\(\int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))^3} \, dx\) [178]

Optimal result

Integrand size = 13, antiderivative size = 47 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {x^2}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {x}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2199, 2188, 29} \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x}{b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {x^2}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

[In]

Int[x^2/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2188

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2199

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[u^(m + 1)*(v^
n/(a*(m + 1))), x] - Dist[b*(n/(a*(m + 1))), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps \begin{align*} \text {integral}& = -\frac {x^2}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac {\int \frac {x}{\coth ^{-1}(\tanh (a+b x))^2} \, dx}{b} \\ & = -\frac {x^2}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {x}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {\int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^2} \\ & = -\frac {x^2}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {x}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b^3} \\ & = -\frac {x^2}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {x}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {3-\frac {b^2 x^2}{\coth ^{-1}(\tanh (a+b x))^2}-\frac {2 b x}{\coth ^{-1}(\tanh (a+b x))}+2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{2 b^3} \]

[In]

Integrate[x^2/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

(3 - (b^2*x^2)/ArcCoth[Tanh[a + b*x]]^2 - (2*b*x)/ArcCoth[Tanh[a + b*x]] + 2*Log[ArcCoth[Tanh[a + b*x]]])/(2*b
^3)

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.15

method	result	size
parallelrisch	\(\frac {-b^{2} x^{2}+2 \ln \left (\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )\right ) \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}-2 b x \,\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{2 b^{3} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}\)	\(54\)
risch	\(\text {Expression too large to display}\)	\(952\)

[In]

int(x^2/arccoth(tanh(b*x+a))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.62 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {16 \, a b x - 3 \, \pi ^{2} + 4 i \, \pi {\left (2 \, b x + 3 \, a\right )} + 12 \, a^{2} + 2 \, {\left (4 \, b^{2} x^{2} + 8 \, a b x - \pi ^{2} + 4 i \, \pi {\left (b x + a\right )} + 4 \, a^{2}\right )} \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{2 \, {\left (4 \, b^{5} x^{2} + 8 \, a b^{4} x - \pi ^{2} b^{3} + 4 \, a^{2} b^{3} + 4 i \, \pi {\left (b^{4} x + a b^{3}\right )}\right )}} \]

[In]

integrate(x^2/arccoth(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

1/2*(16*a*b*x - 3*pi^2 + 4*I*pi*(2*b*x + 3*a) + 12*a^2 + 2*(4*b^2*x^2 + 8*a*b*x - pi^2 + 4*I*pi*(b*x + a) + 4*
a^2)*log(I*pi + 2*b*x + 2*a))/(4*b^5*x^2 + 8*a*b^4*x - pi^2*b^3 + 4*a^2*b^3 + 4*I*pi*(b^4*x + a*b^3))

Sympy [A] (verification not implemented)

Time = 24.64 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.15 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\begin {cases} - \frac {x^{2}}{2 b \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {x}{b^{2} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}} + \frac {\log {\left (\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 \operatorname {acoth}^{3}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2/acoth(tanh(b*x+a))**3,x)

[Out]

Piecewise((-x**2/(2*b*acoth(tanh(a + b*x))**2) - x/(b**2*acoth(tanh(a + b*x))) + log(acoth(tanh(a + b*x)))/b**
3, Ne(b, 0)), (x**3/(3*acoth(tanh(a))**3), True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.77 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.04 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {3 \, \pi ^{2} + 12 i \, \pi a - 12 \, a^{2} - 8 \, {\left (-i \, \pi b + 2 \, a b\right )} x}{2 \, {\left (4 \, b^{5} x^{2} - \pi ^{2} b^{3} - 4 i \, \pi a b^{3} + 4 \, a^{2} b^{3} - 4 \, {\left (i \, \pi b^{4} - 2 \, a b^{4}\right )} x\right )}} + \frac {\log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{b^{3}} \]

[In]

integrate(x^2/arccoth(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-1/2*(3*pi^2 + 12*I*pi*a - 12*a^2 - 8*(-I*pi*b + 2*a*b)*x)/(4*b^5*x^2 - pi^2*b^3 - 4*I*pi*a*b^3 + 4*a^2*b^3 -
4*(I*pi*b^4 - 2*a*b^4)*x) + log(-I*pi + 2*b*x + 2*a)/b^3

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.96 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {8 \, \pi b x - 16 i \, a b x + 3 i \, \pi ^{2} + 12 \, \pi a - 12 i \, a^{2}}{8 i \, b^{5} x^{2} - 8 \, \pi b^{4} x + 16 i \, a b^{4} x - 2 i \, \pi ^{2} b^{3} - 8 \, \pi a b^{3} + 8 i \, a^{2} b^{3}} + \frac {\log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{b^{3}} \]

[In]

integrate(x^2/arccoth(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

-(8*pi*b*x - 16*I*a*b*x + 3*I*pi^2 + 12*pi*a - 12*I*a^2)/(8*I*b^5*x^2 - 8*pi*b^4*x + 16*I*a*b^4*x - 2*I*pi^2*b
^3 - 8*pi*a*b^3 + 8*I*a^2*b^3) + log(I*pi + 2*b*x + 2*a)/b^3

Mupad [B] (verification not implemented)

Time = 3.86 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {\ln \left (\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )\right )}{b^3}-\frac {\frac {b^2\,x^2}{2}+b\,x\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{b^3\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2} \]

[In]

int(x^2/acoth(tanh(a + b*x))^3,x)

[Out]

log(acoth(tanh(a + b*x)))/b^3 - ((b^2*x^2)/2 + b*x*acoth(tanh(a + b*x)))/(b^3*acoth(tanh(a + b*x))^2)