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

Optimal. Leaf size=14 \[ \log (x)-\frac{1}{2} \coth (a+2 \log (x)) \]

[Out]

-Coth[a + 2*Log[x]]/2 + Log[x]

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Rubi [A]  time = 0.0239965, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3473, 8} \[ \log (x)-\frac{1}{2} \coth (a+2 \log (x)) \]

Antiderivative was successfully verified.

[In]

Int[Coth[a + 2*Log[x]]^2/x,x]

[Out]

-Coth[a + 2*Log[x]]/2 + Log[x]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [C]  time = 0.0539936, size = 28, normalized size = 2. \[ -\frac{1}{2} \coth (a+2 \log (x)) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\tanh ^2(a+2 \log (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[a + 2*Log[x]]^2/x,x]

[Out]

-(Coth[a + 2*Log[x]]*Hypergeometric2F1[-1/2, 1, 1/2, Tanh[a + 2*Log[x]]^2])/2

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Maple [B]  time = 0.003, size = 35, normalized size = 2.5 \begin{align*} -{\frac{{\rm coth} \left (a+2\,\ln \left ( x \right ) \right )}{2}}-{\frac{\ln \left ({\rm coth} \left (a+2\,\ln \left ( x \right ) \right )-1 \right ) }{4}}+{\frac{\ln \left ({\rm coth} \left (a+2\,\ln \left ( x \right ) \right )+1 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*coth(a+2*ln(x))-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 = 1.03175, size = 26, normalized size = 1.86 \begin{align*} \frac{1}{2} \, a + \frac{1}{e^{\left (-2 \, a - 4 \, \log \left (x\right )\right )} - 1} + \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.52145, size = 68, normalized size = 4.86 \begin{align*} \frac{{\left (x^{4} e^{\left (2 \, a\right )} - 1\right )} \log \left (x\right ) - 1}{x^{4} e^{\left (2 \, a\right )} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 17.2266, size = 32, normalized size = 2.29 \begin{align*} \begin{cases} \tilde{\infty } \log{\left (x \right )} & \text{for}\: a = \log{\left (- \frac{1}{x^{2}} \right )} \vee a = \log{\left (\frac{1}{x^{2}} \right )} \\\log{\left (x \right )} - \frac{1}{2 \tanh{\left (a + 2 \log{\left (x \right )} \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*log(x), Eq(a, log(x**(-2))) | Eq(a, log(-1/x**2))), (log(x) - 1/(2*tanh(a + 2*log(x))), True))

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Giac [A]  time = 1.11646, size = 28, normalized size = 2. \begin{align*} -\frac{1}{x^{4} e^{\left (2 \, a\right )} - 1} + \frac{1}{4} \, \log \left (x^{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(x^4*e^(2*a) - 1) + 1/4*log(x^4)