\(\int \frac {1}{\log ^2(c x)} \, dx\) [32]

Optimal result

Integrand size = 6, antiderivative size = 18 \[ \int \frac {1}{\log ^2(c x)} \, dx=-\frac {x}{\log (c x)}+\frac {\operatorname {LogIntegral}(c x)}{c} \]

[Out]

Li(c*x)/c-x/ln(c*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2334, 2335} \[ \int \frac {1}{\log ^2(c x)} \, dx=\frac {\operatorname {LogIntegral}(c x)}{c}-\frac {x}{\log (c x)} \]

[In]

Int[Log[c*x]^(-2),x]

[Out]

-(x/Log[c*x]) + LogIntegral[c*x]/c

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]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\log ^2(c x)} \, dx=-\frac {x}{\log (c x)}+\frac {\operatorname {LogIntegral}(c x)}{c} \]

[In]

Integrate[Log[c*x]^(-2),x]

[Out]

-(x/Log[c*x]) + LogIntegral[c*x]/c

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33

method	result	size
risch	\(-\frac {x}{\ln \left (x c \right )}-\frac {\operatorname {Ei}_{1}\left (-\ln \left (x c \right )\right )}{c}\)	\(24\)
derivativedivides	\(\frac {-\frac {x c}{\ln \left (x c \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x c \right )\right )}{c}\)	\(26\)
default	\(\frac {-\frac {x c}{\ln \left (x c \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x c \right )\right )}{c}\)	\(26\)

[In]

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

[Out]

-x/ln(x*c)-1/c*Ei(1,-ln(x*c))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {1}{\log ^2(c x)} \, dx=-\frac {c x - \log \left (c x\right ) \operatorname {log\_integral}\left (c x\right )}{c \log \left (c x\right )} \]

[In]

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

[Out]

-(c*x - log(c*x)*log_integral(c*x))/(c*log(c*x))

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\log ^2(c x)} \, dx=- \frac {x}{\log {\left (c x \right )}} + \frac {\operatorname {li}{\left (c x \right )}}{c} \]

[In]

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

[Out]

-x/log(c*x) + li(c*x)/c

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\log ^2(c x)} \, dx=\frac {\Gamma \left (-1, -\log \left (c x\right )\right )}{c} \]

[In]

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

[Out]

gamma(-1, -log(c*x))/c

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\log ^2(c x)} \, dx=\frac {{\rm Ei}\left (\log \left (c x\right )\right )}{c} - \frac {x}{\log \left (c x\right )} \]

[In]

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

[Out]

Ei(log(c*x))/c - x/log(c*x)

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\log ^2(c x)} \, dx=\frac {\mathrm {logint}\left (c\,x\right )}{c}-\frac {x}{\ln \left (c\,x\right )} \]

[In]

int(1/log(c*x)^2,x)

[Out]

logint(c*x)/c - x/log(c*x)