, dx \\ & = \int \frac {e^{\frac {1-10 x+5 x^2}{(-2+x) x}} (2-2 x)}{(2-x)^2 x^2} \, dx \\ & = \int \left (-\frac {e^{\frac {1-10 x+5 x^2}{(-2+x) x}}}{2 (-2+x)^2}+\frac {e^{\frac {1-10 x+5 x^2}{(-2+x) x}}}{2 x^2}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {e^{\frac {1-10 x+5 x^2}{(-2+x) x}}}{(-2+x)^2} \, dx\right )+\frac {1}{2} \int \frac {e^{\frac {1-10 x+5 x^2}{(-2+x) x}}}{x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.54 \[ \int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{4 x^2-4 x^3+x^4} \, dx=e^{5+\frac {1}{2 (-2+x)}-\frac {1}{2 x}} \]

[In]

Integrate[(E^((1 - 10*x + 5*x^2)/(-2*x + x^2))*(2 - 2*x))/(4*x^2 - 4*x^3 + x^4),x]

[Out]

E^(5 + 1/(2*(-2 + x)) - 1/(2*x))

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.62

method	result	size
gosper	\({\mathrm e}^{\frac {5 x^{2}-10 x +1}{\left (-2+x \right ) x}}\)	\(21\)
risch	\({\mathrm e}^{\frac {5 x^{2}-10 x +1}{\left (-2+x \right ) x}}\)	\(21\)
parallelrisch	\({\mathrm e}^{\frac {5 x^{2}-10 x +1}{\left (-2+x \right ) x}}\)	\(21\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {5 x^{2}-10 x +1}{x^{2}-2 x}}-2 x \,{\mathrm e}^{\frac {5 x^{2}-10 x +1}{x^{2}-2 x}}}{\left (-2+x \right ) x}\)	\(60\)

[In]

int((2-2*x)*exp((5*x^2-10*x+1)/(x^2-2*x))/(x^4-4*x^3+4*x^2),x,method=_RETURNVERBOSE)

[Out]

exp((5*x^2-10*x+1)/(-2+x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.62 \[ \int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{4 x^2-4 x^3+x^4} \, dx=e^{\left (\frac {5 \, x^{2} - 10 \, x + 1}{x^{2} - 2 \, x}\right )} \]

[In]

integrate((2-2*x)*exp((5*x^2-10*x+1)/(x^2-2*x))/(x^4-4*x^3+4*x^2),x, algorithm="fricas")

[Out]

e^((5*x^2 - 10*x + 1)/(x^2 - 2*x))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{4 x^2-4 x^3+x^4} \, dx=e^{\frac {5 x^{2} - 10 x + 1}{x^{2} - 2 x}} \]

[In]

integrate((2-2*x)*exp((5*x**2-10*x+1)/(x**2-2*x))/(x**4-4*x**3+4*x**2),x)

[Out]

exp((5*x**2 - 10*x + 1)/(x**2 - 2*x))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{4 x^2-4 x^3+x^4} \, dx=e^{\left (\frac {1}{2 \, {\left (x - 2\right )}} - \frac {1}{2 \, x} + 5\right )} \]

[In]

integrate((2-2*x)*exp((5*x^2-10*x+1)/(x^2-2*x))/(x^4-4*x^3+4*x^2),x, algorithm="maxima")

[Out]

e^(1/2/(x - 2) - 1/2/x + 5)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (12) = 24\).

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.85 \[ \int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{4 x^2-4 x^3+x^4} \, dx=e^{\left (\frac {5 \, x^{2}}{x^{2} - 2 \, x} - \frac {10 \, x}{x^{2} - 2 \, x} + \frac {1}{x^{2} - 2 \, x}\right )} \]

[In]

integrate((2-2*x)*exp((5*x^2-10*x+1)/(x^2-2*x))/(x^4-4*x^3+4*x^2),x, algorithm="giac")

[Out]

e^(5*x^2/(x^2 - 2*x) - 10*x/(x^2 - 2*x) + 1/(x^2 - 2*x))

Mupad [B] (verification not implemented)

Time = 14.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.46 \[ \int \frac {e^{\frac {1-10 x+5 x^2}{-2 x+x^2}} (2-2 x)}{4 x^2-4 x^3+x^4} \, dx={\mathrm {e}}^{\frac {5\,x}{x-2}}\,{\mathrm {e}}^{-\frac {1}{2\,x-x^2}}\,{\mathrm {e}}^{-\frac {10}{x-2}} \]

[In]

int(-(exp(-(5*x^2 - 10*x + 1)/(2*x - x^2))*(2*x - 2))/(4*x^2 - 4*x^3 + x^4),x)

[Out]

exp((5*x)/(x - 2))*exp(-1/(2*x - x^2))*exp(-10/(x - 2))