3.2.73 \(\int \frac {-8+e^{2+\frac {4-8 x+4 x^2}{4-4 x+x^2}} (8-8 x)+12 x-6 x^2+x^3}{-8+12 x-6 x^2+x^3} \, dx\)

Optimal. Leaf size=24 \[ e^2 \left (-4+e^{\left (1-\frac {x}{2-x}\right )^2}\right )+x \]

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Rubi [F]  time = 0.69, 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 {-8+e^{2+\frac {4-8 x+4 x^2}{4-4 x+x^2}} (8-8 x)+12 x-6 x^2+x^3}{-8+12 x-6 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8 + E^(2 + (4 - 8*x + 4*x^2)/(4 - 4*x + x^2))*(8 - 8*x) + 12*x - 6*x^2 + x^3)/(-8 + 12*x - 6*x^2 + x^3),
x]

[Out]

12/(2 - x)^2 - 12/(2 - x) + x - (3*x^2)/(2 - x)^2 - 8*Defer[Int][E^((2*(6 - 8*x + 3*x^2))/(-2 + x)^2)/(-2 + x)
^3, x] - 8*Defer[Int][E^((2*(6 - 8*x + 3*x^2))/(-2 + x)^2)/(-2 + x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {8}{(-2+x)^3}+\frac {8 e^{\frac {2 \left (6-8 x+3 x^2\right )}{(-2+x)^2}} (-1+x)}{(2-x)^3}+\frac {12 x}{(-2+x)^3}-\frac {6 x^2}{(-2+x)^3}+\frac {x^3}{(-2+x)^3}\right ) \, dx\\ &=\frac {4}{(2-x)^2}-6 \int \frac {x^2}{(-2+x)^3} \, dx+8 \int \frac {e^{\frac {2 \left (6-8 x+3 x^2\right )}{(-2+x)^2}} (-1+x)}{(2-x)^3} \, dx+12 \int \frac {x}{(-2+x)^3} \, dx+\int \frac {x^3}{(-2+x)^3} \, dx\\ &=\frac {4}{(2-x)^2}-\frac {3 x^2}{(2-x)^2}-6 \int \left (\frac {4}{(-2+x)^3}+\frac {4}{(-2+x)^2}+\frac {1}{-2+x}\right ) \, dx+8 \int \left (-\frac {e^{\frac {2 \left (6-8 x+3 x^2\right )}{(-2+x)^2}}}{(-2+x)^3}-\frac {e^{\frac {2 \left (6-8 x+3 x^2\right )}{(-2+x)^2}}}{(-2+x)^2}\right ) \, dx+\int \left (1+\frac {8}{(-2+x)^3}+\frac {12}{(-2+x)^2}+\frac {6}{-2+x}\right ) \, dx\\ &=\frac {12}{(2-x)^2}-\frac {12}{2-x}+x-\frac {3 x^2}{(2-x)^2}-8 \int \frac {e^{\frac {2 \left (6-8 x+3 x^2\right )}{(-2+x)^2}}}{(-2+x)^3} \, dx-8 \int \frac {e^{\frac {2 \left (6-8 x+3 x^2\right )}{(-2+x)^2}}}{(-2+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 20, normalized size = 0.83 \begin {gather*} e^{6+\frac {4}{(-2+x)^2}+\frac {8}{-2+x}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + E^(2 + (4 - 8*x + 4*x^2)/(4 - 4*x + x^2))*(8 - 8*x) + 12*x - 6*x^2 + x^3)/(-8 + 12*x - 6*x^2 +
x^3),x]

[Out]

E^(6 + 4/(-2 + x)^2 + 8/(-2 + x)) + x

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fricas [A]  time = 0.86, size = 25, normalized size = 1.04 \begin {gather*} x + e^{\left (\frac {2 \, {\left (3 \, x^{2} - 8 \, x + 6\right )}}{x^{2} - 4 \, x + 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+8)*exp(1)^2*exp((4*x^2-8*x+4)/(x^2-4*x+4))+x^3-6*x^2+12*x-8)/(x^3-6*x^2+12*x-8),x, algorithm=
"fricas")

[Out]

x + e^(2*(3*x^2 - 8*x + 6)/(x^2 - 4*x + 4))

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giac [B]  time = 0.34, size = 44, normalized size = 1.83 \begin {gather*} x + e^{\left (\frac {6 \, x^{2}}{x^{2} - 4 \, x + 4} - \frac {16 \, x}{x^{2} - 4 \, x + 4} + \frac {12}{x^{2} - 4 \, x + 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+8)*exp(1)^2*exp((4*x^2-8*x+4)/(x^2-4*x+4))+x^3-6*x^2+12*x-8)/(x^3-6*x^2+12*x-8),x, algorithm=
"giac")

[Out]

x + e^(6*x^2/(x^2 - 4*x + 4) - 16*x/(x^2 - 4*x + 4) + 12/(x^2 - 4*x + 4))

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maple [A]  time = 0.06, size = 21, normalized size = 0.88

method	result	size
derivativedivides	\(x -2+{\mathrm e}^{\frac {4}{\left (x -2\right )^{2}}+\frac {8}{x -2}+6}\)	\(21\)
default	\(x -2+{\mathrm e}^{\frac {4}{\left (x -2\right )^{2}}+\frac {8}{x -2}+6}\)	\(21\)
risch	\({\mathrm e}^{\frac {6 x^{2}-16 x +12}{\left (x -2\right )^{2}}}+x\)	\(21\)
norman	\(\frac {x^{3}-12 x +x^{2} {\mathrm e}^{2} {\mathrm e}^{\frac {4 x^{2}-8 x +4}{x^{2}-4 x +4}}+4 \,{\mathrm e}^{2} {\mathrm e}^{\frac {4 x^{2}-8 x +4}{x^{2}-4 x +4}}-4 x \,{\mathrm e}^{2} {\mathrm e}^{\frac {4 x^{2}-8 x +4}{x^{2}-4 x +4}}+16}{\left (x -2\right )^{2}}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x+8)*exp(1)^2*exp((4*x^2-8*x+4)/(x^2-4*x+4))+x^3-6*x^2+12*x-8)/(x^3-6*x^2+12*x-8),x,method=_RETURNVER
BOSE)

[Out]

x-2+exp(4/(x-2)^2+8/(x-2)+6)

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maxima [B]  time = 0.60, size = 85, normalized size = 3.54 \begin {gather*} x - \frac {4 \, {\left (3 \, x - 5\right )}}{x^{2} - 4 \, x + 4} + \frac {12 \, {\left (2 \, x - 3\right )}}{x^{2} - 4 \, x + 4} - \frac {12 \, {\left (x - 1\right )}}{x^{2} - 4 \, x + 4} + \frac {4}{x^{2} - 4 \, x + 4} + e^{\left (\frac {4}{x^{2} - 4 \, x + 4} + \frac {8}{x - 2} + 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+8)*exp(1)^2*exp((4*x^2-8*x+4)/(x^2-4*x+4))+x^3-6*x^2+12*x-8)/(x^3-6*x^2+12*x-8),x, algorithm=
"maxima")

[Out]

x - 4*(3*x - 5)/(x^2 - 4*x + 4) + 12*(2*x - 3)/(x^2 - 4*x + 4) - 12*(x - 1)/(x^2 - 4*x + 4) + 4/(x^2 - 4*x + 4
) + e^(4/(x^2 - 4*x + 4) + 8/(x - 2) + 6)

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mupad [B]  time = 0.17, size = 46, normalized size = 1.92 \begin {gather*} x+{\mathrm {e}}^{\frac {6\,x^2}{x^2-4\,x+4}}\,{\mathrm {e}}^{\frac {12}{x^2-4\,x+4}}\,{\mathrm {e}}^{-\frac {16\,x}{x^2-4\,x+4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*x^2 - 12*x - x^3 + exp(2)*exp((4*x^2 - 8*x + 4)/(x^2 - 4*x + 4))*(8*x - 8) + 8)/(12*x - 6*x^2 + x^3 -
8),x)

[Out]

x + exp((6*x^2)/(x^2 - 4*x + 4))*exp(12/(x^2 - 4*x + 4))*exp(-(16*x)/(x^2 - 4*x + 4))

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sympy [A]  time = 0.26, size = 24, normalized size = 1.00 \begin {gather*} x + e^{2} e^{\frac {4 x^{2} - 8 x + 4}{x^{2} - 4 x + 4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+8)*exp(1)**2*exp((4*x**2-8*x+4)/(x**2-4*x+4))+x**3-6*x**2+12*x-8)/(x**3-6*x**2+12*x-8),x)

[Out]

x + exp(2)*exp((4*x**2 - 8*x + 4)/(x**2 - 4*x + 4))

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