3.1012 \(\int \text {csch}(x) \log (\tanh (x)) \text {sech}(x) \, dx\)

Optimal. Leaf size=9 \[ \frac {1}{2} \log ^2(\tanh (x)) \]

[Out]

1/2*ln(tanh(x))^2

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Rubi [A]  time = 0.03, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2620, 29, 6686} \[ \frac {1}{2} \log ^2(\tanh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]*Log[Tanh[x]]*Sech[x],x]

[Out]

Log[Tanh[x]]^2/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \text {csch}(x) \log (\tanh (x)) \text {sech}(x) \, dx &=\frac {1}{2} \log ^2(\tanh (x))\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 9, normalized size = 1.00 \[ \frac {1}{2} \log ^2(\tanh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]*Log[Tanh[x]]*Sech[x],x]

[Out]

Log[Tanh[x]]^2/2

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fricas [A]  time = 0.42, size = 12, normalized size = 1.33 \[ \frac {1}{2} \, \log \left (\frac {\sinh \relax (x)}{\cosh \relax (x)}\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(sinh(x)/cosh(x))^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(csch(x)*log(tanh(x))*sech(x), x)

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maple [A]  time = 0.17, size = 8, normalized size = 0.89 \[ \frac {\ln \left (\tanh \relax (x )\right )^{2}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)*ln(tanh(x))*sech(x),x)

[Out]

1/2*ln(tanh(x))^2

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maxima [B]  time = 1.48, size = 95, normalized size = 10.56 \[ {\left (\log \left (e^{x} + 1\right ) + \log \left (-e^{x} + 1\right )\right )} \log \left (e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{2} \, \log \left (e^{\left (2 \, x\right )} + 1\right )^{2} - \frac {1}{2} \, \log \left (e^{x} + 1\right )^{2} - \log \left (e^{x} + 1\right ) \log \left (-e^{x} + 1\right ) - \frac {1}{2} \, \log \left (-e^{x} + 1\right )^{2} + {\left (\log \left (e^{\left (-x\right )} + 1\right ) + \log \left (e^{\left (-x\right )} - 1\right ) - \log \left (e^{\left (-2 \, x\right )} + 1\right )\right )} \log \left (\tanh \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(e^x + 1) + log(-e^x + 1))*log(e^(2*x) + 1) - 1/2*log(e^(2*x) + 1)^2 - 1/2*log(e^x + 1)^2 - log(e^x + 1)*l
og(-e^x + 1) - 1/2*log(-e^x + 1)^2 + (log(e^(-x) + 1) + log(e^(-x) - 1) - log(e^(-2*x) + 1))*log(tanh(x))

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mupad [B]  time = 1.76, size = 21, normalized size = 2.33 \[ \frac {{\left (\ln \left ({\mathrm {e}}^{2\,x}-1\right )-\ln \left ({\mathrm {e}}^{2\,x}+1\right )\right )}^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(tanh(x))/(cosh(x)*sinh(x)),x)

[Out]

(log(exp(2*x) - 1) - log(exp(2*x) + 1))^2/2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)*ln(tanh(x))*sech(x),x)

[Out]

Integral(log(tanh(x))*csch(x)*sech(x), x)

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