\(\int x^2 \coth ^{-1}(\tanh (a+b x))^2 \, dx\) [138]

Optimal result

Integrand size = 13, antiderivative size = 42 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {b^2 x^5}{30}-\frac {1}{6} b x^4 \coth ^{-1}(\tanh (a+b x))+\frac {1}{3} x^3 \coth ^{-1}(\tanh (a+b x))^2 \]

[Out]

1/30*b^2*x^5-1/6*b*x^4*arccoth(tanh(b*x+a))+1/3*x^3*arccoth(tanh(b*x+a))^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2199, 30} \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^2 \, dx=-\frac {1}{6} b x^4 \coth ^{-1}(\tanh (a+b x))+\frac {1}{3} x^3 \coth ^{-1}(\tanh (a+b x))^2+\frac {b^2 x^5}{30} \]

[In]

Int[x^2*ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

(b^2*x^5)/30 - (b*x^4*ArcCoth[Tanh[a + b*x]])/6 + (x^3*ArcCoth[Tanh[a + b*x]]^2)/3

Rule 30

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

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 {1}{3} x^3 \coth ^{-1}(\tanh (a+b x))^2-\frac {1}{3} (2 b) \int x^3 \coth ^{-1}(\tanh (a+b x)) \, dx \\ & = -\frac {1}{6} b x^4 \coth ^{-1}(\tanh (a+b x))+\frac {1}{3} x^3 \coth ^{-1}(\tanh (a+b x))^2+\frac {1}{6} b^2 \int x^4 \, dx \\ & = \frac {b^2 x^5}{30}-\frac {1}{6} b x^4 \coth ^{-1}(\tanh (a+b x))+\frac {1}{3} x^3 \coth ^{-1}(\tanh (a+b x))^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {1}{30} x^3 \left (b^2 x^2-5 b x \coth ^{-1}(\tanh (a+b x))+10 \coth ^{-1}(\tanh (a+b x))^2\right ) \]

[In]

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

[Out]

(x^3*(b^2*x^2 - 5*b*x*ArcCoth[Tanh[a + b*x]] + 10*ArcCoth[Tanh[a + b*x]]^2))/30

Maple [A] (verified)

Time = 25.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(\frac {b^{2} x^{5}}{30}-\frac {b \,x^{4} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{6}+\frac {x^{3} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}{3}\)	\(37\)
risch	\(\text {Expression too large to display}\)	\(3418\)

[In]

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

[Out]

1/30*b^2*x^5-1/6*b*x^4*arccoth(tanh(b*x+a))+1/3*x^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 = 48, normalized size of antiderivative = 1.14 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {1}{5} \, b^{2} x^{5} + \frac {1}{2} \, a b x^{4} - \frac {1}{12} \, \pi ^{2} x^{3} + \frac {1}{3} \, a^{2} x^{3} + \frac {1}{12} i \, \pi {\left (3 \, b x^{4} + 4 \, a x^{3}\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {b^{2} x^{5}}{30} - \frac {b x^{4} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{6} + \frac {x^{3} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{3} \]

[In]

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

[Out]

b**2*x**5/30 - b*x**4*acoth(tanh(a + b*x))/6 + x**3*acoth(tanh(a + b*x))**2/3

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {1}{30} \, b^{2} x^{5} - \frac {1}{6} \, b x^{4} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) + \frac {1}{3} \, x^{3} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \]

[In]

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

[Out]

1/30*b^2*x^5 - 1/6*b*x^4*arccoth(tanh(b*x + a)) + 1/3*x^3*arccoth(tanh(b*x + a))^2

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.98 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {1}{5} \, b^{2} x^{5} - \frac {1}{4} \, {\left (-i \, \pi b - 2 \, a b\right )} x^{4} - \frac {1}{12} \, {\left (\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}\right )} x^{3} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int x^2 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {b^2\,x^5}{30}-\frac {b\,x^4\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{6}+\frac {x^3\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{3} \]

[In]

int(x^2*acoth(tanh(a + b*x))^2,x)

[Out]

(x^3*acoth(tanh(a + b*x))^2)/3 + (b^2*x^5)/30 - (b*x^4*acoth(tanh(a + b*x)))/6