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

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

[Out]

ln(t)*ln(1+t)+polylog(2,-t)

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Rubi [A]  time = 0.02, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 1.00 \[ \operatorname {PolyLog}(2,-t)+\log (t) \log (t+1) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[t]/(1 + t),t]

[Out]

Log[t]*Log[1 + t] + PolyLog[2, -t]

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \relax (t)}{t + 1}, t\right ) \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: integrate((t+1)^-1*ln(t),t)

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maple [A]  time = 0.00, size = 13, normalized size = 1.00 \[ \ln \relax (t ) \ln \left (t +1\right )+\dilog \left (t +1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 12, normalized size = 0.92 \[ \log \left (t + 1\right ) \log \relax (t) + {\rm Li}_2\left (-t\right ) \]

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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mupad [B]  time = 0.03, size = 13, normalized size = 1.00 \[ \mathrm {polylog}\left (2,-t\right )+\ln \left (t+1\right )\,\ln \relax (t) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(t)/(t + 1),t)

[Out]

polylog(2, -t) + log(t + 1)*log(t)

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sympy [C]  time = 2.44, size = 58, normalized size = 4.46 \[ \begin {cases} i \pi \log {\left (t + 1 \right )} - \operatorname {Li}_{2}\left (t + 1\right ) & \text {for}\: \left |{t + 1}\right | < 1 \\- i \pi \log {\left (\frac {1}{t + 1} \right )} - \operatorname {Li}_{2}\left (t + 1\right ) & \text {for}\: \frac {1}{\left |{t + 1}\right |} < 1 \\- i \pi {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {t + 1} \right )} + i \pi {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {t + 1} \right )} - \operatorname {Li}_{2}\left (t + 1\right ) & \text {otherwise} \end {cases} \]

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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