3.155 \(\int \frac {\coth (a+2 \log (x))}{x} \, dx\)

Optimal. Leaf size=12 \[ \frac {1}{2} \log (\sinh (a+2 \log (x))) \]

[Out]

1/2*ln(sinh(a+2*ln(x)))

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Rubi [A]  time = 0.01, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3475} \[ \frac {1}{2} \log (\sinh (a+2 \log (x))) \]

Antiderivative was successfully verified.

[In]

Int[Coth[a + 2*Log[x]]/x,x]

[Out]

Log[Sinh[a + 2*Log[x]]]/2

Rule 3475

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

Rubi steps

\begin {align*} \int \frac {\coth (a+2 \log (x))}{x} \, dx &=\operatorname {Subst}(\int \coth (a+2 x) \, dx,x,\log (x))\\ &=\frac {1}{2} \log (\sinh (a+2 \log (x)))\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 1.75 \[ \frac {1}{2} (\log (\tanh (a+2 \log (x)))+\log (\cosh (a+2 \log (x)))) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[a + 2*Log[x]]/x,x]

[Out]

(Log[Cosh[a + 2*Log[x]]] + Log[Tanh[a + 2*Log[x]]])/2

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fricas [A]  time = 0.42, size = 18, normalized size = 1.50 \[ \frac {1}{2} \, \log \left (x^{4} e^{\left (2 \, a\right )} - 1\right ) - \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(a+2*log(x))/x,x, algorithm="fricas")

[Out]

1/2*log(x^4*e^(2*a) - 1) - log(x)

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giac [B]  time = 0.13, size = 21, normalized size = 1.75 \[ -\frac {1}{4} \, \log \left (x^{4}\right ) + \frac {1}{2} \, \log \left ({\left | x^{4} e^{\left (2 \, a\right )} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(a+2*log(x))/x,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.01, size = 26, normalized size = 2.17 \[ -\frac {\ln \left (\coth \left (a +2 \ln \relax (x )\right )-1\right )}{4}-\frac {\ln \left (\coth \left (a +2 \ln \relax (x )\right )+1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a+2*ln(x))/x,x)

[Out]

-1/4*ln(coth(a+2*ln(x))-1)-1/4*ln(coth(a+2*ln(x))+1)

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maxima [A]  time = 0.30, size = 10, normalized size = 0.83 \[ \frac {1}{2} \, \log \left (\sinh \left (a + 2 \, \log \relax (x)\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(a+2*log(x))/x,x, algorithm="maxima")

[Out]

1/2*log(sinh(a + 2*log(x)))

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mupad [B]  time = 1.21, size = 18, normalized size = 1.50 \[ \frac {\ln \left (x^4-{\mathrm {e}}^{-2\,a}\right )}{2}-\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a + 2*log(x))/x,x)

[Out]

log(x^4 - exp(-2*a))/2 - log(x)

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sympy [B]  time = 0.98, size = 27, normalized size = 2.25 \[ \log {\relax (x )} - \frac {\log {\left (\tanh {\left (a + 2 \log {\relax (x )} \right )} + 1 \right )}}{2} + \frac {\log {\left (\tanh {\left (a + 2 \log {\relax (x )} \right )} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(a+2*ln(x))/x,x)

[Out]

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

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