3.2.28 \(\int \frac {2+x}{4-4 x+x^2} \, dx\) [128]

Optimal. Leaf size=16 \[ \frac {4}{2-x}+\log (2-x) \]

[Out]

4/(2-x)+ln(2-x)

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Rubi [A]
time = 0.00, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 45} \begin {gather*} \frac {4}{2-x}+\log (2-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + x)/(4 - 4*x + x^2),x]

[Out]

4/(2 - x) + Log[2 - 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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {2+x}{4-4 x+x^2} \, dx &=\int \frac {2+x}{(-2+x)^2} \, dx\\ &=\int \left (\frac {4}{(-2+x)^2}+\frac {1}{-2+x}\right ) \, dx\\ &=\frac {4}{2-x}+\log (2-x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 12, normalized size = 0.75 \begin {gather*} -\frac {4}{-2+x}+\log (-2+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + x)/(4 - 4*x + x^2),x]

[Out]

-4/(-2 + x) + Log[-2 + x]

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Mathics [A]
time = 1.61, size = 16, normalized size = 1.00 \begin {gather*} \frac {-4+\text {Log}\left [-2+x\right ] \left (-2+x\right )}{-2+x} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(x + 2)/(x^2 - 4*x + 4),x]')

[Out]

(-4 + Log[-2 + x] (-2 + x)) / (-2 + x)

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Maple [A]
time = 0.08, size = 13, normalized size = 0.81

method result size
default \(\ln \left (-2+x \right )-\frac {4}{-2+x}\) \(13\)
norman \(\ln \left (-2+x \right )-\frac {4}{-2+x}\) \(13\)
risch \(\ln \left (-2+x \right )-\frac {4}{-2+x}\) \(13\)
meijerg \(\frac {x}{1-\frac {x}{2}}+\ln \left (1-\frac {x}{2}\right )\) \(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(-2+x)-4/(-2+x)

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Maxima [A]
time = 0.31, size = 12, normalized size = 0.75 \begin {gather*} -\frac {4}{x - 2} + \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(x - 2) + log(x - 2)

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Fricas [A]
time = 0.33, size = 16, normalized size = 1.00 \begin {gather*} \frac {{\left (x - 2\right )} \log \left (x - 2\right ) - 4}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x - 2)*log(x - 2) - 4)/(x - 2)

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Sympy [A]
time = 0.04, size = 8, normalized size = 0.50 \begin {gather*} \log {\left (x - 2 \right )} - \frac {4}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)/(x**2-4*x+4),x)

[Out]

log(x - 2) - 4/(x - 2)

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Giac [A]
time = 0.00, size = 13, normalized size = 0.81 \begin {gather*} -\frac {4}{x-2}+\ln \left |x-2\right | \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)/(x^2-4*x+4),x)

[Out]

-4/(x - 2) + log(abs(x - 2))

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Mupad [B]
time = 0.04, size = 12, normalized size = 0.75 \begin {gather*} \ln \left (x-2\right )-\frac {4}{x-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 2)/(x^2 - 4*x + 4),x)

[Out]

log(x - 2) - 4/(x - 2)

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