\(\int \frac {1-4 \log (x)+\log ^2(x)}{x (-1+\log (x))^4} \, dx\) [148]

Optimal result

Integrand size = 20, antiderivative size = 27 \[ \int \frac {1-4 \log (x)+\log ^2(x)}{x (-1+\log (x))^4} \, dx=-\frac {2}{3 (1-\log (x))^3}+\frac {1}{1-\log (x)}+\frac {1}{(-1+\log (x))^2} \]

[Out]

-2/3/(1-ln(x))^3+1/(1-ln(x))+1/(-1+ln(x))^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {712} \[ \int \frac {1-4 \log (x)+\log ^2(x)}{x (-1+\log (x))^4} \, dx=\frac {1}{(\log (x)-1)^2}+\frac {1}{1-\log (x)}-\frac {2}{3 (1-\log (x))^3} \]

[In]

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

[Out]

-2/(3*(1 - Log[x])^3) + (1 - Log[x])^(-1) + (-1 + Log[x])^(-2)

Rule 712

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

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1-4 x+x^2}{(-1+x)^4} \, dx,x,\log (x)\right ) \\ & = \text {Subst}\left (\int \left (-\frac {2}{(-1+x)^4}-\frac {2}{(-1+x)^3}+\frac {1}{(-1+x)^2}\right ) \, dx,x,\log (x)\right ) \\ & = -\frac {2}{3 (1-\log (x))^3}+\frac {1}{1-\log (x)}+\frac {1}{(-1+\log (x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1-4 \log (x)+\log ^2(x)}{x (-1+\log (x))^4} \, dx=\frac {-4+9 \log (x)-3 \log ^2(x)}{3 (-1+\log (x))^3} \]

[In]

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

[Out]

(-4 + 9*Log[x] - 3*Log[x]^2)/(3*(-1 + Log[x])^3)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74

method	result	size
norman	\(\frac {-\ln \left (x \right )^{2}+3 \ln \left (x \right )-\frac {4}{3}}{\left (-1+\ln \left (x \right )\right )^{3}}\)	\(20\)
risch	\(-\frac {3 \ln \left (x \right )^{2}-9 \ln \left (x \right )+4}{3 \left (-1+\ln \left (x \right )\right )^{3}}\)	\(21\)
parallelrisch	\(\frac {-4-3 \ln \left (x \right )^{2}+9 \ln \left (x \right )}{3 \left (-1+\ln \left (x \right )\right )^{3}}\)	\(21\)
derivativedivides	\(\frac {2}{3 \left (-1+\ln \left (x \right )\right )^{3}}-\frac {1}{-1+\ln \left (x \right )}+\frac {1}{\left (-1+\ln \left (x \right )\right )^{2}}\)	\(24\)
default	\(\frac {2}{3 \left (-1+\ln \left (x \right )\right )^{3}}-\frac {1}{-1+\ln \left (x \right )}+\frac {1}{\left (-1+\ln \left (x \right )\right )^{2}}\)	\(24\)

[In]

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

[Out]

(-ln(x)^2+3*ln(x)-4/3)/(-1+ln(x))^3

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {1-4 \log (x)+\log ^2(x)}{x (-1+\log (x))^4} \, dx=-\frac {3 \, \log \left (x\right )^{2} - 9 \, \log \left (x\right ) + 4}{3 \, {\left (\log \left (x\right )^{3} - 3 \, \log \left (x\right )^{2} + 3 \, \log \left (x\right ) - 1\right )}} \]

[In]

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

[Out]

-1/3*(3*log(x)^2 - 9*log(x) + 4)/(log(x)^3 - 3*log(x)^2 + 3*log(x) - 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {1-4 \log (x)+\log ^2(x)}{x (-1+\log (x))^4} \, dx=\frac {- 3 \log {\left (x \right )}^{2} + 9 \log {\left (x \right )} - 4}{3 \log {\left (x \right )}^{3} - 9 \log {\left (x \right )}^{2} + 9 \log {\left (x \right )} - 3} \]

[In]

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

[Out]

(-3*log(x)**2 + 9*log(x) - 4)/(3*log(x)**3 - 9*log(x)**2 + 9*log(x) - 3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {1-4 \log (x)+\log ^2(x)}{x (-1+\log (x))^4} \, dx=-\frac {3 \, \log \left (x\right )^{2} - 9 \, \log \left (x\right ) + 4}{3 \, {\left (\log \left (x\right )^{3} - 3 \, \log \left (x\right )^{2} + 3 \, \log \left (x\right ) - 1\right )}} \]

[In]

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

[Out]

-1/3*(3*log(x)^2 - 9*log(x) + 4)/(log(x)^3 - 3*log(x)^2 + 3*log(x) - 1)

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {1-4 \log (x)+\log ^2(x)}{x (-1+\log (x))^4} \, dx=-\frac {3 \, \log \left (x\right )^{2} - 9 \, \log \left (x\right ) + 4}{3 \, {\left (\log \left (x\right ) - 1\right )}^{3}} \]

[In]

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

[Out]

-1/3*(3*log(x)^2 - 9*log(x) + 4)/(log(x) - 1)^3

Mupad [B] (verification not implemented)

Time = 1.56 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {1-4 \log (x)+\log ^2(x)}{x (-1+\log (x))^4} \, dx=-\frac {{\ln \left (x\right )}^2-3\,\ln \left (x\right )+\frac {4}{3}}{{\left (\ln \left (x\right )-1\right )}^3} \]

[In]

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

[Out]

-(log(x)^2 - 3*log(x) + 4/3)/(log(x) - 1)^3