3.5 \(\int x \coth ^{-1}(a x) \, dx\)

Optimal. Leaf size=31 \[ -\frac{\tanh ^{-1}(a x)}{2 a^2}+\frac{1}{2} x^2 \coth ^{-1}(a x)+\frac{x}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.0132059, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5917, 321, 206} \[ -\frac{\tanh ^{-1}(a x)}{2 a^2}+\frac{1}{2} x^2 \coth ^{-1}(a x)+\frac{x}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[a*x],x]

[Out]

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

Rule 5917

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
oth[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCoth[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 321

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

Rule 206

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

Rubi steps

\begin{align*} \int x \coth ^{-1}(a x) \, dx &=\frac{1}{2} x^2 \coth ^{-1}(a x)-\frac{1}{2} a \int \frac{x^2}{1-a^2 x^2} \, dx\\ &=\frac{x}{2 a}+\frac{1}{2} x^2 \coth ^{-1}(a x)-\frac{\int \frac{1}{1-a^2 x^2} \, dx}{2 a}\\ &=\frac{x}{2 a}+\frac{1}{2} x^2 \coth ^{-1}(a x)-\frac{\tanh ^{-1}(a x)}{2 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0069993, size = 47, normalized size = 1.52 \[ \frac{\log (1-a x)}{4 a^2}-\frac{\log (a x+1)}{4 a^2}+\frac{1}{2} x^2 \coth ^{-1}(a x)+\frac{x}{2 a} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCoth[a*x],x]

[Out]

x/(2*a) + (x^2*ArcCoth[a*x])/2 + Log[1 - a*x]/(4*a^2) - Log[1 + a*x]/(4*a^2)

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Maple [A]  time = 0.033, size = 39, normalized size = 1.3 \begin{align*}{\frac{{x}^{2}{\rm arccoth} \left (ax\right )}{2}}+{\frac{x}{2\,a}}+{\frac{\ln \left ( ax-1 \right ) }{4\,{a}^{2}}}-{\frac{\ln \left ( ax+1 \right ) }{4\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccoth(a*x),x)

[Out]

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

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Maxima [A]  time = 0.973175, size = 55, normalized size = 1.77 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{arcoth}\left (a x\right ) + \frac{1}{4} \, a{\left (\frac{2 \, x}{a^{2}} - \frac{\log \left (a x + 1\right )}{a^{3}} + \frac{\log \left (a x - 1\right )}{a^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(a*x),x, algorithm="maxima")

[Out]

1/2*x^2*arccoth(a*x) + 1/4*a*(2*x/a^2 - log(a*x + 1)/a^3 + log(a*x - 1)/a^3)

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Fricas [A]  time = 1.61763, size = 78, normalized size = 2.52 \begin{align*} \frac{2 \, a x +{\left (a^{2} x^{2} - 1\right )} \log \left (\frac{a x + 1}{a x - 1}\right )}{4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(a*x),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.918862, size = 32, normalized size = 1.03 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{acoth}{\left (a x \right )}}{2} + \frac{x}{2 a} - \frac{\operatorname{acoth}{\left (a x \right )}}{2 a^{2}} & \text{for}\: a \neq 0 \\\frac{i \pi x^{2}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acoth(a*x),x)

[Out]

Piecewise((x**2*acoth(a*x)/2 + x/(2*a) - acoth(a*x)/(2*a**2), Ne(a, 0)), (I*pi*x**2/4, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{arcoth}\left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(a*x),x, algorithm="giac")

[Out]

integrate(x*arccoth(a*x), x)