3.61 \(\int \frac{1}{\log (1+x)} \, dx\)

Optimal. Leaf size=4 \[ \text{LogIntegral}(x+1) \]

[Out]

LogIntegral[1 + x]

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Rubi [A]  time = 0.0037252, antiderivative size = 4, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2389, 2298} \[ \text{LogIntegral}(x+1) \]

Antiderivative was successfully verified.

[In]

Int[Log[1 + x]^(-1),x]

[Out]

LogIntegral[1 + 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]

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 (1+x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\log (x)} \, dx,x,1+x\right )\\ &=\text{li}(1+x)\\ \end{align*}

Mathematica [A]  time = 0.0101527, size = 5, normalized size = 1.25 \[ \text{ExpIntegralEi}(\log (x+1)) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[1 + x]^(-1),x]

[Out]

ExpIntegralEi[Log[1 + x]]

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Maple [B]  time = 0.002, size = 11, normalized size = 2.8 \begin{align*} -{\it Ei} \left ( 1,-\ln \left ( 1+x \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/ln(1+x),x)

[Out]

-Ei(1,-ln(1+x))

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Maxima [A]  time = 1.03548, size = 7, normalized size = 1.75 \begin{align*}{\rm Ei}\left (\log \left (x + 1\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/log(1+x),x, algorithm="maxima")

[Out]

Ei(log(x + 1))

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Fricas [A]  time = 1.85379, size = 28, normalized size = 7. \begin{align*} \logintegral \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/log(1+x),x, algorithm="fricas")

[Out]

log_integral(x + 1)

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Sympy [A]  time = 0.471035, size = 3, normalized size = 0.75 \begin{align*} \operatorname{li}{\left (x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/ln(1+x),x)

[Out]

li(x + 1)

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Giac [A]  time = 1.08028, size = 7, normalized size = 1.75 \begin{align*}{\rm Ei}\left (\log \left (x + 1\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/log(1+x),x, algorithm="giac")

[Out]

Ei(log(x + 1))