\(\int \frac {1}{\log ^2(c (d+e x))} \, dx\) [6]

Optimal result

Integrand size = 10, antiderivative size = 36 \[ \int \frac {1}{\log ^2(c (d+e x))} \, dx=-\frac {d+e x}{e \log (c (d+e x))}+\frac {\operatorname {LogIntegral}(c (d+e x))}{c e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2436, 2334, 2335} \[ \int \frac {1}{\log ^2(c (d+e x))} \, dx=\frac {\operatorname {LogIntegral}(c (d+e x))}{c e}-\frac {d+e x}{e \log (c (d+e x))} \]

[In]

Int[Log[c*(d + e*x)]^(-2),x]

[Out]

-((d + e*x)/(e*Log[c*(d + e*x)])) + LogIntegral[c*(d + e*x)]/(c*e)

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 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 ^2(c x)} \, dx,x,d+e x\right )}{e} \\ & = -\frac {d+e x}{e \log (c (d+e x))}+\frac {\text {Subst}\left (\int \frac {1}{\log (c x)} \, dx,x,d+e x\right )}{e} \\ & = -\frac {d+e x}{e \log (c (d+e x))}+\frac {\text {li}(c (d+e x))}{c e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\log ^2(c (d+e x))} \, dx=-\frac {d+e x}{e \log (c (d+e x))}+\frac {\operatorname {LogIntegral}(c (d+e x))}{c e} \]

[In]

Integrate[Log[c*(d + e*x)]^(-2),x]

[Out]

-((d + e*x)/(e*Log[c*(d + e*x)])) + LogIntegral[c*(d + e*x)]/(c*e)

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.19

method	result	size
risch	\(-\frac {e x +d}{\ln \left (c \left (e x +d \right )\right ) e}-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{c e}\)	\(43\)
derivativedivides	\(\frac {-\frac {c e x +c d}{\ln \left (c e x +c d \right )}-\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{c e}\)	\(45\)
default	\(\frac {-\frac {c e x +c d}{\ln \left (c e x +c d \right )}-\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{c e}\)	\(45\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\log ^2(c (d+e x))} \, dx=-\frac {c e x + c d - \log \left (c e x + c d\right ) \operatorname {log\_integral}\left (c e x + c d\right )}{c e \log \left (c e x + c d\right )} \]

[In]

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

[Out]

-(c*e*x + c*d - log(c*e*x + c*d)*log_integral(c*e*x + c*d))/(c*e*log(c*e*x + c*d))

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\log ^2(c (d+e x))} \, dx=\frac {- d - e x}{e \log {\left (c \left (d + e x\right ) \right )}} + \frac {\operatorname {li}{\left (c d + c e x \right )}}{c e} \]

[In]

integrate(1/ln(c*(e*x+d))**2,x)

[Out]

(-d - e*x)/(e*log(c*(d + e*x))) + li(c*d + c*e*x)/(c*e)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.56 \[ \int \frac {1}{\log ^2(c (d+e x))} \, dx=\frac {\Gamma \left (-1, -\log \left (c e x + c d\right )\right )}{c e} \]

[In]

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

[Out]

gamma(-1, -log(c*e*x + c*d))/(c*e)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\log ^2(c (d+e x))} \, dx=\frac {{\rm Ei}\left (\log \left ({\left (e x + d\right )} c\right )\right )}{c e} - \frac {e x + d}{e \log \left ({\left (e x + d\right )} c\right )} \]

[In]

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

[Out]

Ei(log((e*x + d)*c))/(c*e) - (e*x + d)/(e*log((e*x + d)*c))

Mupad [B] (verification not implemented)

Time = 1.36 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\log ^2(c (d+e x))} \, dx=\frac {\mathrm {logint}\left (c\,\left (d+e\,x\right )\right )}{c\,e}-\frac {d+e\,x}{e\,\ln \left (c\,\left (d+e\,x\right )\right )} \]

[In]

int(1/log(c*(d + e*x))^2,x)

[Out]

logint(c*(d + e*x))/(c*e) - (d + e*x)/(e*log(c*(d + e*x)))