\(\int \frac {60-50 x+(-10+10 x) \log (-1+x)}{-4+8 x-5 x^2+x^3} \, dx\) [8332]

Optimal result

Integrand size = 31, antiderivative size = 19 \[ \int \frac {60-50 x+(-10+10 x) \log (-1+x)}{-4+8 x-5 x^2+x^3} \, dx=2+\frac {10 (2 x-\log (-1+x))}{-2+x} \]

[Out]

2*(2*x-ln(-1+x))/(1/5*x-2/5)+2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {6874, 78, 2442, 36, 31} \[ \int \frac {60-50 x+(-10+10 x) \log (-1+x)}{-4+8 x-5 x^2+x^3} \, dx=\frac {10 \log (x-1)}{2-x}-\frac {40}{2-x} \]

[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 78

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 2442

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 6874

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

Rubi steps \begin{align*} \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{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(39\) vs. \(2(19)=38\).

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.05 \[ \int \frac {60-50 x+(-10+10 x) \log (-1+x)}{-4+8 x-5 x^2+x^3} \, dx=10 \left (2 \text {arctanh}(3-2 x)+\log (1-x)-\log (2-x)+\frac {4-\log (-1+x)}{-2+x}\right ) \]

[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))

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {60-50 x+(-10+10 x) \log (-1+x)}{-4+8 x-5 x^2+x^3} \, dx=-\frac {10 \, {\left (\log \left (x - 1\right ) - 4\right )}}{x - 2} \]

[In]

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

[Out]

-10*(log(x - 1) - 4)/(x - 2)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {60-50 x+(-10+10 x) \log (-1+x)}{-4+8 x-5 x^2+x^3} \, dx=- \frac {10 \log {\left (x - 1 \right )}}{x - 2} + \frac {40}{x - 2} \]

[In]

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

[Out]

-10*log(x - 1)/(x - 2) + 40/(x - 2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {60-50 x+(-10+10 x) \log (-1+x)}{-4+8 x-5 x^2+x^3} \, dx=-\frac {10 \, {\left ({\left (6 \, x - 11\right )} \log \left (x - 1\right ) - 10\right )}}{x - 2} - \frac {60}{x - 2} + 60 \, \log \left (x - 1\right ) \]

[In]

integrate(((10*x-10)*log(-1+x)-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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {60-50 x+(-10+10 x) \log (-1+x)}{-4+8 x-5 x^2+x^3} \, dx=-\frac {10 \, \log \left (x - 1\right )}{x - 2} + \frac {40}{x - 2} \]

[In]

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

[Out]

-10*log(x - 1)/(x - 2) + 40/(x - 2)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {60-50 x+(-10+10 x) \log (-1+x)}{-4+8 x-5 x^2+x^3} \, dx=-\frac {10\,\ln \left (x-1\right )-40}{x-2} \]

[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)