\(\int \frac {-1+\log (x)}{\log ^2(x)} \, dx\) [3673]

Optimal result

Integrand size = 9, antiderivative size = 12 \[ \int \frac {-1+\log (x)}{\log ^2(x)} \, dx=\frac {x}{\log (x)}+\log (3 \log (5)) \]

[Out]

ln(3*ln(5))+x/ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.50, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2407, 2334, 2335} \[ \int \frac {-1+\log (x)}{\log ^2(x)} \, dx=\frac {x}{\log (x)} \]

[In]

Int[(-1 + Log[x])/Log[x]^2,x]

[Out]

x/Log[x]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2407

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.50 \[ \int \frac {-1+\log (x)}{\log ^2(x)} \, dx=\frac {x}{\log (x)} \]

[In]

Integrate[(-1 + Log[x])/Log[x]^2,x]

[Out]

x/Log[x]

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {x}{\ln \left (x \right )}\)	\(7\)
norman	\(\frac {x}{\ln \left (x \right )}\)	\(7\)
risch	\(\frac {x}{\ln \left (x \right )}\)	\(7\)
parallelrisch	\(\frac {x}{\ln \left (x \right )}\)	\(7\)
parts	\(\frac {x}{\ln \left (x \right )}\)	\(7\)

[In]

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

[Out]

x/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.50 \[ \int \frac {-1+\log (x)}{\log ^2(x)} \, dx=\frac {x}{\log \left (x\right )} \]

[In]

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

[Out]

x/log(x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.25 \[ \int \frac {-1+\log (x)}{\log ^2(x)} \, dx=\frac {x}{\log {\left (x \right )}} \]

[In]

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

[Out]

x/log(x)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.22 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {-1+\log (x)}{\log ^2(x)} \, dx={\rm Ei}\left (\log \left (x\right )\right ) - \Gamma \left (-1, -\log \left (x\right )\right ) \]

[In]

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

[Out]

Ei(log(x)) - gamma(-1, -log(x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.50 \[ \int \frac {-1+\log (x)}{\log ^2(x)} \, dx=\frac {x}{\log \left (x\right )} \]

[In]

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

[Out]

x/log(x)

Mupad [B] (verification not implemented)

Time = 8.91 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.50 \[ \int \frac {-1+\log (x)}{\log ^2(x)} \, dx=\frac {x}{\ln \left (x\right )} \]

[In]

int((log(x) - 1)/log(x)^2,x)

[Out]

x/log(x)