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

Optimal. Leaf size=81 \[ -\frac {\coth ^{-1}(a x)^2}{4 a^4}+\frac {x \coth ^{-1}(a x)}{2 a^3}+\frac {x^2}{12 a^2}+\frac {\log \left (1-a^2 x^2\right )}{3 a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^2+\frac {x^3 \coth ^{-1}(a x)}{6 a} \]

[Out]

1/12*x^2/a^2+1/2*x*arccoth(a*x)/a^3+1/6*x^3*arccoth(a*x)/a-1/4*arccoth(a*x)^2/a^4+1/4*x^4*arccoth(a*x)^2+1/3*l
n(-a^2*x^2+1)/a^4

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Rubi [A]  time = 0.16, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5917, 5981, 266, 43, 5911, 260, 5949} \[ \frac {x^2}{12 a^2}+\frac {\log \left (1-a^2 x^2\right )}{3 a^4}+\frac {x \coth ^{-1}(a x)}{2 a^3}-\frac {\coth ^{-1}(a x)^2}{4 a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^2+\frac {x^3 \coth ^{-1}(a x)}{6 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^2/(12*a^2) + (x*ArcCoth[a*x])/(2*a^3) + (x^3*ArcCoth[a*x])/(6*a) - ArcCoth[a*x]^2/(4*a^4) + (x^4*ArcCoth[a*x
]^2)/4 + Log[1 - a^2*x^2]/(3*a^4)

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 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 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 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 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 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]

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]

Rubi steps

\begin {align*} \int x^3 \coth ^{-1}(a x)^2 \, dx &=\frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{2} a \int \frac {x^4 \coth ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=\frac {1}{4} x^4 \coth ^{-1}(a x)^2+\frac {\int x^2 \coth ^{-1}(a x) \, dx}{2 a}-\frac {\int \frac {x^2 \coth ^{-1}(a x)}{1-a^2 x^2} \, dx}{2 a}\\ &=\frac {x^3 \coth ^{-1}(a x)}{6 a}+\frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{6} \int \frac {x^3}{1-a^2 x^2} \, dx+\frac {\int \coth ^{-1}(a x) \, dx}{2 a^3}-\frac {\int \frac {\coth ^{-1}(a x)}{1-a^2 x^2} \, dx}{2 a^3}\\ &=\frac {x \coth ^{-1}(a x)}{2 a^3}+\frac {x^3 \coth ^{-1}(a x)}{6 a}-\frac {\coth ^{-1}(a x)^2}{4 a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{12} \operatorname {Subst}\left (\int \frac {x}{1-a^2 x} \, dx,x,x^2\right )-\frac {\int \frac {x}{1-a^2 x^2} \, dx}{2 a^2}\\ &=\frac {x \coth ^{-1}(a x)}{2 a^3}+\frac {x^3 \coth ^{-1}(a x)}{6 a}-\frac {\coth ^{-1}(a x)^2}{4 a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^2+\frac {\log \left (1-a^2 x^2\right )}{4 a^4}-\frac {1}{12} \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}{12 a^2}+\frac {x \coth ^{-1}(a x)}{2 a^3}+\frac {x^3 \coth ^{-1}(a x)}{6 a}-\frac {\coth ^{-1}(a x)^2}{4 a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^2+\frac {\log \left (1-a^2 x^2\right )}{3 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 62, normalized size = 0.77 \[ \frac {3 \left (a^4 x^4-1\right ) \coth ^{-1}(a x)^2+a^2 x^2+4 \log \left (1-a^2 x^2\right )+2 a x \left (a^2 x^2+3\right ) \coth ^{-1}(a x)}{12 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x^2 + 2*a*x*(3 + a^2*x^2)*ArcCoth[a*x] + 3*(-1 + a^4*x^4)*ArcCoth[a*x]^2 + 4*Log[1 - a^2*x^2])/(12*a^4)

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fricas [A]  time = 0.53, size = 81, normalized size = 1.00 \[ \frac {4 \, a^{2} x^{2} + 3 \, {\left (a^{4} x^{4} - 1\right )} \log \left (\frac {a x + 1}{a x - 1}\right )^{2} + 4 \, {\left (a^{3} x^{3} + 3 \, a x\right )} \log \left (\frac {a x + 1}{a x - 1}\right ) + 16 \, \log \left (a^{2} x^{2} - 1\right )}{48 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(4*a^2*x^2 + 3*(a^4*x^4 - 1)*log((a*x + 1)/(a*x - 1))^2 + 4*(a^3*x^3 + 3*a*x)*log((a*x + 1)/(a*x - 1)) +
16*log(a^2*x^2 - 1))/a^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {arcoth}\left (a x\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arccoth(a*x)^2, x)

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maple [B]  time = 0.06, size = 176, normalized size = 2.17 \[ \frac {x^{4} \mathrm {arccoth}\left (a x \right )^{2}}{4}+\frac {x^{3} \mathrm {arccoth}\left (a x \right )}{6 a}+\frac {x \,\mathrm {arccoth}\left (a x \right )}{2 a^{3}}+\frac {\mathrm {arccoth}\left (a x \right ) \ln \left (a x -1\right )}{4 a^{4}}-\frac {\mathrm {arccoth}\left (a x \right ) \ln \left (a x +1\right )}{4 a^{4}}+\frac {\ln \left (a x -1\right )^{2}}{16 a^{4}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8 a^{4}}+\frac {\ln \left (a x +1\right )^{2}}{16 a^{4}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8 a^{4}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{8 a^{4}}+\frac {x^{2}}{12 a^{2}}+\frac {\ln \left (a x -1\right )}{3 a^{4}}+\frac {\ln \left (a x +1\right )}{3 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arccoth(a*x)^2+1/6*x^3*arccoth(a*x)/a+1/2*x*arccoth(a*x)/a^3+1/4/a^4*arccoth(a*x)*ln(a*x-1)-1/4/a^4*ar
ccoth(a*x)*ln(a*x+1)+1/16/a^4*ln(a*x-1)^2-1/8/a^4*ln(a*x-1)*ln(1/2+1/2*a*x)+1/16/a^4*ln(a*x+1)^2+1/8/a^4*ln(-1
/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/8/a^4*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/12*x^2/a^2+1/3/a^4*ln(a*x-1)+1/3*ln(a*x+1)/a^
4

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maxima [A]  time = 0.32, size = 118, normalized size = 1.46 \[ \frac {1}{4} \, x^{4} \operatorname {arcoth}\left (a x\right )^{2} + \frac {1}{12} \, a {\left (\frac {2 \, {\left (a^{2} x^{3} + 3 \, x\right )}}{a^{4}} - \frac {3 \, \log \left (a x + 1\right )}{a^{5}} + \frac {3 \, \log \left (a x - 1\right )}{a^{5}}\right )} \operatorname {arcoth}\left (a x\right ) + \frac {4 \, a^{2} x^{2} - 2 \, {\left (3 \, \log \left (a x - 1\right ) - 8\right )} \log \left (a x + 1\right ) + 3 \, \log \left (a x + 1\right )^{2} + 3 \, \log \left (a x - 1\right )^{2} + 16 \, \log \left (a x - 1\right )}{48 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arccoth(a*x)^2 + 1/12*a*(2*(a^2*x^3 + 3*x)/a^4 - 3*log(a*x + 1)/a^5 + 3*log(a*x - 1)/a^5)*arccoth(a*x)
+ 1/48*(4*a^2*x^2 - 2*(3*log(a*x - 1) - 8)*log(a*x + 1) + 3*log(a*x + 1)^2 + 3*log(a*x - 1)^2 + 16*log(a*x -
1))/a^4

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mupad [B]  time = 1.26, size = 65, normalized size = 0.80 \[ \frac {x^4\,{\mathrm {acoth}\left (a\,x\right )}^2}{4}+\frac {\frac {\ln \left (a^2\,x^2-1\right )}{3}+\frac {a^2\,x^2}{12}-\frac {{\mathrm {acoth}\left (a\,x\right )}^2}{4}+\frac {a^3\,x^3\,\mathrm {acoth}\left (a\,x\right )}{6}+\frac {a\,x\,\mathrm {acoth}\left (a\,x\right )}{2}}{a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acoth(a*x)^2,x)

[Out]

(x^4*acoth(a*x)^2)/4 + (log(a^2*x^2 - 1)/3 + (a^2*x^2)/12 - acoth(a*x)^2/4 + (a^3*x^3*acoth(a*x))/6 + (a*x*aco
th(a*x))/2)/a^4

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sympy [A]  time = 1.44, size = 90, normalized size = 1.11 \[ \begin {cases} \frac {x^{4} \operatorname {acoth}^{2}{\left (a x \right )}}{4} + \frac {x^{3} \operatorname {acoth}{\left (a x \right )}}{6 a} + \frac {x^{2}}{12 a^{2}} + \frac {x \operatorname {acoth}{\left (a x \right )}}{2 a^{3}} + \frac {2 \log {\left (a x + 1 \right )}}{3 a^{4}} - \frac {\operatorname {acoth}^{2}{\left (a x \right )}}{4 a^{4}} - \frac {2 \operatorname {acoth}{\left (a x \right )}}{3 a^{4}} & \text {for}\: a \neq 0 \\- \frac {\pi ^{2} x^{4}}{16} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**4*acoth(a*x)**2/4 + x**3*acoth(a*x)/(6*a) + x**2/(12*a**2) + x*acoth(a*x)/(2*a**3) + 2*log(a*x +
1)/(3*a**4) - acoth(a*x)**2/(4*a**4) - 2*acoth(a*x)/(3*a**4), Ne(a, 0)), (-pi**2*x**4/16, True))

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