3.156 \(\int \frac{\log (t)}{1+t} \, dt\)

Optimal. Leaf size=13 \[ \text{PolyLog}(2,-t)+\log (t) \log (t+1) \]

[Out]

Log[t]*Log[1 + t] + PolyLog[2, -t]

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Rubi [A]  time = 0.0153927, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2317, 2391} \[ \text{PolyLog}(2,-t)+\log (t) \log (t+1) \]

Antiderivative was successfully verified.

[In]

Int[Log[t]/(1 + t),t]

[Out]

Log[t]*Log[1 + t] + PolyLog[2, -t]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\log (t)}{1+t} \, dt &=\log (t) \log (1+t)-\int \frac{\log (1+t)}{t} \, dt\\ &=\log (t) \log (1+t)+\text{Li}_2(-t)\\ \end{align*}

Mathematica [A]  time = 0.0015629, size = 13, normalized size = 1. \[ \text{PolyLog}(2,-t)+\log (t) \log (t+1) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[t]/(1 + t),t]

[Out]

Log[t]*Log[1 + t] + PolyLog[2, -t]

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Maple [A]  time = 0.002, size = 13, normalized size = 1. \begin{align*}{\it dilog} \left ( 1+t \right ) +\ln \left ( t \right ) \ln \left ( 1+t \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

dilog(1+t)+ln(t)*ln(1+t)

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Maxima [A]  time = 0.943187, size = 16, normalized size = 1.23 \begin{align*} \log \left (t + 1\right ) \log \left (t\right ) +{\rm Li}_2\left (-t\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(t + 1)*log(t) + dilog(-t)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left (t\right )}{t + 1}, t\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(t)/(t + 1), t)

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Sympy [C]  time = 2.34793, size = 58, normalized size = 4.46 \begin{align*} \begin{cases} i \pi \log{\left (t + 1 \right )} - \operatorname{Li}_{2}\left (t + 1\right ) & \text{for}\: \left |{t + 1}\right | < 1 \\- i \pi \log{\left (\frac{1}{t + 1} \right )} - \operatorname{Li}_{2}\left (t + 1\right ) & \text{for}\: \frac{1}{\left |{t + 1}\right |} < 1 \\- i \pi{G_{2, 2}^{2, 0}\left (\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle |{t + 1} \right )} + i \pi{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{t + 1} \right )} - \operatorname{Li}_{2}\left (t + 1\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*pi*log(t + 1) - polylog(2, t + 1), Abs(t + 1) < 1), (-I*pi*log(1/(t + 1)) - polylog(2, t + 1), 1/
Abs(t + 1) < 1), (-I*pi*meijerg(((), (1, 1)), ((0, 0), ()), t + 1) + I*pi*meijerg(((1, 1), ()), ((), (0, 0)),
t + 1) - polylog(2, t + 1), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (t\right )}{t + 1}\,{d t} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(t)/(t + 1), t)