3.49.93 \(\int \frac {2 x-x^2}{2 e^4 (4-8 x+4 x^2)} \, dx\)

Optimal. Leaf size=19 \[ e^2-\frac {x^2}{8 e^4 (-1+x)} \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 27, 683} \begin {gather*} \frac {1}{8 e^4 (1-x)}-\frac {x}{8 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(8*E^4*(1 - x)) - x/(8*E^4)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.74 \begin {gather*} -\frac {\frac {1}{-1+x}+x}{8 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((-1 + x)^(-1) + x)/E^4

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fricas [A]  time = 0.74, size = 22, normalized size = 1.16 \begin {gather*} -\frac {{\left (x^{2} - x + 1\right )} e^{\left (-\log \relax (2) - 4\right )}}{4 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(x^2 - x + 1)*e^(-log(2) - 4)/(x - 1)

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giac [A]  time = 0.20, size = 16, normalized size = 0.84 \begin {gather*} -\frac {1}{4} \, {\left (x + \frac {1}{x - 1}\right )} e^{\left (-\log \relax (2) - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(x + 1/(x - 1))*e^(-log(2) - 4)

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maple [A]  time = 0.18, size = 15, normalized size = 0.79

method	result	size
norman	\(-\frac {x^{2} {\mathrm e}^{-4}}{8 \left (x -1\right )}\)	\(15\)
risch	\(-\frac {{\mathrm e}^{-4} x}{8}-\frac {{\mathrm e}^{-4}}{8 \left (x -1\right )}\)	\(16\)
gosper	\(-\frac {x^{2} {\mathrm e}^{-4}}{8 \left (x -1\right )}\)	\(18\)
default	\(\frac {{\mathrm e}^{-4} \left (-x -\frac {1}{x -1}\right )}{8}\)	\(21\)
meijerg	\(\frac {{\mathrm e}^{-4-\ln \relax (2)} \left (-\frac {x \left (-3 x +6\right )}{3 \left (1-x \right )}-2 \ln \left (1-x \right )\right )}{4}+\frac {{\mathrm e}^{-4-\ln \relax (2)} \left (\frac {x}{1-x}+\ln \left (1-x \right )\right )}{2}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*x^2/exp(4)/(x-1)

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maxima [A]  time = 0.35, size = 11, normalized size = 0.58 \begin {gather*} -\frac {1}{8} \, {\left (x + \frac {1}{x - 1}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(x + 1/(x - 1))*e^(-4)

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mupad [B]  time = 3.36, size = 17, normalized size = 0.89 \begin {gather*} -\frac {x\,{\mathrm {e}}^{-4}}{8}-\frac {{\mathrm {e}}^{-4}}{8\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(- log(2) - 4)*(2*x - x^2))/(4*x^2 - 8*x + 4),x)

[Out]

- (x*exp(-4))/8 - exp(-4)/(8*(x - 1))

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sympy [A]  time = 0.13, size = 20, normalized size = 1.05 \begin {gather*} - \frac {x}{8 e^{4}} - \frac {1}{8 x e^{4} - 8 e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*exp(-4)/8 - 1/(8*x*exp(4) - 8*exp(4))

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