3.208 \(\int \log (a \tanh (x)) \, dx\)

Optimal. Leaf size=41 \[ x \log (a \tanh (x))+\frac {1}{2} \text {Li}_2\left (-e^{2 x}\right )-\frac {\text {Li}_2\left (e^{2 x}\right )}{2}+2 x \tanh ^{-1}\left (e^{2 x}\right ) \]

[Out]

2*x*arctanh(exp(2*x))+x*ln(a*tanh(x))+1/2*polylog(2,-exp(2*x))-1/2*polylog(2,exp(2*x))

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Rubi [A]  time = 0.04, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2548, 5461, 4182, 2279, 2391} \[ \frac {1}{2} \text {PolyLog}\left (2,-e^{2 x}\right )-\frac {1}{2} \text {PolyLog}\left (2,e^{2 x}\right )+x \log (a \tanh (x))+2 x \tanh ^{-1}\left (e^{2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[a*Tanh[x]],x]

[Out]

2*x*ArcTanh[E^(2*x)] + x*Log[a*Tanh[x]] + PolyLog[2, -E^(2*x)]/2 - PolyLog[2, E^(2*x)]/2

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rubi steps

\begin {align*} \int \log (a \tanh (x)) \, dx &=x \log (a \tanh (x))-\int x \text {csch}(x) \text {sech}(x) \, dx\\ &=x \log (a \tanh (x))-2 \int x \text {csch}(2 x) \, dx\\ &=2 x \tanh ^{-1}\left (e^{2 x}\right )+x \log (a \tanh (x))+\int \log \left (1-e^{2 x}\right ) \, dx-\int \log \left (1+e^{2 x}\right ) \, dx\\ &=2 x \tanh ^{-1}\left (e^{2 x}\right )+x \log (a \tanh (x))+\frac {1}{2} \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 x}\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 x}\right )\\ &=2 x \tanh ^{-1}\left (e^{2 x}\right )+x \log (a \tanh (x))+\frac {1}{2} \text {Li}_2\left (-e^{2 x}\right )-\frac {\text {Li}_2\left (e^{2 x}\right )}{2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 49, normalized size = 1.20 \[ -\frac {1}{2} \log (1-\tanh (x)) \log (a \tanh (x))+\frac {1}{2} \log (\tanh (x)+1) \log (a \tanh (x))+\frac {1}{2} \text {Li}_2(-\tanh (x))-\frac {\text {Li}_2(\tanh (x))}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[a*Tanh[x]],x]

[Out]

-1/2*(Log[1 - Tanh[x]]*Log[a*Tanh[x]]) + (Log[a*Tanh[x]]*Log[1 + Tanh[x]])/2 + PolyLog[2, -Tanh[x]]/2 - PolyLo
g[2, Tanh[x]]/2

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fricas [C]  time = 0.50, size = 102, normalized size = 2.49 \[ x \log \left (\frac {a \sinh \relax (x)}{\cosh \relax (x)}\right ) - x \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + x \log \left (i \, \cosh \relax (x) + i \, \sinh \relax (x) + 1\right ) + x \log \left (-i \, \cosh \relax (x) - i \, \sinh \relax (x) + 1\right ) - x \log \left (-\cosh \relax (x) - \sinh \relax (x) + 1\right ) - {\rm Li}_2\left (\cosh \relax (x) + \sinh \relax (x)\right ) + {\rm Li}_2\left (i \, \cosh \relax (x) + i \, \sinh \relax (x)\right ) + {\rm Li}_2\left (-i \, \cosh \relax (x) - i \, \sinh \relax (x)\right ) - {\rm Li}_2\left (-\cosh \relax (x) - \sinh \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*tanh(x)),x, algorithm="fricas")

[Out]

x*log(a*sinh(x)/cosh(x)) - x*log(cosh(x) + sinh(x) + 1) + x*log(I*cosh(x) + I*sinh(x) + 1) + x*log(-I*cosh(x)
- I*sinh(x) + 1) - x*log(-cosh(x) - sinh(x) + 1) - dilog(cosh(x) + sinh(x)) + dilog(I*cosh(x) + I*sinh(x)) + d
ilog(-I*cosh(x) - I*sinh(x)) - dilog(-cosh(x) - sinh(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log \left (a \tanh \relax (x)\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*tanh(x)),x, algorithm="giac")

[Out]

integrate(log(a*tanh(x)), x)

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maple [B]  time = 0.16, size = 70, normalized size = 1.71 \[ \frac {\ln \left (a \tanh \relax (x )\right ) \ln \left (\frac {a \tanh \relax (x )+a}{a}\right )}{2}-\frac {\ln \left (a \tanh \relax (x )\right ) \ln \left (-\frac {a \tanh \relax (x )-a}{a}\right )}{2}+\frac {\dilog \left (\frac {a \tanh \relax (x )+a}{a}\right )}{2}-\frac {\dilog \left (-\frac {a \tanh \relax (x )-a}{a}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(a*tanh(x)),x)

[Out]

-1/2*ln(a*tanh(x))*ln(-(a*tanh(x)-a)/a)-1/2*dilog(-(a*tanh(x)-a)/a)+1/2*ln(a*tanh(x))*ln((a*tanh(x)+a)/a)+1/2*
dilog((a*tanh(x)+a)/a)

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maxima [A]  time = 1.02, size = 56, normalized size = 1.37 \[ x \log \left (a \tanh \relax (x)\right ) + x \log \left (e^{\left (2 \, x\right )} + 1\right ) - x \log \left (e^{x} + 1\right ) - x \log \left (-e^{x} + 1\right ) + \frac {1}{2} \, {\rm Li}_2\left (-e^{\left (2 \, x\right )}\right ) - {\rm Li}_2\left (-e^{x}\right ) - {\rm Li}_2\left (e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*tanh(x)),x, algorithm="maxima")

[Out]

x*log(a*tanh(x)) + x*log(e^(2*x) + 1) - x*log(e^x + 1) - x*log(-e^x + 1) + 1/2*dilog(-e^(2*x)) - dilog(-e^x) -
dilog(e^x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \ln \left (a\,\mathrm {tanh}\relax (x)\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a*tanh(x)),x)

[Out]

int(log(a*tanh(x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\left (a \tanh {\relax (x )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(a*tanh(x)),x)

[Out]

Integral(log(a*tanh(x)), x)

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