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

Optimal. Leaf size=54 \[ \frac{\log \left (1-a^2 x^2\right )}{2 a^2}-\frac{\coth ^{-1}(a x)^2}{2 a^2}+\frac{1}{2} x^2 \coth ^{-1}(a x)^2+\frac{x \coth ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.0783636, antiderivative size = 54, 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, 5981, 5911, 260, 5949} \[ \frac{\log \left (1-a^2 x^2\right )}{2 a^2}-\frac{\coth ^{-1}(a x)^2}{2 a^2}+\frac{1}{2} x^2 \coth ^{-1}(a x)^2+\frac{x \coth ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*ArcCoth[a*x])/a - ArcCoth[a*x]^2/(2*a^2) + (x^2*ArcCoth[a*x]^2)/2 + Log[1 - a^2*x^2]/(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 5981

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

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

Mathematica [A]  time = 0.012074, size = 43, normalized size = 0.8 \[ \frac{\log \left (1-a^2 x^2\right )+\left (a^2 x^2-1\right ) \coth ^{-1}(a x)^2+2 a x \coth ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.056, size = 155, normalized size = 2.9 \begin{align*}{\frac{{x}^{2} \left ({\rm arccoth} \left (ax\right ) \right ) ^{2}}{2}}+{\frac{x{\rm arccoth} \left (ax\right )}{a}}+{\frac{{\rm arccoth} \left (ax\right )\ln \left ( ax-1 \right ) }{2\,{a}^{2}}}-{\frac{{\rm arccoth} \left (ax\right )\ln \left ( ax+1 \right ) }{2\,{a}^{2}}}+{\frac{ \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{8\,{a}^{2}}}-{\frac{\ln \left ( ax-1 \right ) }{4\,{a}^{2}}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{\ln \left ( ax-1 \right ) }{2\,{a}^{2}}}+{\frac{\ln \left ( ax+1 \right ) }{2\,{a}^{2}}}-{\frac{\ln \left ( ax+1 \right ) }{4\,{a}^{2}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{1}{4\,{a}^{2}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{8\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arccoth(a*x)^2+x*arccoth(a*x)/a+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)+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 [B]  time = 0.986487, size = 131, normalized size = 2.43 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{arcoth}\left (a x\right )^{2} + \frac{1}{2} \, 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 )} \operatorname{arcoth}\left (a x\right ) - \frac{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 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.50581, size = 143, normalized size = 2.65 \begin{align*} \frac{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} + 4 \, \log \left (a^{2} x^{2} - 1\right )}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.52711, size = 60, normalized size = 1.11 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{acoth}^{2}{\left (a x \right )}}{2} + \frac{x \operatorname{acoth}{\left (a x \right )}}{a} + \frac{\log{\left (a x + 1 \right )}}{a^{2}} - \frac{\operatorname{acoth}^{2}{\left (a x \right )}}{2 a^{2}} - \frac{\operatorname{acoth}{\left (a x \right )}}{a^{2}} & \text{for}\: a \neq 0 \\- \frac{\pi ^{2} x^{2}}{8} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*acoth(a*x)**2/2 + x*acoth(a*x)/a + log(a*x + 1)/a**2 - acoth(a*x)**2/(2*a**2) - acoth(a*x)/a**
2, Ne(a, 0)), (-pi**2*x**2/8, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccoth(a*x)^2, x)