3.2.56 \(\int \frac {\log (t)}{1+t} \, dt\) [156]

Optimal. Leaf size=13 \[ \log (t) \log (1+t)+\text {Li}_2(-t) \]

[Out]

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

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Rubi [A]
time = 0.01, 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 = {2354, 2438} \begin {gather*} \text {Li}_2(-t)+\log (t) \log (t+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[t]/(1 + t),t]

[Out]

Log[t]*Log[1 + t] + PolyLog[2, -t]

Rule 2354

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 2438

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 \begin {gather*} \log (t) \log (1+t)+\text {Li}_2(-t) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[t]/(1 + t),t]

[Out]

Log[t]*Log[1 + t] + PolyLog[2, -t]

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 4 in optimal.
time = 2.94, size = 115, normalized size = 8.85 \begin {gather*} \text {Piecewise}\left [\left \{\left \{-\text {polylog}\left [2,1+t\right ],\text {Abs}\left [1+t\right ]<1\text {\&\&}\frac {1}{\text {Abs}\left [1+t\right ]}<1\right \},\left \{I \text {Pi} \text {Log}\left [1+t\right ]-\text {polylog}\left [2,1+t\right ],\text {Abs}\left [1+t\right ]<1\right \},\left \{-I \text {Pi} \text {Log}\left [\frac {1}{1+t}\right ]-\text {polylog}\left [2,1+t\right ],\frac {1}{\text {Abs}\left [1+t\right ]}<1\right \}\right \},-I \text {Pi} \text {meijerg}\left [\left \{\left \{\right \},\left \{1,1\right \}\right \},\left \{\left \{0,0\right \},\left \{\right \}\right \},1+t\right ]+I \text {Pi} \text {meijerg}\left [\left \{\left \{1,1\right \},\left \{\right \}\right \},\left \{\left \{\right \},\left \{0,0\right \}\right \},1+t\right ]-\text {polylog}\left [2,1+t\right ]\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[Log[t]/(t + 1),t]')

[Out]

Piecewise[{{-polylog[2, 1 + t], Abs[1 + t] < 1 && 1 / Abs[1 + t] < 1}, {I Pi Log[1 + t] - polylog[2, 1 + t], A
bs[1 + t] < 1}, {-I Pi Log[1 / (1 + t)] - polylog[2, 1 + t], 1 / Abs[1 + t] < 1}}, -I Pi meijerg[{{}, {1, 1}},
{{0, 0}, {}}, 1 + t] + I Pi meijerg[{{1, 1}, {}}, {{}, {0, 0}}, 1 + t] - polylog[2, 1 + t]]

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Maple [A]
time = 0.07, size = 13, normalized size = 1.00

method	result	size
default	\(\dilog \left (1+t \right )+\ln \left (t \right ) \ln \left (1+t \right )\)	\(13\)
risch	\(\dilog \left (1+t \right )+\ln \left (t \right ) \ln \left (1+t \right )\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.34, size = 12, normalized size = 0.92 \begin {gather*} \log \left (t + 1\right ) \log \left (t\right ) + {\rm Li}_2\left (-t\right ) \end {gather*}

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 = 0.35, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 [A]
time = 0.95, size = 73, normalized size = 5.62 \begin {gather*} \begin {cases} - \operatorname {Li}_{2}\left (t + 1\right ) & \text {for}\: \frac {1}{\left |{t + 1}\right |} < 1 \wedge \left |{t + 1}\right | < 1 \\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-polylog(2, t + 1), (Abs(t + 1) < 1) & (1/Abs(t + 1) < 1)), (I*pi*log(t + 1) - polylog(2, t + 1), A
bs(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] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Could not integrate

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Mupad [B]
time = 0.03, size = 13, normalized size = 1.00 \begin {gather*} \mathrm {polylog}\left (2,-t\right )+\ln \left (t+1\right )\,\ln \left (t\right ) \end {gather*}

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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