3.1.6 \(\int \frac {1}{\log ^2(c (d+e x))} \, dx\) [6]

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

[Out]

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

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Rubi [A]
time = 0.01, 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 = {2436, 2334, 2335} \begin {gather*} \frac {\text {li}(c (d+e x))}{c e}-\frac {d+e x}{e \log (c (d+e x))} \end {gather*}

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 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 ^2(c (d+e x))} \, dx &=\frac {\text {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 {\text {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 = 0.03, size = 37, normalized size = 1.03 \begin {gather*} \frac {\text {Ei}(\log (c (d+e x)))}{c e}-\frac {d+e x}{e \log (c (d+e x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(d + e*x)]^(-2),x]

[Out]

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

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Maple [A]
time = 0.24, size = 45, normalized size = 1.25

method	result	size
risch	\(-\frac {e x +d}{\ln \left (c \left (e x +d \right )\right ) e}-\frac {\expIntegral \left (1, -\ln \left (c e x +c d \right )\right )}{c e}\)	\(43\)
derivativedivides	\(\frac {-\frac {c e x +c d}{\ln \left (c e x +c d \right )}-\expIntegral \left (1, -\ln \left (c e x +c d \right )\right )}{c e}\)	\(45\)
default	\(\frac {-\frac {c e x +c d}{\ln \left (c e x +c d \right )}-\expIntegral \left (1, -\ln \left (c e x +c d \right )\right )}{c e}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.33, size = 20, normalized size = 0.56 \begin {gather*} \frac {e^{\left (-1\right )} \Gamma \left (-1, -\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))^2,x, algorithm="maxima")

[Out]

e^(-1)*gamma(-1, -log(c*x*e + c*d))/c

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Fricas [A]
time = 0.34, size = 50, normalized size = 1.39 \begin {gather*} -\frac {{\left (c x e + c d - \log \left (c x e + c d\right ) \operatorname {log\_integral}\left (c x e + c d\right )\right )} e^{\left (-1\right )}}{c \log \left (c x e + c d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.36, size = 29, normalized size = 0.81 \begin {gather*} \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} \end {gather*}

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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Giac [A]
time = 3.92, size = 38, normalized size = 1.06 \begin {gather*} \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 )} \end {gather*}

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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Mupad [B]
time = 0.23, size = 36, normalized size = 1.00 \begin {gather*} \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 )} \end {gather*}

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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