### 3.209 $$\int \log (a \tanh ^2(x)) \, dx$$

Optimal. Leaf size=37 $\text{PolyLog}\left (2,-e^{2 x}\right )-\text{PolyLog}\left (2,e^{2 x}\right )+x \log \left (a \tanh ^2(x)\right )+4 x \tanh ^{-1}\left (e^{2 x}\right )$

[Out]

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

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Rubi [A]  time = 0.0470444, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 7, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.857, Rules used = {2548, 12, 5461, 4182, 2279, 2391} $\text{PolyLog}\left (2,-e^{2 x}\right )-\text{PolyLog}\left (2,e^{2 x}\right )+x \log \left (a \tanh ^2(x)\right )+4 x \tanh ^{-1}\left (e^{2 x}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[Log[a*Tanh[x]^2],x]

[Out]

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

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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 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]

Rubi steps

\begin{align*} \int \log \left (a \tanh ^2(x)\right ) \, dx &=x \log \left (a \tanh ^2(x)\right )-\int 2 x \text{csch}(x) \text{sech}(x) \, dx\\ &=x \log \left (a \tanh ^2(x)\right )-2 \int x \text{csch}(x) \text{sech}(x) \, dx\\ &=x \log \left (a \tanh ^2(x)\right )-4 \int x \text{csch}(2 x) \, dx\\ &=4 x \tanh ^{-1}\left (e^{2 x}\right )+x \log \left (a \tanh ^2(x)\right )+2 \int \log \left (1-e^{2 x}\right ) \, dx-2 \int \log \left (1+e^{2 x}\right ) \, dx\\ &=4 x \tanh ^{-1}\left (e^{2 x}\right )+x \log \left (a \tanh ^2(x)\right )+\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 x}\right )-\operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 x}\right )\\ &=4 x \tanh ^{-1}\left (e^{2 x}\right )+x \log \left (a \tanh ^2(x)\right )+\text{Li}_2\left (-e^{2 x}\right )-\text{Li}_2\left (e^{2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0095761, size = 47, normalized size = 1.27 $\text{PolyLog}(2,-\tanh (x))-\text{PolyLog}(2,\tanh (x))-\frac{1}{2} \log (1-\tanh (x)) \log \left (a \tanh ^2(x)\right )+\frac{1}{2} \log (\tanh (x)+1) \log \left (a \tanh ^2(x)\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Log[a*Tanh[x]^2],x]

[Out]

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

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Maple [A]  time = 0.03, size = 47, normalized size = 1.3 \begin{align*} -{\frac{\ln \left ( \tanh \left ( x \right ) -1 \right ) \ln \left ( a \left ( \tanh \left ( x \right ) \right ) ^{2} \right ) }{2}}+{\it dilog} \left ( \tanh \left ( x \right ) \right ) +\ln \left ( \tanh \left ( x \right ) -1 \right ) \ln \left ( \tanh \left ( x \right ) \right ) +{\frac{\ln \left ( \tanh \left ( x \right ) +1 \right ) \ln \left ( a \left ( \tanh \left ( x \right ) \right ) ^{2} \right ) }{2}}+{\it dilog} \left ( \tanh \left ( x \right ) +1 \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(ln(a*tanh(x)^2),x)

[Out]

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

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Maxima [A]  time = 1.54846, size = 77, normalized size = 2.08 \begin{align*} x \log \left (a \tanh \left (x\right )^{2}\right ) + 2 \, x \log \left (e^{\left (2 \, x\right )} + 1\right ) - 2 \, x \log \left (e^{x} + 1\right ) - 2 \, x \log \left (-e^{x} + 1\right ) +{\rm Li}_2\left (-e^{\left (2 \, x\right )}\right ) - 2 \,{\rm Li}_2\left (-e^{x}\right ) - 2 \,{\rm Li}_2\left (e^{x}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*tanh(x)^2),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 1.969, size = 454, normalized size = 12.27 \begin{align*} x \log \left (\frac{a \cosh \left (x\right )^{2} + a \sinh \left (x\right )^{2} - a}{\cosh \left (x\right )^{2} + \sinh \left (x\right )^{2} + 1}\right ) - 2 \, x \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + 2 \, x \log \left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right ) + 1\right ) + 2 \, x \log \left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right ) + 1\right ) - 2 \, x \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) - 2 \,{\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + 2 \,{\rm Li}_2\left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) + 2 \,{\rm Li}_2\left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right )\right ) - 2 \,{\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*tanh(x)^2),x, algorithm="fricas")

[Out]

x*log((a*cosh(x)^2 + a*sinh(x)^2 - a)/(cosh(x)^2 + sinh(x)^2 + 1)) - 2*x*log(cosh(x) + sinh(x) + 1) + 2*x*log(
I*cosh(x) + I*sinh(x) + 1) + 2*x*log(-I*cosh(x) - I*sinh(x) + 1) - 2*x*log(-cosh(x) - sinh(x) + 1) - 2*dilog(c
osh(x) + sinh(x)) + 2*dilog(I*cosh(x) + I*sinh(x)) + 2*dilog(-I*cosh(x) - I*sinh(x)) - 2*dilog(-cosh(x) - sinh
(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log{\left (a \tanh ^{2}{\left (x \right )} \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(a*tanh(x)**2),x)

[Out]

Integral(log(a*tanh(x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log \left (a \tanh \left (x\right )^{2}\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*tanh(x)^2),x, algorithm="giac")

[Out]

integrate(log(a*tanh(x)^2), x)