\(\int \frac {1}{\log ^3(c (d+e x))} \, dx\) [7]

Optimal result

Integrand size = 10, antiderivative size = 63 \[ \int \frac {1}{\log ^3(c (d+e x))} \, dx=-\frac {d+e x}{2 e \log ^2(c (d+e x))}-\frac {d+e x}{2 e \log (c (d+e x))}+\frac {\operatorname {LogIntegral}(c (d+e x))}{2 c e} \]

[Out]

1/2*Li(c*(e*x+d))/c/e+1/2*(-e*x-d)/e/ln(c*(e*x+d))^2+1/2*(-e*x-d)/e/ln(c*(e*x+d))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2436, 2334, 2335} \[ \int \frac {1}{\log ^3(c (d+e x))} \, dx=\frac {\operatorname {LogIntegral}(c (d+e x))}{2 c e}-\frac {d+e x}{2 e \log ^2(c (d+e x))}-\frac {d+e x}{2 e \log (c (d+e x))} \]

[In]

Int[Log[c*(d + e*x)]^(-3),x]

[Out]

-1/2*(d + e*x)/(e*Log[c*(d + e*x)]^2) - (d + e*x)/(2*e*Log[c*(d + e*x)]) + LogIntegral[c*(d + e*x)]/(2*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 ^3(c x)} \, dx,x,d+e x\right )}{e} \\ & = -\frac {d+e x}{2 e \log ^2(c (d+e x))}+\frac {\text {Subst}\left (\int \frac {1}{\log ^2(c x)} \, dx,x,d+e x\right )}{2 e} \\ & = -\frac {d+e x}{2 e \log ^2(c (d+e x))}-\frac {d+e x}{2 e \log (c (d+e x))}+\frac {\text {Subst}\left (\int \frac {1}{\log (c x)} \, dx,x,d+e x\right )}{2 e} \\ & = -\frac {d+e x}{2 e \log ^2(c (d+e x))}-\frac {d+e x}{2 e \log (c (d+e x))}+\frac {\text {li}(c (d+e x))}{2 c e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\log ^3(c (d+e x))} \, dx=\frac {-\frac {(d+e x) (1+\log (c (d+e x)))}{\log ^2(c (d+e x))}+\frac {\operatorname {LogIntegral}(c (d+e x))}{c}}{2 e} \]

[In]

Integrate[Log[c*(d + e*x)]^(-3),x]

[Out]

(-(((d + e*x)*(1 + Log[c*(d + e*x)]))/Log[c*(d + e*x)]^2) + LogIntegral[c*(d + e*x)]/c)/(2*e)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02

method	result	size
risch	\(-\frac {\ln \left (c \left (e x +d \right )\right ) x e +d \ln \left (c \left (e x +d \right )\right )+e x +d}{2 e \ln \left (c \left (e x +d \right )\right )^{2}}-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{2 c e}\)	\(64\)
derivativedivides	\(\frac {-\frac {c e x +c d}{2 \ln \left (c e x +c d \right )^{2}}-\frac {c e x +c d}{2 \ln \left (c e x +c d \right )}-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{2}}{c e}\)	\(66\)
default	\(\frac {-\frac {c e x +c d}{2 \ln \left (c e x +c d \right )^{2}}-\frac {c e x +c d}{2 \ln \left (c e x +c d \right )}-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{2}}{c e}\)	\(66\)

[In]

int(1/ln(c*(e*x+d))^3,x,method=_RETURNVERBOSE)

[Out]

-1/2*(ln(c*(e*x+d))*x*e+d*ln(c*(e*x+d))+e*x+d)/e/ln(c*(e*x+d))^2-1/2/c/e*Ei(1,-ln(c*e*x+c*d))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\log ^3(c (d+e x))} \, dx=-\frac {c e x - \log \left (c e x + c d\right )^{2} \operatorname {log\_integral}\left (c e x + c d\right ) + c d + {\left (c e x + c d\right )} \log \left (c e x + c d\right )}{2 \, c e \log \left (c e x + c d\right )^{2}} \]

[In]

integrate(1/log(c*(e*x+d))^3,x, algorithm="fricas")

[Out]

-1/2*(c*e*x - log(c*e*x + c*d)^2*log_integral(c*e*x + c*d) + c*d + (c*e*x + c*d)*log(c*e*x + c*d))/(c*e*log(c*
e*x + c*d)^2)

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\log ^3(c (d+e x))} \, dx=\frac {- d - e x + \left (- d - e x\right ) \log {\left (c \left (d + e x\right ) \right )}}{2 e \log {\left (c \left (d + e x\right ) \right )}^{2}} + \frac {\operatorname {li}{\left (c d + c e x \right )}}{2 c e} \]

[In]

integrate(1/ln(c*(e*x+d))**3,x)

[Out]

(-d - e*x + (-d - e*x)*log(c*(d + e*x)))/(2*e*log(c*(d + e*x))**2) + li(c*d + c*e*x)/(2*c*e)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.33 \[ \int \frac {1}{\log ^3(c (d+e x))} \, dx=-\frac {\Gamma \left (-2, -\log \left (c e x + c d\right )\right )}{c e} \]

[In]

integrate(1/log(c*(e*x+d))^3,x, algorithm="maxima")

[Out]

-gamma(-2, -log(c*e*x + c*d))/(c*e)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\log ^3(c (d+e x))} \, dx=\frac {{\rm Ei}\left (\log \left ({\left (e x + d\right )} c\right )\right )}{2 \, c e} - \frac {e x + d}{2 \, e \log \left ({\left (e x + d\right )} c\right )} - \frac {e x + d}{2 \, e \log \left ({\left (e x + d\right )} c\right )^{2}} \]

[In]

integrate(1/log(c*(e*x+d))^3,x, algorithm="giac")

[Out]

1/2*Ei(log((e*x + d)*c))/(c*e) - 1/2*(e*x + d)/(e*log((e*x + d)*c)) - 1/2*(e*x + d)/(e*log((e*x + d)*c)^2)

Mupad [B] (verification not implemented)

Time = 1.50 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\log ^3(c (d+e x))} \, dx=\frac {\mathrm {logint}\left (c\,\left (d+e\,x\right )\right )}{2\,c\,e}-\frac {\frac {c\,d}{2}+\ln \left (c\,\left (d+e\,x\right )\right )\,\left (\frac {c\,d}{2}+\frac {c\,e\,x}{2}\right )+\frac {c\,e\,x}{2}}{c\,e\,{\ln \left (c\,\left (d+e\,x\right )\right )}^2} \]

[In]

int(1/log(c*(d + e*x))^3,x)

[Out]

logint(c*(d + e*x))/(2*c*e) - ((c*d)/2 + log(c*(d + e*x))*((c*d)/2 + (c*e*x)/2) + (c*e*x)/2)/(c*e*log(c*(d + e
*x))^2)