\(\int x^3 \coth ^{-1}(a x)^2 \, dx\) [14]

Optimal result

Integrand size = 10, antiderivative size = 81 \[ \int x^3 \coth ^{-1}(a x)^2 \, dx=\frac {x^2}{12 a^2}+\frac {x \coth ^{-1}(a x)}{2 a^3}+\frac {x^3 \coth ^{-1}(a x)}{6 a}-\frac {\coth ^{-1}(a x)^2}{4 a^4}+\frac {1}{4} x^4 \coth ^{-1}(a x)^2+\frac {\log \left (1-a^2 x^2\right )}{3 a^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77 \[ \int x^3 \coth ^{-1}(a x)^2 \, dx=\frac {a^2 x^2+2 a x \left (3+a^2 x^2\right ) \coth ^{-1}(a x)+3 \left (-1+a^4 x^4\right ) \coth ^{-1}(a x)^2+4 \log \left (1-a^2 x^2\right )}{12 a^4} \] Input:

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

Output:

(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)
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.42, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6453, 6543, 6453, 243, 49, 2009, 6543, 6437, 240, 6511}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 6453

\(\displaystyle \frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{2} a \int \frac {x^4 \coth ^{-1}(a x)}{1-a^2 x^2}dx\)

\(\Big \downarrow \) 6543

\(\displaystyle \frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{2} a \left (\frac {\int \frac {x^2 \coth ^{-1}(a x)}{1-a^2 x^2}dx}{a^2}-\frac {\int x^2 \coth ^{-1}(a x)dx}{a^2}\right )\)

\(\Big \downarrow \) 6453

\(\displaystyle \frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{2} a \left (\frac {\int \frac {x^2 \coth ^{-1}(a x)}{1-a^2 x^2}dx}{a^2}-\frac {\frac {1}{3} x^3 \coth ^{-1}(a x)-\frac {1}{3} a \int \frac {x^3}{1-a^2 x^2}dx}{a^2}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{2} a \left (\frac {\int \frac {x^2 \coth ^{-1}(a x)}{1-a^2 x^2}dx}{a^2}-\frac {\frac {1}{3} x^3 \coth ^{-1}(a x)-\frac {1}{6} a \int \frac {x^2}{1-a^2 x^2}dx^2}{a^2}\right )\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{2} a \left (\frac {\int \frac {x^2 \coth ^{-1}(a x)}{1-a^2 x^2}dx}{a^2}-\frac {\frac {1}{3} x^3 \coth ^{-1}(a x)-\frac {1}{6} a \int \left (-\frac {1}{a^2}-\frac {1}{a^2 \left (a^2 x^2-1\right )}\right )dx^2}{a^2}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{2} a \left (\frac {\int \frac {x^2 \coth ^{-1}(a x)}{1-a^2 x^2}dx}{a^2}-\frac {\frac {1}{3} x^3 \coth ^{-1}(a x)-\frac {1}{6} a \left (-\frac {x^2}{a^2}-\frac {\log \left (1-a^2 x^2\right )}{a^4}\right )}{a^2}\right )\)

\(\Big \downarrow \) 6543

\(\displaystyle \frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{2} a \left (\frac {\frac {\int \frac {\coth ^{-1}(a x)}{1-a^2 x^2}dx}{a^2}-\frac {\int \coth ^{-1}(a x)dx}{a^2}}{a^2}-\frac {\frac {1}{3} x^3 \coth ^{-1}(a x)-\frac {1}{6} a \left (-\frac {x^2}{a^2}-\frac {\log \left (1-a^2 x^2\right )}{a^4}\right )}{a^2}\right )\)

\(\Big \downarrow \) 6437

\(\displaystyle \frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{2} a \left (\frac {\frac {\int \frac {\coth ^{-1}(a x)}{1-a^2 x^2}dx}{a^2}-\frac {x \coth ^{-1}(a x)-a \int \frac {x}{1-a^2 x^2}dx}{a^2}}{a^2}-\frac {\frac {1}{3} x^3 \coth ^{-1}(a x)-\frac {1}{6} a \left (-\frac {x^2}{a^2}-\frac {\log \left (1-a^2 x^2\right )}{a^4}\right )}{a^2}\right )\)

\(\Big \downarrow \) 240

\(\displaystyle \frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{2} a \left (\frac {\frac {\int \frac {\coth ^{-1}(a x)}{1-a^2 x^2}dx}{a^2}-\frac {\frac {\log \left (1-a^2 x^2\right )}{2 a}+x \coth ^{-1}(a x)}{a^2}}{a^2}-\frac {\frac {1}{3} x^3 \coth ^{-1}(a x)-\frac {1}{6} a \left (-\frac {x^2}{a^2}-\frac {\log \left (1-a^2 x^2\right )}{a^4}\right )}{a^2}\right )\)

\(\Big \downarrow \) 6511

\(\displaystyle \frac {1}{4} x^4 \coth ^{-1}(a x)^2-\frac {1}{2} a \left (\frac {\frac {\coth ^{-1}(a x)^2}{2 a^3}-\frac {\frac {\log \left (1-a^2 x^2\right )}{2 a}+x \coth ^{-1}(a x)}{a^2}}{a^2}-\frac {\frac {1}{3} x^3 \coth ^{-1}(a x)-\frac {1}{6} a \left (-\frac {x^2}{a^2}-\frac {\log \left (1-a^2 x^2\right )}{a^4}\right )}{a^2}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 49

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}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 240

Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x 
^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6437

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

rule 6453

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : 
> Simp[x^(m + 1)*((a + b*ArcCoth[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m 
+ 1))   Int[x^(m + n)*((a + b*ArcCoth[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x 
], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 
] && IntegerQ[m])) && NeQ[m, -1]
 

rule 6511

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symb 
ol] :> 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 6543

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89

method	result	size
parallelrisch	\(-\frac {-3 x^{4} a^{4} \operatorname {arccoth}\left (x a \right )^{2}-2 x^{3} a^{3} \operatorname {arccoth}\left (x a \right )-1-a^{2} x^{2}-6 x a \,\operatorname {arccoth}\left (x a \right )+3 \operatorname {arccoth}\left (x a \right )^{2}-8 \ln \left (x a -1\right )-8 \,\operatorname {arccoth}\left (x a \right )}{12 a^{4}}\)	\(72\)
risch	\(\frac {\left (a^{4} x^{4}-1\right ) \ln \left (x a +1\right )^{2}}{16 a^{4}}-\frac {\left (3 x^{4} \ln \left (x a -1\right ) a^{4}-2 x^{3} a^{3}-6 x a -3 \ln \left (x a -1\right )\right ) \ln \left (x a +1\right )}{24 a^{4}}+\frac {x^{4} \ln \left (x a -1\right )^{2}}{16}-\frac {x^{3} \ln \left (x a -1\right )}{12 a}+\frac {x^{2}}{12 a^{2}}-\frac {x \ln \left (x a -1\right )}{4 a^{3}}-\frac {\ln \left (x a -1\right )^{2}}{16 a^{4}}+\frac {\ln \left (a^{2} x^{2}-1\right )}{3 a^{4}}\)	\(145\)
parts	\(\frac {x^{4} \operatorname {arccoth}\left (x a \right )^{2}}{4}+\frac {\frac {x^{3} a^{3} \operatorname {arccoth}\left (x a \right )}{3}+x a \,\operatorname {arccoth}\left (x a \right )+\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a -1\right )}{2}-\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a +1\right )}{2}-\frac {\ln \left (x a -1\right ) \ln \left (\frac {x a}{2}+\frac {1}{2}\right )}{4}+\frac {\ln \left (x a -1\right )^{2}}{8}-\frac {\left (\ln \left (x a +1\right )-\ln \left (\frac {x a}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {x a}{2}+\frac {1}{2}\right )}{4}+\frac {\ln \left (x a +1\right )^{2}}{8}+\frac {a^{2} x^{2}}{6}+\frac {2 \ln \left (x a -1\right )}{3}+\frac {2 \ln \left (x a +1\right )}{3}}{2 a^{4}}\)	\(146\)
derivativedivides	\(\frac {\frac {x^{4} a^{4} \operatorname {arccoth}\left (x a \right )^{2}}{4}+\frac {x^{3} a^{3} \operatorname {arccoth}\left (x a \right )}{6}+\frac {x a \,\operatorname {arccoth}\left (x a \right )}{2}+\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a -1\right )}{4}-\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a +1\right )}{4}-\frac {\ln \left (x a -1\right ) \ln \left (\frac {x a}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (x a -1\right )^{2}}{16}-\frac {\left (\ln \left (x a +1\right )-\ln \left (\frac {x a}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {x a}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (x a +1\right )^{2}}{16}+\frac {a^{2} x^{2}}{12}+\frac {\ln \left (x a -1\right )}{3}+\frac {\ln \left (x a +1\right )}{3}}{a^{4}}\)	\(148\)
default	\(\frac {\frac {x^{4} a^{4} \operatorname {arccoth}\left (x a \right )^{2}}{4}+\frac {x^{3} a^{3} \operatorname {arccoth}\left (x a \right )}{6}+\frac {x a \,\operatorname {arccoth}\left (x a \right )}{2}+\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a -1\right )}{4}-\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a +1\right )}{4}-\frac {\ln \left (x a -1\right ) \ln \left (\frac {x a}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (x a -1\right )^{2}}{16}-\frac {\left (\ln \left (x a +1\right )-\ln \left (\frac {x a}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {x a}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (x a +1\right )^{2}}{16}+\frac {a^{2} x^{2}}{12}+\frac {\ln \left (x a -1\right )}{3}+\frac {\ln \left (x a +1\right )}{3}}{a^{4}}\)	\(148\)

Input:

int(x^3*arccoth(x*a)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int x^3 \coth ^{-1}(a x)^2 \, dx=\frac {4 \, a^{2} x^{2} + 3 \, {\left (a^{4} x^{4} - 1\right )} \log \left (\frac {a x + 1}{a x - 1}\right )^{2} + 4 \, {\left (a^{3} x^{3} + 3 \, a x\right )} \log \left (\frac {a x + 1}{a x - 1}\right ) + 16 \, \log \left (a^{2} x^{2} - 1\right )}{48 \, a^{4}} \] Input:

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

Output:

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
 

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.11 \[ \int x^3 \coth ^{-1}(a x)^2 \, dx=\begin {cases} \frac {x^{4} \operatorname {acoth}^{2}{\left (a x \right )}}{4} + \frac {x^{3} \operatorname {acoth}{\left (a x \right )}}{6 a} + \frac {x^{2}}{12 a^{2}} + \frac {x \operatorname {acoth}{\left (a x \right )}}{2 a^{3}} + \frac {2 \log {\left (a x + 1 \right )}}{3 a^{4}} - \frac {\operatorname {acoth}^{2}{\left (a x \right )}}{4 a^{4}} - \frac {2 \operatorname {acoth}{\left (a x \right )}}{3 a^{4}} & \text {for}\: a \neq 0 \\- \frac {\pi ^{2} x^{4}}{16} & \text {otherwise} \end {cases} \] Input:

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

Output:

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))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.46 \[ \int x^3 \coth ^{-1}(a x)^2 \, dx=\frac {1}{4} \, x^{4} \operatorname {arcoth}\left (a x\right )^{2} + \frac {1}{12} \, a {\left (\frac {2 \, {\left (a^{2} x^{3} + 3 \, x\right )}}{a^{4}} - \frac {3 \, \log \left (a x + 1\right )}{a^{5}} + \frac {3 \, \log \left (a x - 1\right )}{a^{5}}\right )} \operatorname {arcoth}\left (a x\right ) + \frac {4 \, a^{2} x^{2} - 2 \, {\left (3 \, \log \left (a x - 1\right ) - 8\right )} \log \left (a x + 1\right ) + 3 \, \log \left (a x + 1\right )^{2} + 3 \, \log \left (a x - 1\right )^{2} + 16 \, \log \left (a x - 1\right )}{48 \, a^{4}} \] Input:

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

Output:

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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (69) = 138\).

Time = 0.13 (sec) , antiderivative size = 335, normalized size of antiderivative = 4.14 \[ \int x^3 \coth ^{-1}(a x)^2 \, dx=\frac {1}{6} \, {\left (\frac {3 \, {\left (\frac {{\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {a x + 1}{a x - 1}\right )} \log \left (\frac {a x + 1}{a x - 1}\right )^{2}}{\frac {{\left (a x + 1\right )}^{4} a^{5}}{{\left (a x - 1\right )}^{4}} - \frac {4 \, {\left (a x + 1\right )}^{3} a^{5}}{{\left (a x - 1\right )}^{3}} + \frac {6 \, {\left (a x + 1\right )}^{2} a^{5}}{{\left (a x - 1\right )}^{2}} - \frac {4 \, {\left (a x + 1\right )} a^{5}}{a x - 1} + a^{5}} + \frac {2 \, {\left (\frac {3 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - \frac {3 \, {\left (a x + 1\right )}}{a x - 1} + 2\right )} \log \left (\frac {a x + 1}{a x - 1}\right )}{\frac {{\left (a x + 1\right )}^{3} a^{5}}{{\left (a x - 1\right )}^{3}} - \frac {3 \, {\left (a x + 1\right )}^{2} a^{5}}{{\left (a x - 1\right )}^{2}} + \frac {3 \, {\left (a x + 1\right )} a^{5}}{a x - 1} - a^{5}} + \frac {2 \, {\left (a x + 1\right )}}{{\left (\frac {{\left (a x + 1\right )}^{2} a^{5}}{{\left (a x - 1\right )}^{2}} - \frac {2 \, {\left (a x + 1\right )} a^{5}}{a x - 1} + a^{5}\right )} {\left (a x - 1\right )}} - \frac {4 \, \log \left (\frac {a x + 1}{a x - 1} - 1\right )}{a^{5}} + \frac {4 \, \log \left (\frac {a x + 1}{a x - 1}\right )}{a^{5}}\right )} a \] Input:

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

Output:

1/6*(3*((a*x + 1)^3/(a*x - 1)^3 + (a*x + 1)/(a*x - 1))*log((a*x + 1)/(a*x 
- 1))^2/((a*x + 1)^4*a^5/(a*x - 1)^4 - 4*(a*x + 1)^3*a^5/(a*x - 1)^3 + 6*( 
a*x + 1)^2*a^5/(a*x - 1)^2 - 4*(a*x + 1)*a^5/(a*x - 1) + a^5) + 2*(3*(a*x 
+ 1)^2/(a*x - 1)^2 - 3*(a*x + 1)/(a*x - 1) + 2)*log((a*x + 1)/(a*x - 1))/( 
(a*x + 1)^3*a^5/(a*x - 1)^3 - 3*(a*x + 1)^2*a^5/(a*x - 1)^2 + 3*(a*x + 1)* 
a^5/(a*x - 1) - a^5) + 2*(a*x + 1)/(((a*x + 1)^2*a^5/(a*x - 1)^2 - 2*(a*x 
+ 1)*a^5/(a*x - 1) + a^5)*(a*x - 1)) - 4*log((a*x + 1)/(a*x - 1) - 1)/a^5 
+ 4*log((a*x + 1)/(a*x - 1))/a^5)*a
 

Mupad [B] (verification not implemented)

Time = 3.82 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80 \[ \int x^3 \coth ^{-1}(a x)^2 \, dx=\frac {x^4\,{\mathrm {acoth}\left (a\,x\right )}^2}{4}+\frac {\frac {\ln \left (a^2\,x^2-1\right )}{3}+\frac {a^2\,x^2}{12}-\frac {{\mathrm {acoth}\left (a\,x\right )}^2}{4}+\frac {a^3\,x^3\,\mathrm {acoth}\left (a\,x\right )}{6}+\frac {a\,x\,\mathrm {acoth}\left (a\,x\right )}{2}}{a^4} \] Input:

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

Output:

(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*acoth(a*x))/2)/a^4
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int x^3 \coth ^{-1}(a x)^2 \, dx=\frac {3 \mathit {acoth} \left (a x \right )^{2} a^{4} x^{4}-3 \mathit {acoth} \left (a x \right )^{2}-2 \mathit {acoth} \left (a x \right ) a^{3} x^{3}-6 \mathit {acoth} \left (a x \right ) a x -8 \mathit {acoth} \left (a x \right )+8 \,\mathrm {log}\left (a^{2} x -a \right )+a^{2} x^{2}}{12 a^{4}} \] Input:

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

Output:

(3*acoth(a*x)**2*a**4*x**4 - 3*acoth(a*x)**2 - 2*acoth(a*x)*a**3*x**3 - 6* 
acoth(a*x)*a*x - 8*acoth(a*x) + 8*log(a**2*x - a) + a**2*x**2)/(12*a**4)