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

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

[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))

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Rubi [A]
time = 0.02, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2436, 2334, 2335} \begin {gather*} \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))} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {1}{\log ^3(c (d+e x))} \, dx &=\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*}

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Mathematica [A]
time = 0.04, size = 48, normalized size = 0.76 \begin {gather*} \frac {\text {Ei}(\log (c (d+e x)))-\frac {c (d+e x) (1+\log (c (d+e x)))}{\log ^2(c (d+e x))}}{2 c e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(d + e*x)]^(-3),x]

[Out]

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

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Maple [A]
time = 0.31, size = 66, normalized size = 1.05

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 21, normalized size = 0.33 \begin {gather*} -\frac {e^{\left (-1\right )} \Gamma \left (-2, -\log \left (c x e + c d\right )\right )}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 72, normalized size = 1.14 \begin {gather*} -\frac {{\left (c x e - \log \left (c x e + c d\right )^{2} \operatorname {log\_integral}\left (c x e + c d\right ) + c d + {\left (c x e + c d\right )} \log \left (c x e + c d\right )\right )} e^{\left (-1\right )}}{2 \, c \log \left (c x e + c d\right )^{2}} \end {gather*}

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

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Sympy [A]
time = 0.37, size = 48, normalized size = 0.76 \begin {gather*} \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 {gather*}

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 = 4.65, size = 60, normalized size = 0.95 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.28, size = 64, normalized size = 1.02 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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