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

Optimal. Leaf size=63 \[ \frac{\text{li}(c (d+e x))}{2 c e}-\frac{d+e x}{2 e \log ^2(c (d+e x))}-\frac{d+e x}{2 e \log (c (d+e x))} \]

[Out]

-(d + e*x)/(2*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)

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Rubi [A]  time = 0.024063, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {2389, 2297, 2298} \[ \frac{\text{li}(c (d+e x))}{2 c e}-\frac{d+e x}{2 e \log ^2(c (d+e x))}-\frac{d+e x}{2 e \log (c (d+e x))} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(d + e*x)]^(-3),x]

[Out]

-(d + e*x)/(2*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 2389

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]

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 (d+e x))} \, dx &=\frac{\operatorname{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{\operatorname{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{\operatorname{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]  time = 0.0161225, size = 47, normalized size = 0.75 \[ \frac{\frac{\text{li}(c (d+e x))}{c}-\frac{(d+e x) (\log (c (d+e x))+1)}{\log ^2(c (d+e x))}}{2 e} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.061, size = 85, normalized size = 1.4 \begin{align*} -{\frac{x}{2\, \left ( \ln \left ( cex+cd \right ) \right ) ^{2}}}-{\frac{d}{2\,e \left ( \ln \left ( cex+cd \right ) \right ) ^{2}}}-{\frac{x}{2\,\ln \left ( cex+cd \right ) }}-{\frac{d}{2\,e\ln \left ( cex+cd \right ) }}-{\frac{{\it Ei} \left ( 1,-\ln \left ( cex+cd \right ) \right ) }{2\,ce}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.12941, size = 28, normalized size = 0.44 \begin{align*} -\frac{\Gamma \left (-2, -\log \left (c e x + c d\right )\right )}{c e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(-2, -log(c*e*x + c*d))/(c*e)

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Fricas [A]  time = 1.88756, size = 169, normalized size = 2.68 \begin{align*} -\frac{c e x - \log \left (c e x + c d\right )^{2} \logintegral \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.917525, size = 48, normalized size = 0.76 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.27196, size = 81, normalized size = 1.29 \begin{align*} \frac{{\rm Ei}\left (\log \left ({\left (x e + d\right )} c\right )\right ) e^{\left (-1\right )}}{2 \, c} - \frac{{\left (x e + d\right )} e^{\left (-1\right )}}{2 \, \log \left ({\left (x e + d\right )} c\right )} - \frac{{\left (x e + d\right )} e^{\left (-1\right )}}{2 \, \log \left ({\left (x e + d\right )} c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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