3.2.53 \(\int x \coth (a+2 \log (x)) \, dx\) [153]

Optimal. Leaf size=23 \[ \frac {x^2}{2}-e^{-a} \tanh ^{-1}\left (e^a x^2\right ) \]

[Out]

1/2*x^2-arctanh(exp(a)*x^2)/exp(a)

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Rubi [A]
time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {5657, 470, 281, 212} \begin {gather*} \frac {x^2}{2}-e^{-a} \tanh ^{-1}\left (e^a x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Coth[a + 2*Log[x]],x]

[Out]

x^2/2 - ArcTanh[E^a*x^2]/E^a

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 5657

Int[Coth[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*((-1 - E^(2*a*d)*x^
(2*b*d))^p/(1 - E^(2*a*d)*x^(2*b*d))^p), x] /; FreeQ[{a, b, d, e, m, p}, x]

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 26, normalized size = 1.13 \begin {gather*} \frac {x^2}{2}+\tanh ^{-1}\left (x^2 (\cosh (a)+\sinh (a))\right ) (-\cosh (a)+\sinh (a)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Coth[a + 2*Log[x]],x]

[Out]

x^2/2 + ArcTanh[x^2*(Cosh[a] + Sinh[a])]*(-Cosh[a] + Sinh[a])

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Maple [A]
time = 0.57, size = 37, normalized size = 1.61

method	result	size
risch	\(\frac {x^{2}}{2}-\frac {{\mathrm e}^{-a} \ln \left ({\mathrm e}^{a} x^{2}+1\right )}{2}+\frac {{\mathrm e}^{-a} \ln \left ({\mathrm e}^{a} x^{2}-1\right )}{2}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*coth(a+2*ln(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 36, normalized size = 1.57 \begin {gather*} \frac {1}{2} \, x^{2} - \frac {1}{2} \, e^{\left (-a\right )} \log \left (x^{2} e^{a} + 1\right ) + \frac {1}{2} \, e^{\left (-a\right )} \log \left (x^{2} e^{a} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 33, normalized size = 1.43 \begin {gather*} \frac {1}{2} \, {\left (x^{2} e^{a} - \log \left (x^{2} e^{a} + 1\right ) + \log \left (x^{2} e^{a} - 1\right )\right )} e^{\left (-a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*coth(a + 2*log(x)), x)

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Giac [A]
time = 0.41, size = 37, normalized size = 1.61 \begin {gather*} \frac {1}{2} \, x^{2} - \frac {1}{2} \, e^{\left (-a\right )} \log \left (x^{2} e^{a} + 1\right ) + \frac {1}{2} \, e^{\left (-a\right )} \log \left ({\left | x^{2} e^{a} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.23, size = 25, normalized size = 1.09 \begin {gather*} \frac {x^2}{2}-\frac {\mathrm {atanh}\left (x^2\,\sqrt {{\mathrm {e}}^{2\,a}}\right )}{\sqrt {{\mathrm {e}}^{2\,a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*coth(a + 2*log(x)),x)

[Out]

x^2/2 - atanh(x^2*exp(2*a)^(1/2))/exp(2*a)^(1/2)

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