3.218 \(\int \log (a \text {csch}(x)) \, dx\)

Optimal. Leaf size=38 \[ x \log (a \text {csch}(x))+\frac {\text {Li}_2\left (e^{2 x}\right )}{2}-\frac {x^2}{2}+x \log \left (1-e^{2 x}\right ) \]

[Out]

-1/2*x^2+x*ln(1-exp(2*x))+x*ln(a*csch(x))+1/2*polylog(2,exp(2*x))

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Rubi [A]  time = 0.05, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2548, 3716, 2190, 2279, 2391} \[ \frac {1}{2} \text {PolyLog}\left (2,e^{2 x}\right )+x \log (a \text {csch}(x))-\frac {x^2}{2}+x \log \left (1-e^{2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^2/2 + x*Log[1 - E^(2*x)] + x*Log[a*Csch[x]] + PolyLog[2, E^(2*x)]/2

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, 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]

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 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 37, normalized size = 0.97 \[ \frac {1}{2} \left (x \left (2 \log (a \text {csch}(x))+x+2 \log \left (1-e^{-2 x}\right )\right )-\text {Li}_2\left (e^{-2 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(x + 2*Log[1 - E^(-2*x)] + 2*Log[a*Csch[x]]) - PolyLog[2, E^(-2*x)])/2

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fricas [B]  time = 0.47, size = 76, normalized size = 2.00 \[ -\frac {1}{2} \, x^{2} + x \log \left (\frac {2 \, {\left (a \cosh \relax (x) + a \sinh \relax (x)\right )}}{\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} - 1}\right ) + x \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + x \log \left (-\cosh \relax (x) - \sinh \relax (x) + 1\right ) + {\rm Li}_2\left (\cosh \relax (x) + \sinh \relax (x)\right ) + {\rm Li}_2\left (-\cosh \relax (x) - \sinh \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(a*csch(x)), x)

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maple [C]  time = 0.95, size = 293, normalized size = 7.71 \[ -\frac {i \pi x \,\mathrm {csgn}\left (i a \right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{2 x}-1}\right ) \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{x}}{{\mathrm e}^{2 x}-1}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i a \right ) \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{x}}{{\mathrm e}^{2 x}-1}\right )^{2}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 x}-1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{2 x}-1}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 x}-1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{2 x}-1}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{2 x}-1}\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{2 x}-1}\right )^{3}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{2 x}-1}\right ) \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{x}}{{\mathrm e}^{2 x}-1}\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{x}}{{\mathrm e}^{2 x}-1}\right )^{3}}{2}-\frac {x^{2}}{2}+x \ln \relax (a )+x \ln \left ({\mathrm e}^{x}\right )-\ln \left ({\mathrm e}^{2 x}-1\right ) \ln \left ({\mathrm e}^{x}\right )+\ln \left ({\mathrm e}^{x}+1\right ) \ln \left ({\mathrm e}^{x}\right )+\ln \relax (2) x +\dilog \left ({\mathrm e}^{x}+1\right )-\dilog \left ({\mathrm e}^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*ln(exp(x))+1/2*I*Pi*csgn(I*exp(x))*csgn(I*exp(x)/(exp(2*x)-1))^2*x+1/2*I*Pi*csgn(I*exp(x)/(exp(2*x)-1))*csgn
(I*a/(exp(2*x)-1)*exp(x))^2*x-1/2*I*Pi*csgn(I*exp(x))*csgn(I/(exp(2*x)-1))*csgn(I*exp(x)/(exp(2*x)-1))*x-1/2*I
*Pi*csgn(I*a)*csgn(I*exp(x)/(exp(2*x)-1))*csgn(I*a/(exp(2*x)-1)*exp(x))*x-1/2*I*Pi*csgn(I*exp(x)/(exp(2*x)-1))
^3*x+1/2*I*Pi*csgn(I*a)*csgn(I*a/(exp(2*x)-1)*exp(x))^2*x+ln(2)*x+x*ln(a)-1/2*x^2-1/2*I*Pi*csgn(I*a/(exp(2*x)-
1)*exp(x))^3*x+1/2*I*Pi*csgn(I/(exp(2*x)-1))*csgn(I*exp(x)/(exp(2*x)-1))^2*x-ln(exp(2*x)-1)*ln(exp(x))-dilog(e
xp(x))+dilog(exp(x)+1)+ln(exp(x)+1)*ln(exp(x))

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maxima [A]  time = 0.65, size = 37, normalized size = 0.97 \[ -\frac {1}{2} \, x^{2} + x \log \left (a \operatorname {csch}\relax (x)\right ) + x \log \left (e^{x} + 1\right ) + x \log \left (-e^{x} + 1\right ) + {\rm Li}_2\left (-e^{x}\right ) + {\rm Li}_2\left (e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \ln \left (\frac {a}{\mathrm {sinh}\relax (x)}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a/sinh(x)),x)

[Out]

int(log(a/sinh(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(a*csch(x)), x)

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