3.84.32 \(\int \frac {60-50 x+(-10+10 x) \log (-1+x)}{-4+8 x-5 x^2+x^3} \, dx\)

Optimal. Leaf size=19 \[ 2+\frac {10 (2 x-\log (-1+x))}{-2+x} \]

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Rubi [A]  time = 0.08, antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {6742, 77, 2395, 36, 31} \begin {gather*} \frac {10 \log (x-1)}{2-x}-\frac {40}{2-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(60 - 50*x + (-10 + 10*x)*Log[-1 + x])/(-4 + 8*x - 5*x^2 + x^3),x]

[Out]

-40/(2 - x) + (10*Log[-1 + x])/(2 - x)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {10 (-6+5 x)}{(-2+x)^2 (-1+x)}+\frac {10 \log (-1+x)}{(-2+x)^2}\right ) \, dx\\ &=-\left (10 \int \frac {-6+5 x}{(-2+x)^2 (-1+x)} \, dx\right )+10 \int \frac {\log (-1+x)}{(-2+x)^2} \, dx\\ &=\frac {10 \log (-1+x)}{2-x}-10 \int \left (\frac {1}{1-x}+\frac {4}{(-2+x)^2}+\frac {1}{-2+x}\right ) \, dx+10 \int \frac {1}{(-2+x) (-1+x)} \, dx\\ &=-\frac {40}{2-x}+10 \log (1-x)-10 \log (2-x)+\frac {10 \log (-1+x)}{2-x}+10 \int \frac {1}{-2+x} \, dx-10 \int \frac {1}{-1+x} \, dx\\ &=-\frac {40}{2-x}+\frac {10 \log (-1+x)}{2-x}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.04, size = 39, normalized size = 2.05 \begin {gather*} 10 \left (2 \tanh ^{-1}(3-2 x)+\log (1-x)-\log (2-x)+\frac {4-\log (-1+x)}{-2+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(60 - 50*x + (-10 + 10*x)*Log[-1 + x])/(-4 + 8*x - 5*x^2 + x^3),x]

[Out]

10*(2*ArcTanh[3 - 2*x] + Log[1 - x] - Log[2 - x] + (4 - Log[-1 + x])/(-2 + x))

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fricas [A]  time = 0.76, size = 13, normalized size = 0.68 \begin {gather*} -\frac {10 \, {\left (\log \left (x - 1\right ) - 4\right )}}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-10)*log(x-1)-50*x+60)/(x^3-5*x^2+8*x-4),x, algorithm="fricas")

[Out]

-10*(log(x - 1) - 4)/(x - 2)

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giac [A]  time = 0.14, size = 19, normalized size = 1.00 \begin {gather*} -\frac {10 \, \log \left (x - 1\right )}{x - 2} + \frac {40}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-10)*log(x-1)-50*x+60)/(x^3-5*x^2+8*x-4),x, algorithm="giac")

[Out]

-10*log(x - 1)/(x - 2) + 40/(x - 2)

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maple [A]  time = 0.05, size = 15, normalized size = 0.79

method	result	size
norman	\(\frac {-10 \ln \left (x -1\right )+40}{x -2}\)	\(15\)
risch	\(-\frac {10 \ln \left (x -1\right )}{x -2}+\frac {40}{x -2}\)	\(20\)
derivativedivides	\(-\frac {10 \ln \left (x -1\right ) \left (x -1\right )}{x -2}+\frac {40}{x -2}+10 \ln \left (x -1\right )\)	\(29\)
default	\(-\frac {10 \ln \left (x -1\right ) \left (x -1\right )}{x -2}+\frac {40}{x -2}+10 \ln \left (x -1\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x-10)*ln(x-1)-50*x+60)/(x^3-5*x^2+8*x-4),x,method=_RETURNVERBOSE)

[Out]

(-10*ln(x-1)+40)/(x-2)

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maxima [A]  time = 0.40, size = 33, normalized size = 1.74 \begin {gather*} -\frac {10 \, {\left ({\left (6 \, x - 11\right )} \log \left (x - 1\right ) - 10\right )}}{x - 2} - \frac {60}{x - 2} + 60 \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-10)*log(x-1)-50*x+60)/(x^3-5*x^2+8*x-4),x, algorithm="maxima")

[Out]

-10*((6*x - 11)*log(x - 1) - 10)/(x - 2) - 60/(x - 2) + 60*log(x - 1)

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mupad [B]  time = 0.12, size = 15, normalized size = 0.79 \begin {gather*} -\frac {10\,\ln \left (x-1\right )-40}{x-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x - 1)*(10*x - 10) - 50*x + 60)/(8*x - 5*x^2 + x^3 - 4),x)

[Out]

-(10*log(x - 1) - 40)/(x - 2)

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sympy [A]  time = 0.15, size = 14, normalized size = 0.74 \begin {gather*} - \frac {10 \log {\left (x - 1 \right )}}{x - 2} + \frac {40}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-10)*ln(x-1)-50*x+60)/(x**3-5*x**2+8*x-4),x)

[Out]

-10*log(x - 1)/(x - 2) + 40/(x - 2)

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