\(\int \frac {1}{\log (c (d+e x))} \, dx\) [5]

Optimal result

Integrand size = 10, antiderivative size = 15 \[ \int \frac {1}{\log (c (d+e x))} \, dx=\frac {\operatorname {LogIntegral}(c (d+e x))}{c e} \]

[Out]

Li(c*(e*x+d))/c/e

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2436, 2335} \[ \int \frac {1}{\log (c (d+e x))} \, dx=\frac {\operatorname {LogIntegral}(c (d+e x))}{c e} \]

[In]

Int[Log[c*(d + e*x)]^(-1),x]

[Out]

LogIntegral[c*(d + e*x)]/(c*e)

Rule 2335

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

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\log (c (d+e x))} \, dx=\frac {\operatorname {LogIntegral}(c (d+e x))}{c e} \]

[In]

Integrate[Log[c*(d + e*x)]^(-1),x]

[Out]

LogIntegral[c*(d + e*x)]/(c*e)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47

method	result	size
derivativedivides	\(-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{c e}\)	\(22\)
default	\(-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{c e}\)	\(22\)
risch	\(-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{c e}\)	\(22\)

[In]

int(1/ln(c*(e*x+d)),x,method=_RETURNVERBOSE)

[Out]

-1/c/e*Ei(1,-ln(c*e*x+c*d))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\log (c (d+e x))} \, dx=\frac {\operatorname {log\_integral}\left (c e x + c d\right )}{c e} \]

[In]

integrate(1/log(c*(e*x+d)),x, algorithm="fricas")

[Out]

log_integral(c*e*x + c*d)/(c*e)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\log (c (d+e x))} \, dx=\frac {\operatorname {li}{\left (c d + c e x \right )}}{c e} \]

[In]

integrate(1/ln(c*(e*x+d)),x)

[Out]

li(c*d + c*e*x)/(c*e)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \frac {1}{\log (c (d+e x))} \, dx=\frac {{\rm Ei}\left (\log \left (c e x + c d\right )\right )}{c e} \]

[In]

integrate(1/log(c*(e*x+d)),x, algorithm="maxima")

[Out]

Ei(log(c*e*x + c*d))/(c*e)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\log (c (d+e x))} \, dx=\frac {{\rm Ei}\left (\log \left ({\left (e x + d\right )} c\right )\right )}{c e} \]

[In]

integrate(1/log(c*(e*x+d)),x, algorithm="giac")

[Out]

Ei(log((e*x + d)*c))/(c*e)

Mupad [B] (verification not implemented)

Time = 1.37 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\log (c (d+e x))} \, dx=\frac {\mathrm {logint}\left (c\,\left (d+e\,x\right )\right )}{c\,e} \]

[In]

int(1/log(c*(d + e*x)),x)

[Out]

logint(c*(d + e*x))/(c*e)