3.20 \(\int \frac{\coth ^{-1}(a x)^2}{x^3} \, dx\)

Optimal. Leaf size=61 \[ -\frac{1}{2} a^2 \log \left (1-a^2 x^2\right )+a^2 \log (x)+\frac{1}{2} a^2 \coth ^{-1}(a x)^2-\frac{\coth ^{-1}(a x)^2}{2 x^2}-\frac{a \coth ^{-1}(a x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.0985716, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {5917, 5983, 266, 36, 29, 31, 5949} \[ -\frac{1}{2} a^2 \log \left (1-a^2 x^2\right )+a^2 \log (x)+\frac{1}{2} a^2 \coth ^{-1}(a x)^2-\frac{\coth ^{-1}(a x)^2}{2 x^2}-\frac{a \coth ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]^2/x^3,x]

[Out]

-((a*ArcCoth[a*x])/x) + (a^2*ArcCoth[a*x]^2)/2 - ArcCoth[a*x]^2/(2*x^2) + a^2*Log[x] - (a^2*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 5983

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcCoth[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcCoth[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[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]

Rule 5949

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}(a x)^2}{x^3} \, dx &=-\frac{\coth ^{-1}(a x)^2}{2 x^2}+a \int \frac{\coth ^{-1}(a x)}{x^2 \left (1-a^2 x^2\right )} \, dx\\ &=-\frac{\coth ^{-1}(a x)^2}{2 x^2}+a \int \frac{\coth ^{-1}(a x)}{x^2} \, dx+a^3 \int \frac{\coth ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=-\frac{a \coth ^{-1}(a x)}{x}+\frac{1}{2} a^2 \coth ^{-1}(a x)^2-\frac{\coth ^{-1}(a x)^2}{2 x^2}+a^2 \int \frac{1}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac{a \coth ^{-1}(a x)}{x}+\frac{1}{2} a^2 \coth ^{-1}(a x)^2-\frac{\coth ^{-1}(a x)^2}{2 x^2}+\frac{1}{2} a^2 \operatorname{Subst}\left (\int \frac{1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{a \coth ^{-1}(a x)}{x}+\frac{1}{2} a^2 \coth ^{-1}(a x)^2-\frac{\coth ^{-1}(a x)^2}{2 x^2}+\frac{1}{2} a^2 \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{2} a^4 \operatorname{Subst}\left (\int \frac{1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac{a \coth ^{-1}(a x)}{x}+\frac{1}{2} a^2 \coth ^{-1}(a x)^2-\frac{\coth ^{-1}(a x)^2}{2 x^2}+a^2 \log (x)-\frac{1}{2} a^2 \log \left (1-a^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0170421, size = 57, normalized size = 0.93 \[ -\frac{1}{2} a^2 \log \left (1-a^2 x^2\right )+\frac{\left (a^2 x^2-1\right ) \coth ^{-1}(a x)^2}{2 x^2}+a^2 \log (x)-\frac{a \coth ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a*x]^2/x^3,x]

[Out]

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

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Maple [B]  time = 0.055, size = 164, normalized size = 2.7 \begin{align*} -{\frac{ \left ({\rm arccoth} \left (ax\right ) \right ) ^{2}}{2\,{x}^{2}}}-{\frac{a{\rm arccoth} \left (ax\right )}{x}}-{\frac{{a}^{2}{\rm arccoth} \left (ax\right )\ln \left ( ax-1 \right ) }{2}}+{\frac{{a}^{2}{\rm arccoth} \left (ax\right )\ln \left ( ax+1 \right ) }{2}}-{\frac{{a}^{2} \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{8}}+{\frac{{a}^{2}\ln \left ( ax-1 \right ) }{4}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{{a}^{2}\ln \left ( ax-1 \right ) }{2}}+{a}^{2}\ln \left ( ax \right ) -{\frac{{a}^{2}\ln \left ( ax+1 \right ) }{2}}+{\frac{{a}^{2}\ln \left ( ax+1 \right ) }{4}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }-{\frac{{a}^{2}}{4}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{{a}^{2} \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x)^2/x^3,x)

[Out]

-1/2*arccoth(a*x)^2/x^2-a*arccoth(a*x)/x-1/2*a^2*arccoth(a*x)*ln(a*x-1)+1/2*a^2*arccoth(a*x)*ln(a*x+1)-1/8*a^2
*ln(a*x-1)^2+1/4*a^2*ln(a*x-1)*ln(1/2+1/2*a*x)-1/2*a^2*ln(a*x-1)+a^2*ln(a*x)-1/2*a^2*ln(a*x+1)+1/4*a^2*ln(-1/2
*a*x+1/2)*ln(a*x+1)-1/4*a^2*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/8*a^2*ln(a*x+1)^2

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Maxima [A]  time = 0.971504, size = 130, normalized size = 2.13 \begin{align*} \frac{1}{8} \,{\left (2 \,{\left (\log \left (a x - 1\right ) - 2\right )} \log \left (a x + 1\right ) - \log \left (a x + 1\right )^{2} - \log \left (a x - 1\right )^{2} - 4 \, \log \left (a x - 1\right ) + 8 \, \log \left (x\right )\right )} a^{2} + \frac{1}{2} \,{\left (a \log \left (a x + 1\right ) - a \log \left (a x - 1\right ) - \frac{2}{x}\right )} a \operatorname{arcoth}\left (a x\right ) - \frac{\operatorname{arcoth}\left (a x\right )^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^2/x^3,x, algorithm="maxima")

[Out]

1/8*(2*(log(a*x - 1) - 2)*log(a*x + 1) - log(a*x + 1)^2 - log(a*x - 1)^2 - 4*log(a*x - 1) + 8*log(x))*a^2 + 1/
2*(a*log(a*x + 1) - a*log(a*x - 1) - 2/x)*a*arccoth(a*x) - 1/2*arccoth(a*x)^2/x^2

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Fricas [A]  time = 1.89944, size = 181, normalized size = 2.97 \begin{align*} -\frac{4 \, a^{2} x^{2} \log \left (a^{2} x^{2} - 1\right ) - 8 \, a^{2} x^{2} \log \left (x\right ) + 4 \, a x \log \left (\frac{a x + 1}{a x - 1}\right ) -{\left (a^{2} x^{2} - 1\right )} \log \left (\frac{a x + 1}{a x - 1}\right )^{2}}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^2/x^3,x, algorithm="fricas")

[Out]

-1/8*(4*a^2*x^2*log(a^2*x^2 - 1) - 8*a^2*x^2*log(x) + 4*a*x*log((a*x + 1)/(a*x - 1)) - (a^2*x^2 - 1)*log((a*x
+ 1)/(a*x - 1))^2)/x^2

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Sympy [A]  time = 1.0102, size = 56, normalized size = 0.92 \begin{align*} a^{2} \log{\left (x \right )} - a^{2} \log{\left (a x + 1 \right )} + \frac{a^{2} \operatorname{acoth}^{2}{\left (a x \right )}}{2} + a^{2} \operatorname{acoth}{\left (a x \right )} - \frac{a \operatorname{acoth}{\left (a x \right )}}{x} - \frac{\operatorname{acoth}^{2}{\left (a x \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x)**2/x**3,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (a x\right )^{2}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^2/x^3,x, algorithm="giac")

[Out]

integrate(arccoth(a*x)^2/x^3, x)