\(\int \frac {-1+\log (x)}{2 x^2} \, dx\) [5860]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 11, antiderivative size = 9 \[ \int \frac {-1+\log (x)}{2 x^2} \, dx=-\frac {\log (x)}{2 x} \]

[Out]

-1/2*ln(x)/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 2340} \[ \int \frac {-1+\log (x)}{2 x^2} \, dx=-\frac {\log (x)}{2 x} \]

[In]

Int[(-1 + Log[x])/(2*x^2),x]

[Out]

-1/2*Log[x]/x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2340

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[b*(d*x)^(m + 1)*(Log[c*x^n]/(d
*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {-1+\log (x)}{2 x^2} \, dx=-\frac {\log (x)}{2 x} \]

[In]

Integrate[(-1 + Log[x])/(2*x^2),x]

[Out]

-1/2*Log[x]/x

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89

method result size
default \(-\frac {\ln \left (x \right )}{2 x}\) \(8\)
norman \(-\frac {\ln \left (x \right )}{2 x}\) \(8\)
risch \(-\frac {\ln \left (x \right )}{2 x}\) \(8\)
parallelrisch \(-\frac {\ln \left (x \right )}{2 x}\) \(8\)
parts \(-\frac {\ln \left (x \right )}{2 x}\) \(8\)

[In]

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

[Out]

-1/2*ln(x)/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {-1+\log (x)}{2 x^2} \, dx=-\frac {\log \left (x\right )}{2 \, x} \]

[In]

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

[Out]

-1/2*log(x)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {-1+\log (x)}{2 x^2} \, dx=- \frac {\log {\left (x \right )}}{2 x} \]

[In]

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

[Out]

-log(x)/(2*x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (7) = 14\).

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.67 \[ \int \frac {-1+\log (x)}{2 x^2} \, dx=-\frac {\log \left (x\right ) + 1}{2 \, x} + \frac {1}{2 \, x} \]

[In]

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

[Out]

-1/2*(log(x) + 1)/x + 1/2/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {-1+\log (x)}{2 x^2} \, dx=-\frac {\log \left (x\right )}{2 \, x} \]

[In]

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

[Out]

-1/2*log(x)/x

Mupad [B] (verification not implemented)

Time = 12.29 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {-1+\log (x)}{2 x^2} \, dx=-\frac {\ln \left (x\right )}{2\,x} \]

[In]

int((log(x)/2 - 1/2)/x^2,x)

[Out]

-log(x)/(2*x)