3.8 \(\int \frac{\coth ^{-1}(a x)}{x^2} \, dx\)

Optimal. Leaf size=30 \[ -\frac{1}{2} a \log \left (1-a^2 x^2\right )+a \log (x)-\frac{\coth ^{-1}(a x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.020794, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {5917, 266, 36, 29, 31} \[ -\frac{1}{2} a \log \left (1-a^2 x^2\right )+a \log (x)-\frac{\coth ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]/x^2,x]

[Out]

-(ArcCoth[a*x]/x) + a*Log[x] - (a*Log[1 - a^2*x^2])/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 266

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}(a x)}{x^2} \, dx &=-\frac{\coth ^{-1}(a x)}{x}+a \int \frac{1}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac{\coth ^{-1}(a x)}{x}+\frac{1}{2} a \operatorname{Subst}\left (\int \frac{1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{\coth ^{-1}(a x)}{x}+\frac{1}{2} a \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{2} a^3 \operatorname{Subst}\left (\int \frac{1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac{\coth ^{-1}(a x)}{x}+a \log (x)-\frac{1}{2} a \log \left (1-a^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0075953, size = 30, normalized size = 1. \[ -\frac{1}{2} a \log \left (1-a^2 x^2\right )+a \log (x)-\frac{\coth ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a*x]/x^2,x]

[Out]

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

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Maple [A]  time = 0.036, size = 35, normalized size = 1.2 \begin{align*} -{\frac{{\rm arccoth} \left (ax\right )}{x}}-{\frac{a\ln \left ( ax-1 \right ) }{2}}+a\ln \left ( ax \right ) -{\frac{a\ln \left ( ax+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x)/x^2,x)

[Out]

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

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Maxima [A]  time = 0.974439, size = 41, normalized size = 1.37 \begin{align*} -\frac{1}{2} \, a{\left (\log \left (a^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} - \frac{\operatorname{arcoth}\left (a x\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.70012, size = 99, normalized size = 3.3 \begin{align*} -\frac{a x \log \left (a^{2} x^{2} - 1\right ) - 2 \, a x \log \left (x\right ) + \log \left (\frac{a x + 1}{a x - 1}\right )}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.610749, size = 26, normalized size = 0.87 \begin{align*} a \log{\left (x \right )} - a \log{\left (a x + 1 \right )} + a \operatorname{acoth}{\left (a x \right )} - \frac{\operatorname{acoth}{\left (a x \right )}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x)/x**2,x)

[Out]

a*log(x) - a*log(a*x + 1) + a*acoth(a*x) - acoth(a*x)/x

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (a x\right )}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/x^2,x, algorithm="giac")

[Out]

integrate(arccoth(a*x)/x^2, x)