3.638 \(\int \frac{\log (-1+x)}{x^3} \, dx\)

Optimal. Leaf size=35 \[ -\frac{\log (x-1)}{2 x^2}+\frac{1}{2 x}+\frac{1}{2} \log (1-x)-\frac{\log (x)}{2} \]

[Out]

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

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Rubi [A]  time = 0.0145775, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2395, 44} \[ -\frac{\log (x-1)}{2 x^2}+\frac{1}{2 x}+\frac{1}{2} \log (1-x)-\frac{\log (x)}{2} \]

Antiderivative was successfully verified.

[In]

Int[Log[-1 + x]/x^3,x]

[Out]

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

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0102805, size = 27, normalized size = 0.77 \[ \frac{1}{2} \left (-\frac{\log (x-1)}{x^2}+\frac{1}{x}+\log (1-x)-\log (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[-1 + x]/x^3,x]

[Out]

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

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Maple [A]  time = 0.011, size = 26, normalized size = 0.7 \begin{align*} -{\frac{\ln \left ( x \right ) }{2}}+{\frac{1}{2\,x}}+{\frac{\ln \left ( -1+x \right ) \left ( -1+x \right ) \left ( 1+x \right ) }{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(-1+x)/x^3,x)

[Out]

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

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Maxima [A]  time = 0.944461, size = 34, normalized size = 0.97 \begin{align*} \frac{1}{2 \, x} - \frac{\log \left (x - 1\right )}{2 \, x^{2}} + \frac{1}{2} \, \log \left (x - 1\right ) - \frac{1}{2} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-1+x)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.37088, size = 68, normalized size = 1.94 \begin{align*} -\frac{x^{2} \log \left (x\right ) -{\left (x^{2} - 1\right )} \log \left (x - 1\right ) - x}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-1+x)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.127624, size = 26, normalized size = 0.74 \begin{align*} - \frac{\log{\left (x \right )}}{2} + \frac{\log{\left (x - 1 \right )}}{2} + \frac{1}{2 x} - \frac{\log{\left (x - 1 \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(-1+x)/x**3,x)

[Out]

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

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Giac [A]  time = 1.0585, size = 36, normalized size = 1.03 \begin{align*} \frac{1}{2 \, x} - \frac{\log \left (x - 1\right )}{2 \, x^{2}} + \frac{1}{2} \, \log \left ({\left | x - 1 \right |}\right ) - \frac{1}{2} \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-1+x)/x^3,x, algorithm="giac")

[Out]

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