3.39 \(\int \frac{1}{\log ^3(c x)} \, dx\)

Optimal. Leaf size=34 \[ \frac{\text{li}(c x)}{2 c}-\frac{x}{2 \log ^2(c x)}-\frac{x}{2 \log (c x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0087665, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2297, 2298} \[ \frac{\text{li}(c x)}{2 c}-\frac{x}{2 \log ^2(c x)}-\frac{x}{2 \log (c x)} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*x]^(-3),x]

[Out]

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

Rule 2297

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 2298

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

Rubi steps

\begin{align*} \int \frac{1}{\log ^3(c x)} \, dx &=-\frac{x}{2 \log ^2(c x)}+\frac{1}{2} \int \frac{1}{\log ^2(c x)} \, dx\\ &=-\frac{x}{2 \log ^2(c x)}-\frac{x}{2 \log (c x)}+\frac{1}{2} \int \frac{1}{\log (c x)} \, dx\\ &=-\frac{x}{2 \log ^2(c x)}-\frac{x}{2 \log (c x)}+\frac{\text{li}(c x)}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.0051097, size = 34, normalized size = 1. \[ \frac{\text{li}(c x)}{2 c}-\frac{x}{2 \log ^2(c x)}-\frac{x}{2 \log (c x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*x]^(-3),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.035, size = 33, normalized size = 1. \begin{align*} -{\frac{x}{2\, \left ( \ln \left ( cx \right ) \right ) ^{2}}}-{\frac{x}{2\,\ln \left ( cx \right ) }}-{\frac{{\it Ei} \left ( 1,-\ln \left ( cx \right ) \right ) }{2\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/ln(c*x)^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.1648, size = 18, normalized size = 0.53 \begin{align*} -\frac{\Gamma \left (-2, -\log \left (c x\right )\right )}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(-2, -log(c*x))/c

________________________________________________________________________________________

Fricas [A]  time = 0.663737, size = 99, normalized size = 2.91 \begin{align*} -\frac{c x \log \left (c x\right ) - \log \left (c x\right )^{2} \logintegral \left (c x\right ) + c x}{2 \, c \log \left (c x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.519021, size = 26, normalized size = 0.76 \begin{align*} \frac{- x \log{\left (c x \right )} - x}{2 \log{\left (c x \right )}^{2}} + \frac{\operatorname{li}{\left (c x \right )}}{2 c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/ln(c*x)**3,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.11764, size = 39, normalized size = 1.15 \begin{align*} \frac{{\rm Ei}\left (\log \left (c x\right )\right )}{2 \, c} - \frac{x}{2 \, \log \left (c x\right )} - \frac{x}{2 \, \log \left (c x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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