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

Optimal. Leaf size=36 \[ \frac {\text {li}(c (d+e x))}{c e}-\frac {d+e x}{e \log (c (d+e x))} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2389, 2297, 2298} \[ \frac {\text {li}(c (d+e x))}{c e}-\frac {d+e x}{e \log (c (d+e x))} \]

Antiderivative was successfully verified.

[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 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]

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]

Rubi steps

\begin {align*} \int \frac {1}{\log ^2(c (d+e x))} \, dx &=\frac {\operatorname {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 {\operatorname {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*}

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Mathematica [A]  time = 0.01, size = 36, normalized size = 1.00 \[ \frac {\text {li}(c (d+e x))}{c e}-\frac {d+e x}{e \log (c (d+e x))} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.50, size = 47, normalized size = 1.31 \[ -\frac {c e x + c d - \log \left (c e x + c d\right ) \operatorname {log\_integral}\left (c e x + c d\right )}{c e \log \left (c e x + c d\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.18, size = 38, normalized size = 1.06 \[ \frac {{\rm Ei}\left (\log \left ({\left (x e + d\right )} c\right )\right ) e^{\left (-1\right )}}{c} - \frac {{\left (x e + d\right )} e^{\left (-1\right )}}{\log \left ({\left (x e + d\right )} c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 54, normalized size = 1.50 \[ -\frac {x}{\ln \left (c e x +c d \right )}-\frac {\Ei \left (1, -\ln \left (c e x +c d \right )\right )}{c e}-\frac {d}{e \ln \left (c e x +c d \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.98, size = 20, normalized size = 0.56 \[ \frac {\Gamma \left (-1, -\log \left (c e x + c d\right )\right )}{c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

gamma(-1, -log(c*e*x + c*d))/(c*e)

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mupad [B]  time = 0.23, size = 36, normalized size = 1.00 \[ \frac {\mathrm {logint}\left (c\,\left (d+e\,x\right )\right )}{c\,e}-\frac {d+e\,x}{e\,\ln \left (c\,\left (d+e\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [A]  time = 0.80, size = 29, normalized size = 0.81 \[ \frac {- d - e x}{e \log {\left (c \left (d + e x\right ) \right )}} + \frac {\operatorname {li}{\left (c d + c e x \right )}}{c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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