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

Optimal. Leaf size=13 \[ e^{5+\frac {1}{-2 x+x^2}} \]

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Rubi [F]  time = 1.01, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 20, normalized size = 1.54 \begin {gather*} e^{5+\frac {1}{2 (-2+x)}-\frac {1}{2 x}} \end {gather*}

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.62, size = 21, normalized size = 1.62 \begin {gather*} e^{\left (\frac {5 \, x^{2} - 10 \, x + 1}{x^{2} - 2 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x+2)*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))

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giac [B]  time = 0.20, size = 37, normalized size = 2.85 \begin {gather*} e^{\left (\frac {5 \, x^{2}}{x^{2} - 2 \, x} - \frac {10 \, x}{x^{2} - 2 \, x} + \frac {1}{x^{2} - 2 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x+2)*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))

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maple [A]  time = 0.11, size = 21, normalized size = 1.62

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x+2)*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)/(x-2)/x)

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maxima [A]  time = 0.45, size = 15, normalized size = 1.15 \begin {gather*} e^{\left (\frac {1}{2 \, {\left (x - 2\right )}} - \frac {1}{2 \, x} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x+2)*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)

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mupad [B]  time = 7.42, size = 32, normalized size = 2.46 \begin {gather*} {\mathrm {e}}^{\frac {5\,x}{x-2}}\,{\mathrm {e}}^{-\frac {1}{2\,x-x^2}}\,{\mathrm {e}}^{-\frac {10}{x-2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.16, size = 17, normalized size = 1.31 \begin {gather*} e^{\frac {5 x^{2} - 10 x + 1}{x^{2} - 2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x+2)*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))

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