3.11.57 \(\int \frac {2-x}{4 x-4 x^2+e^2 (-4+4 x)+(-4+4 x) \log (-1+x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {1}{4} \log \left (-e^2+x-\log (-1+x)\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6741, 12, 6684} \begin {gather*} \frac {1}{4} \log \left (-x+\log (x-1)+e^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[E^2 - x + Log[-1 + x]]/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 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 18, normalized size = 1.00 \begin {gather*} \frac {1}{4} \log \left (-e^2+x-\log (-1+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[-E^2 + x - Log[-1 + x]]/4

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fricas [A]  time = 0.56, size = 13, normalized size = 0.72 \begin {gather*} \frac {1}{4} \, \log \left (-x + e^{2} + \log \left (x - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(-x + e^2 + log(x - 1))

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giac [A]  time = 0.38, size = 13, normalized size = 0.72 \begin {gather*} \frac {1}{4} \, \log \left (-x + e^{2} + \log \left (x - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(-x + e^2 + log(x - 1))

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maple [A]  time = 0.06, size = 14, normalized size = 0.78

method	result	size
norman	\(\frac {\ln \left (-x +{\mathrm e}^{2}+\ln \left (x -1\right )\right )}{4}\)	\(14\)
risch	\(\frac {\ln \left (-x +{\mathrm e}^{2}+\ln \left (x -1\right )\right )}{4}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*ln(-x+exp(2)+ln(x-1))

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maxima [A]  time = 0.49, size = 13, normalized size = 0.72 \begin {gather*} \frac {1}{4} \, \log \left (-x + e^{2} + \log \left (x - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(-x + e^2 + log(x - 1))

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mupad [B]  time = 0.71, size = 15, normalized size = 0.83 \begin {gather*} \frac {\ln \left (x-\ln \left (x-1\right )-{\mathrm {e}}^2\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - 2)/(4*x - 4*x^2 + log(x - 1)*(4*x - 4) + exp(2)*(4*x - 4)),x)

[Out]

log(x - log(x - 1) - exp(2))/4

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sympy [A]  time = 0.18, size = 12, normalized size = 0.67 \begin {gather*} \frac {\log {\left (- x + \log {\left (x - 1 \right )} + e^{2} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-x)/((4*x-4)*ln(x-1)+(4*x-4)*exp(2)-4*x**2+4*x),x)

[Out]

log(-x + log(x - 1) + exp(2))/4

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