\(\int \frac {1}{\log (t)} \, dt\) [170]

Optimal result

Integrand size = 4, antiderivative size = 2 \[ \int \frac {1}{\log (t)} \, dt=\operatorname {LogIntegral}(t) \]

[Out]

Li(t)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 2, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2335} \[ \int \frac {1}{\log (t)} \, dt=\operatorname {LogIntegral}(t) \]

[In]

Int[Log[t]^(-1),t]

[Out]

LogIntegral[t]

Rule 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rubi steps \begin{align*} \text {integral}& = \operatorname {LogIntegral}(t) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 2, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\log (t)} \, dt=\operatorname {LogIntegral}(t) \]

[In]

Integrate[Log[t]^(-1),t]

[Out]

LogIntegral[t]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(8\) vs. \(2(2)=4\).

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 4.50

method	result	size
default	\(-\operatorname {Ei}_{1}\left (-\ln \left (t \right )\right )\)	\(9\)
risch	\(-\operatorname {Ei}_{1}\left (-\ln \left (t \right )\right )\)	\(9\)

[In]

int(1/ln(t),t,method=_RETURNVERBOSE)

[Out]

-Ei(1,-ln(t))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 2, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\log (t)} \, dt=\operatorname {log\_integral}\left (t\right ) \]

[In]

integrate(1/log(t),t, algorithm="fricas")

[Out]

log_integral(t)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 2, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\log (t)} \, dt=\operatorname {li}{\left (t \right )} \]

[In]

integrate(1/ln(t),t)

[Out]

li(t)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 3, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\log (t)} \, dt={\rm Ei}\left (\log \left (t\right )\right ) \]

[In]

integrate(1/log(t),t, algorithm="maxima")

[Out]

Ei(log(t))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 3, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\log (t)} \, dt={\rm Ei}\left (\log \left (t\right )\right ) \]

[In]

integrate(1/log(t),t, algorithm="giac")

[Out]

Ei(log(t))

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 2, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\log (t)} \, dt=\mathrm {logint}\left (t\right ) \]

[In]

int(1/log(t),t)

[Out]

logint(t)