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

Optimal. Leaf size=25 \[ \frac {x-\log \left (8 e^2 x \left (-x+x^2\right )\right )}{-1+\log (4)} \]

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Rubi [A]  time = 0.06, antiderivative size = 39, normalized size of antiderivative = 1.56, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1984, 1593, 893} \begin {gather*} -\frac {x}{1-\log (4)}+\frac {\log (1-x)}{1-\log (4)}+\frac {2 \log (x)}{1-\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/(1 - Log[4])) + Log[1 - x]/(1 - Log[4]) + (2*Log[x])/(1 - Log[4])

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1984

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
QuadraticQ[{u, v}, x] &&  !QuadraticMatchQ[{u, v}, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.84 \begin {gather*} \frac {x-\log (1-x)-2 \log (x)}{-1+\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x - Log[1 - x] - 2*Log[x])/(-1 + Log[4])

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fricas [A]  time = 0.54, size = 21, normalized size = 0.84 \begin {gather*} \frac {x - \log \left (x - 1\right ) - 2 \, \log \relax (x)}{2 \, \log \relax (2) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x - log(x - 1) - 2*log(x))/(2*log(2) - 1)

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giac [A]  time = 0.21, size = 39, normalized size = 1.56 \begin {gather*} \frac {x}{2 \, \log \relax (2) - 1} - \frac {\log \left ({\left | x - 1 \right |}\right )}{2 \, \log \relax (2) - 1} - \frac {2 \, \log \left ({\left | x \right |}\right )}{2 \, \log \relax (2) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(2*log(2) - 1) - log(abs(x - 1))/(2*log(2) - 1) - 2*log(abs(x))/(2*log(2) - 1)

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maple [A]  time = 0.16, size = 22, normalized size = 0.88

method	result	size
default	\(\frac {x -\ln \left (x -1\right )-2 \ln \relax (x )}{2 \ln \relax (2)-1}\)	\(22\)
norman	\(\frac {x}{2 \ln \relax (2)-1}-\frac {2 \ln \relax (x )}{2 \ln \relax (2)-1}-\frac {\ln \left (x -1\right )}{2 \ln \relax (2)-1}\)	\(38\)
risch	\(\frac {x}{2 \ln \relax (2)-1}-\frac {2 \ln \relax (x )}{2 \ln \relax (2)-1}-\frac {\ln \left (x -1\right )}{2 \ln \relax (2)-1}\)	\(38\)
meijerg	\(\frac {\frac {x \left (2 \ln \relax (2)-1\right )}{1-2 \ln \relax (2)}-\ln \left (1+\frac {x \left (2 \ln \relax (2)-1\right )}{1-2 \ln \relax (2)}\right )}{1-2 \ln \relax (2)}+\frac {4 \ln \left (1+\frac {x \left (2 \ln \relax (2)-1\right )}{1-2 \ln \relax (2)}\right )}{1-2 \ln \relax (2)}-\frac {2 \left (2 \ln \relax (2)-1\right ) \left (\ln \relax (x )+\ln \left (2 \ln \relax (2)-1\right )+\ln \left (-\frac {1}{1-2 \ln \relax (2)}\right )+i \pi -\ln \left (1+\frac {x \left (2 \ln \relax (2)-1\right )}{1-2 \ln \relax (2)}\right )\right )}{\left (1-2 \ln \relax (2)\right )^{2}}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(2*ln(2)-1)*(x-ln(x-1)-2*ln(x))

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maxima [A]  time = 0.37, size = 37, normalized size = 1.48 \begin {gather*} \frac {x}{2 \, \log \relax (2) - 1} - \frac {\log \left (x - 1\right )}{2 \, \log \relax (2) - 1} - \frac {2 \, \log \relax (x)}{2 \, \log \relax (2) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(2*log(2) - 1) - log(x - 1)/(2*log(2) - 1) - 2*log(x)/(2*log(2) - 1)

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mupad [B]  time = 0.11, size = 37, normalized size = 1.48 \begin {gather*} \frac {x}{2\,\ln \relax (2)-1}-\frac {\ln \left (x-1\right )}{2\,\ln \relax (2)-1}-\frac {2\,\ln \relax (x)}{2\,\ln \relax (2)-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(2*log(2) - 1) - log(x - 1)/(2*log(2) - 1) - (2*log(x))/(2*log(2) - 1)

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sympy [B]  time = 0.70, size = 73, normalized size = 2.92 \begin {gather*} \frac {x}{-1 + 2 \log {\relax (2 )}} - \frac {2 \log {\left (x - \frac {2}{-1 + 2 \log {\relax (2 )}} - 2 + \frac {4 \log {\relax (2 )}}{-1 + 2 \log {\relax (2 )}} \right )}}{-1 + 2 \log {\relax (2 )}} - \frac {\log {\left (x - \frac {1}{-1 + 2 \log {\relax (2 )}} - 2 + \frac {2 \log {\relax (2 )}}{-1 + 2 \log {\relax (2 )}} \right )}}{-1 + 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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