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

Optimal. Leaf size=25 \[ \frac {\log \left (1-a^2 x^2\right )}{2 a}+x \coth ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5911, 260} \[ \frac {\log \left (1-a^2 x^2\right )}{2 a}+x \coth ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x],x]

[Out]

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 5911

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCoth[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcCoth[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 25, normalized size = 1.00 \[ \frac {\log \left (1-a^2 x^2\right )}{2 a}+x \coth ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a*x],x]

[Out]

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

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fricas [A]  time = 0.61, size = 33, normalized size = 1.32 \[ \frac {a x \log \left (\frac {a x + 1}{a x - 1}\right ) + \log \left (a^{2} x^{2} - 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arcoth}\left (a x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(a*x), x)

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maple [A]  time = 0.04, size = 23, normalized size = 0.92 \[ x \,\mathrm {arccoth}\left (a x \right )+\frac {\ln \left (a^{2} x^{2}-1\right )}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x),x)

[Out]

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

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maxima [A]  time = 0.30, size = 25, normalized size = 1.00 \[ \frac {2 \, a x \operatorname {arcoth}\left (a x\right ) + \log \left (-a^{2} x^{2} + 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.15, size = 22, normalized size = 0.88 \[ x\,\mathrm {acoth}\left (a\,x\right )+\frac {\ln \left (a^2\,x^2-1\right )}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a*x),x)

[Out]

x*acoth(a*x) + log(a^2*x^2 - 1)/(2*a)

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sympy [A]  time = 0.33, size = 27, normalized size = 1.08 \[ \begin {cases} x \operatorname {acoth}{\left (a x \right )} + \frac {\log {\left (a x + 1 \right )}}{a} - \frac {\operatorname {acoth}{\left (a x \right )}}{a} & \text {for}\: a \neq 0 \\\frac {i \pi x}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x),x)

[Out]

Piecewise((x*acoth(a*x) + log(a*x + 1)/a - acoth(a*x)/a, Ne(a, 0)), (I*pi*x/2, True))

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