3.11.13 \(\int \text {csch}(2 x) \log (\tanh (x)) \, dx\) [1013]

Optimal. Leaf size=9 \[ \frac {1}{4} \log ^2(\tanh (x)) \]

[Out]

1/4*ln(tanh(x))^2

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Rubi [A]
time = 0.02, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3855, 6818} \begin {gather*} \frac {1}{4} \log ^2(\tanh (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csch[2*x]*Log[Tanh[x]],x]

[Out]

Log[Tanh[x]]^2/4

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 6818

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}(2 x) \log (\tanh (x)) \, dx &=\frac {1}{4} \log ^2(\tanh (x))\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 9, normalized size = 1.00 \begin {gather*} \frac {1}{4} \log ^2(\tanh (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csch[2*x]*Log[Tanh[x]],x]

[Out]

Log[Tanh[x]]^2/4

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Maple [A]
time = 1.36, size = 8, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {\ln \left (\tanh \left (x \right )\right )^{2}}{4}\)	\(8\)
default	\(\frac {\ln \left (\tanh \left (x \right )\right )^{2}}{4}\)	\(8\)
risch	\(\frac {\ln \left (1+{\mathrm e}^{2 x}\right )^{2}}{4}-\frac {\ln \left ({\mathrm e}^{2 x}-1\right ) \ln \left (1+{\mathrm e}^{2 x}\right )}{2}+\frac {\ln \left ({\mathrm e}^{2 x}-1\right )^{2}}{4}+\frac {i \ln \left (1+{\mathrm e}^{2 x}\right ) \pi \,\mathrm {csgn}\left (\frac {i}{1+{\mathrm e}^{2 x}}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}-1\right )}{1+{\mathrm e}^{2 x}}\right )}{4}-\frac {i \ln \left (1+{\mathrm e}^{2 x}\right ) \pi \,\mathrm {csgn}\left (\frac {i}{1+{\mathrm e}^{2 x}}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}-1\right )}{1+{\mathrm e}^{2 x}}\right )^{2}}{4}-\frac {i \ln \left (1+{\mathrm e}^{2 x}\right ) \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}-1\right )}{1+{\mathrm e}^{2 x}}\right )^{2}}{4}+\frac {i \ln \left (1+{\mathrm e}^{2 x}\right ) \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}-1\right )}{1+{\mathrm e}^{2 x}}\right )^{3}}{4}-\frac {i \ln \left ({\mathrm e}^{2 x}-1\right ) \pi \,\mathrm {csgn}\left (\frac {i}{1+{\mathrm e}^{2 x}}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}-1\right )}{1+{\mathrm e}^{2 x}}\right )}{4}+\frac {i \ln \left ({\mathrm e}^{2 x}-1\right ) \pi \,\mathrm {csgn}\left (\frac {i}{1+{\mathrm e}^{2 x}}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}-1\right )}{1+{\mathrm e}^{2 x}}\right )^{2}}{4}+\frac {i \ln \left ({\mathrm e}^{2 x}-1\right ) \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}-1\right )}{1+{\mathrm e}^{2 x}}\right )^{2}}{4}-\frac {i \ln \left ({\mathrm e}^{2 x}-1\right ) \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}-1\right )}{1+{\mathrm e}^{2 x}}\right )^{3}}{4}\)	\(372\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(2*x)*ln(tanh(x)),x,method=_RETURNVERBOSE)

[Out]

1/4*ln(tanh(x))^2

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Maxima [A]
time = 0.26, size = 7, normalized size = 0.78 \begin {gather*} \frac {1}{4} \, \log \left (\tanh \left (x\right )\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(tanh(x))^2

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Fricas [A]
time = 0.36, size = 12, normalized size = 1.33 \begin {gather*} \frac {1}{4} \, \log \left (\frac {\sinh \left (x\right )}{\cosh \left (x\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(sinh(x)/cosh(x))^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \log {\left (\tanh {\left (x \right )} \right )} \operatorname {csch}{\left (2 x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(tanh(x))*csch(2*x), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 20 vs. \(2 (7) = 14\).
time = 0.39, size = 20, normalized size = 2.22 \begin {gather*} \frac {1}{4} \, \log \left (\frac {e^{\left (2 \, x\right )} - 1}{e^{\left (2 \, x\right )} + 1}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log((e^(2*x) - 1)/(e^(2*x) + 1))^2

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Mupad [B]
time = 1.67, size = 21, normalized size = 2.33 \begin {gather*} \frac {{\left (\ln \left ({\mathrm {e}}^{2\,x}-1\right )-\ln \left ({\mathrm {e}}^{2\,x}+1\right )\right )}^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(tanh(x))/sinh(2*x),x)

[Out]

(log(exp(2*x) - 1) - log(exp(2*x) + 1))^2/4

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