\(\int x^2 \coth ^{-1}(a x) \, dx\) [4]

Optimal result

Integrand size = 8, antiderivative size = 40 \[ \int x^2 \coth ^{-1}(a x) \, dx=\frac {x^2}{6 a}+\frac {1}{3} x^3 \coth ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{6 a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6038, 272, 45} \[ \int x^2 \coth ^{-1}(a x) \, dx=\frac {\log \left (1-a^2 x^2\right )}{6 a^3}+\frac {1}{3} x^3 \coth ^{-1}(a x)+\frac {x^2}{6 a} \]

[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 45

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 272

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 6038

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] - Dist[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
]

Rubi steps \begin{align*} \text {integral}& = \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 \text {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 \text {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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int x^2 \coth ^{-1}(a x) \, dx=\frac {x^2}{6 a}+\frac {1}{3} x^3 \coth ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{6 a^3} \]

[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)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.95

method	result	size
parts	\(\frac {x^{3} \operatorname {arccoth}\left (a x \right )}{3}+\frac {a \left (\frac {x^{2}}{2 a^{2}}+\frac {\ln \left (a^{2} x^{2}-1\right )}{2 a^{4}}\right )}{3}\)	\(38\)
parallelrisch	\(-\frac {-2 a^{3} x^{3} \operatorname {arccoth}\left (a x \right )-a^{2} x^{2}-2 \ln \left (a x -1\right )-2 \,\operatorname {arccoth}\left (a x \right )}{6 a^{3}}\)	\(41\)
derivativedivides	\(\frac {\frac {a^{3} x^{3} \operatorname {arccoth}\left (a x \right )}{3}+\frac {a^{2} x^{2}}{6}+\frac {\ln \left (a x -1\right )}{6}+\frac {\ln \left (a x +1\right )}{6}}{a^{3}}\)	\(42\)
default	\(\frac {\frac {a^{3} x^{3} \operatorname {arccoth}\left (a x \right )}{3}+\frac {a^{2} x^{2}}{6}+\frac {\ln \left (a x -1\right )}{6}+\frac {\ln \left (a x +1\right )}{6}}{a^{3}}\)	\(42\)
risch	\(\frac {x^{3} \ln \left (a x +1\right )}{6}-\frac {\ln \left (a x -1\right ) x^{3}}{6}+\frac {x^{2}}{6 a}+\frac {\ln \left (a^{2} x^{2}-1\right )}{6 a^{3}}\)	\(47\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.10 \[ \int x^2 \coth ^{-1}(a x) \, dx=\frac {a^{3} x^{3} \log \left (\frac {a x + 1}{a x - 1}\right ) + a^{2} x^{2} + \log \left (a^{2} x^{2} - 1\right )}{6 \, a^{3}} \]

[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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.15 \[ \int x^2 \coth ^{-1}(a x) \, dx=\begin {cases} \frac {x^{3} \operatorname {acoth}{\left (a x \right )}}{3} + \frac {x^{2}}{6 a} + \frac {\log {\left (a x + 1 \right )}}{3 a^{3}} - \frac {\operatorname {acoth}{\left (a x \right )}}{3 a^{3}} & \text {for}\: a \neq 0 \\\frac {i \pi x^{3}}{6} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.88 \[ \int x^2 \coth ^{-1}(a x) \, dx=\frac {1}{3} \, x^{3} \operatorname {arcoth}\left (a x\right ) + \frac {1}{6} \, a {\left (\frac {x^{2}}{a^{2}} + \frac {\log \left (a^{2} x^{2} - 1\right )}{a^{4}}\right )} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (34) = 68\).

Time = 0.28 (sec) , antiderivative size = 206, normalized size of antiderivative = 5.15 \[ \int x^2 \coth ^{-1}(a x) \, dx=\frac {1}{3} \, a {\left (\frac {\log \left (\frac {{\left | a x + 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{4}} - \frac {\log \left ({\left | \frac {a x + 1}{a x - 1} - 1 \right |}\right )}{a^{4}} + \frac {{\left (\frac {3 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + 1\right )} \log \left (-\frac {\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} + 1}{\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} - 1}\right )}{a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{3}} + \frac {2 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{2}}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.40 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.88 \[ \int x^2 \coth ^{-1}(a x) \, dx=\frac {\frac {\ln \left (a^2\,x^2-1\right )}{6}+\frac {a^2\,x^2}{6}}{a^3}+\frac {x^3\,\mathrm {acoth}\left (a\,x\right )}{3} \]

[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