\(\int x^2 \coth ^{-1}(\tanh (a+b x))^3 \, dx\) [149]

Optimal result

Integrand size = 13, antiderivative size = 53 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {\coth ^{-1}(\tanh (a+b x))^6}{60 b^3} \]

[Out]

1/4*x^2*arccoth(tanh(b*x+a))^4/b-1/10*x*arccoth(tanh(b*x+a))^5/b^2+1/60*arccoth(tanh(b*x+a))^6/b^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, 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, 30} \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^6}{60 b^3}-\frac {x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b} \]

[In]

Int[x^2*ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

(x^2*ArcCoth[Tanh[a + b*x]]^4)/(4*b) - (x*ArcCoth[Tanh[a + b*x]]^5)/(10*b^2) + ArcCoth[Tanh[a + b*x]]^6/(60*b^
3)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 \coth ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {\int x \coth ^{-1}(\tanh (a+b x))^4 \, dx}{2 b} \\ & = \frac {x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {\int \coth ^{-1}(\tanh (a+b x))^5 \, dx}{10 b^2} \\ & = \frac {x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {\text {Subst}\left (\int x^5 \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{10 b^3} \\ & = \frac {x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {\coth ^{-1}(\tanh (a+b x))^6}{60 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.02 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {1}{60} x^3 \left (b^3 x^3-6 b^2 x^2 \coth ^{-1}(\tanh (a+b x))+15 b x \coth ^{-1}(\tanh (a+b x))^2-20 \coth ^{-1}(\tanh (a+b x))^3\right ) \]

[In]

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

[Out]

-1/60*(x^3*(b^3*x^3 - 6*b^2*x^2*ArcCoth[Tanh[a + b*x]] + 15*b*x*ArcCoth[Tanh[a + b*x]]^2 - 20*ArcCoth[Tanh[a +
b*x]]^3))

Maple [F(-1)]

Timed out.

\[\int x^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3}d x\]

[In]

int(x^2*arccoth(tanh(b*x+a))^3,x)

[Out]

int(x^2*arccoth(tanh(b*x+a))^3,x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.66 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {1}{6} \, b^{3} x^{6} + \frac {3}{5} \, a b^{2} x^{5} + \frac {3}{4} \, a^{2} b x^{4} - \frac {1}{24} i \, \pi ^{3} x^{3} + \frac {1}{3} \, a^{3} x^{3} - \frac {1}{16} \, \pi ^{2} {\left (3 \, b x^{4} + 4 \, a x^{3}\right )} + \frac {1}{20} i \, \pi {\left (6 \, b^{2} x^{5} + 15 \, a b x^{4} + 10 \, a^{2} x^{3}\right )} \]

[In]

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

[Out]

1/6*b^3*x^6 + 3/5*a*b^2*x^5 + 3/4*a^2*b*x^4 - 1/24*I*pi^3*x^3 + 1/3*a^3*x^3 - 1/16*pi^2*(3*b*x^4 + 4*a*x^3) +
1/20*I*pi*(6*b^2*x^5 + 15*a*b*x^4 + 10*a^2*x^3)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^3 \, dx=- \frac {b^{3} x^{6}}{60} + \frac {b^{2} x^{5} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{10} - \frac {b x^{4} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{4} + \frac {x^{3} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3} \]

[In]

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

[Out]

-b**3*x**6/60 + b**2*x**5*acoth(tanh(a + b*x))/10 - b*x**4*acoth(tanh(a + b*x))**2/4 + x**3*acoth(tanh(a + b*x
))**3/3

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.02 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {1}{4} \, b x^{4} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} + \frac {1}{3} \, x^{3} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {1}{60} \, {\left (b^{2} x^{6} - 6 \, b x^{5} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )\right )} b \]

[In]

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

[Out]

-1/4*b*x^4*arccoth(tanh(b*x + a))^2 + 1/3*x^3*arccoth(tanh(b*x + a))^3 - 1/60*(b^2*x^6 - 6*b*x^5*arccoth(tanh(
b*x + a)))*b

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.45 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {1}{6} \, b^{3} x^{6} - \frac {3}{10} \, {\left (-i \, \pi b^{2} - 2 \, a b^{2}\right )} x^{5} - \frac {3}{16} \, {\left (\pi ^{2} b - 4 i \, \pi a b - 4 \, a^{2} b\right )} x^{4} - \frac {1}{24} \, {\left (i \, \pi ^{3} + 6 \, \pi ^{2} a - 12 i \, \pi a^{2} - 8 \, a^{3}\right )} x^{3} \]

[In]

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

[Out]

1/6*b^3*x^6 - 3/10*(-I*pi*b^2 - 2*a*b^2)*x^5 - 3/16*(pi^2*b - 4*I*pi*a*b - 4*a^2*b)*x^4 - 1/24*(I*pi^3 + 6*pi^
2*a - 12*I*pi*a^2 - 8*a^3)*x^3

Mupad [B] (verification not implemented)

Time = 3.85 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {b^3\,x^6}{60}+\frac {b^2\,x^5\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{10}-\frac {b\,x^4\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{4}+\frac {x^3\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{3} \]

[In]

int(x^2*acoth(tanh(a + b*x))^3,x)

[Out]

(x^3*acoth(tanh(a + b*x))^3)/3 - (b^3*x^6)/60 - (b*x^4*acoth(tanh(a + b*x))^2)/4 + (b^2*x^5*acoth(tanh(a + b*x
)))/10