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

Optimal. Leaf size=40 \[ \frac {\log \left (1-a^2 x^2\right )}{6 a^3}+\frac {1}{3} x^3 \coth ^{-1}(a x)+\frac {x^2}{6 a} \]

[Out]

1/6*x^2/a+1/3*x^3*arccoth(a*x)+1/6*ln(-a^2*x^2+1)/a^3

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Rubi [A]  time = 0.03, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5917, 266, 43} \[ \frac {\log \left (1-a^2 x^2\right )}{6 a^3}+\frac {x^2}{6 a}+\frac {1}{3} x^3 \coth ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^2/(6*a) + (x^3*ArcCoth[a*x])/3 + Log[1 - a^2*x^2]/(6*a^3)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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]

Rubi steps

\begin {align*} \int x^2 \coth ^{-1}(a x) \, dx &=\frac {1}{3} x^3 \coth ^{-1}(a x)-\frac {1}{3} a \int \frac {x^3}{1-a^2 x^2} \, dx\\ &=\frac {1}{3} x^3 \coth ^{-1}(a x)-\frac {1}{6} a \operatorname {Subst}\left (\int \frac {x}{1-a^2 x} \, dx,x,x^2\right )\\ &=\frac {1}{3} x^3 \coth ^{-1}(a x)-\frac {1}{6} a \operatorname {Subst}\left (\int \left (-\frac {1}{a^2}-\frac {1}{a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {x^2}{6 a}+\frac {1}{3} x^3 \coth ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{6 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 40, normalized size = 1.00 \[ \frac {\log \left (1-a^2 x^2\right )}{6 a^3}+\frac {1}{3} x^3 \coth ^{-1}(a x)+\frac {x^2}{6 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^2/(6*a) + (x^3*ArcCoth[a*x])/3 + Log[1 - a^2*x^2]/(6*a^3)

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fricas [A]  time = 0.52, size = 44, normalized size = 1.10 \[ \frac {a^{3} x^{3} \log \left (\frac {a x + 1}{a x - 1}\right ) + a^{2} x^{2} + \log \left (a^{2} x^{2} - 1\right )}{6 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(a^3*x^3*log((a*x + 1)/(a*x - 1)) + a^2*x^2 + log(a^2*x^2 - 1))/a^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {arcoth}\left (a x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arccoth(a*x), x)

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maple [A]  time = 0.03, size = 41, normalized size = 1.02 \[ \frac {x^{3} \mathrm {arccoth}\left (a x \right )}{3}+\frac {x^{2}}{6 a}+\frac {\ln \left (a x -1\right )}{6 a^{3}}+\frac {\ln \left (a x +1\right )}{6 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccoth(a*x)+1/6*x^2/a+1/6/a^3*ln(a*x-1)+1/6/a^3*ln(a*x+1)

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maxima [A]  time = 0.31, size = 35, normalized size = 0.88 \[ \frac {1}{3} \, x^{3} \operatorname {arcoth}\left (a x\right ) + \frac {1}{6} \, a {\left (\frac {x^{2}}{a^{2}} + \frac {\log \left (a^{2} x^{2} - 1\right )}{a^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccoth(a*x) + 1/6*a*(x^2/a^2 + log(a^2*x^2 - 1)/a^4)

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mupad [B]  time = 1.22, size = 35, normalized size = 0.88 \[ \frac {\frac {\ln \left (a^2\,x^2-1\right )}{6}+\frac {a^2\,x^2}{6}}{a^3}+\frac {x^3\,\mathrm {acoth}\left (a\,x\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(a^2*x^2 - 1)/6 + (a^2*x^2)/6)/a^3 + (x^3*acoth(a*x))/3

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sympy [A]  time = 0.67, size = 46, normalized size = 1.15 \[ \begin {cases} \frac {x^{3} \operatorname {acoth}{\left (a x \right )}}{3} + \frac {x^{2}}{6 a} + \frac {\log {\left (a x + 1 \right )}}{3 a^{3}} - \frac {\operatorname {acoth}{\left (a x \right )}}{3 a^{3}} & \text {for}\: a \neq 0 \\\frac {i \pi x^{3}}{6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**3*acoth(a*x)/3 + x**2/(6*a) + log(a*x + 1)/(3*a**3) - acoth(a*x)/(3*a**3), Ne(a, 0)), (I*pi*x**3
/6, True))

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