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

Optimal. Leaf size=53 \[ \frac {\coth ^{-1}(\tanh (a+b x))^6}{60 b^3}-\frac {x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b} \]

[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]  time = 0.03, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ \frac {\coth ^{-1}(\tanh (a+b x))^6}{60 b^3}-\frac {x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b} \]

Antiderivative was successfully verified.

[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 2157

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 2168

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

Antiderivative was successfully verified.

[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))

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 52, normalized size = 0.98 \[ \frac {1}{6} \, b^{3} x^{6} + \frac {3}{5} \, a b^{2} x^{5} - \frac {3}{16} \, {\left (\pi ^{2} b - 4 \, a^{2} b\right )} x^{4} - \frac {1}{12} \, {\left (3 \, \pi ^{2} a - 4 \, a^{3}\right )} x^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 - 3/16*(pi^2*b - 4*a^2*b)*x^4 - 1/12*(3*pi^2*a - 4*a^3)*x^3

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 1.13, size = 18111, normalized size = 341.72 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

maxima [A]  time = 0.52, size = 54, normalized size = 1.02 \[ -\frac {1}{4} \, b x^{4} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} + \frac {1}{3} \, x^{3} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {1}{60} \, {\left (b^{2} x^{6} - 6 \, b x^{5} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

mupad [B]  time = 1.21, size = 53, normalized size = 1.00 \[ -\frac {b^3\,x^6}{60}+\frac {b^2\,x^5\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{10}-\frac {b\,x^4\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{4}+\frac {x^3\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [A]  time = 2.38, size = 60, normalized size = 1.13 \[ \begin {cases} \frac {x^{2} \operatorname {acoth}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{4 b} - \frac {x \operatorname {acoth}^{5}{\left (\tanh {\left (a + b x \right )} \right )}}{10 b^{2}} + \frac {\operatorname {acoth}^{6}{\left (\tanh {\left (a + b x \right )} \right )}}{60 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \operatorname {acoth}^{3}{\left (\tanh {\relax (a )} \right )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*acoth(tanh(a + b*x))**4/(4*b) - x*acoth(tanh(a + b*x))**5/(10*b**2) + acoth(tanh(a + b*x))**6/
(60*b**3), Ne(b, 0)), (x**3*acoth(tanh(a))**3/3, True))

________________________________________________________________________________________