\(\int x^3 \coth ^{-1}(\tanh (a+b x))^2 \, dx\) [137]

Optimal result

Integrand size = 13, antiderivative size = 42 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {b^2 x^6}{60}-\frac {1}{10} b x^5 \coth ^{-1}(\tanh (a+b x))+\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^2 \]

[Out]

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

Rubi [A] (verified)

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

[In]

Int[x^3*ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

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

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}{4} x^4 \coth ^{-1}(\tanh (a+b x))^2-\frac {1}{2} b \int x^4 \coth ^{-1}(\tanh (a+b x)) \, dx \\ & = -\frac {1}{10} b x^5 \coth ^{-1}(\tanh (a+b x))+\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^2+\frac {1}{10} b^2 \int x^5 \, dx \\ & = \frac {b^2 x^6}{60}-\frac {1}{10} b x^5 \coth ^{-1}(\tanh (a+b x))+\frac {1}{4} x^4 \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 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {1}{60} x^4 \left (b^2 x^2-6 b x \coth ^{-1}(\tanh (a+b x))+15 \coth ^{-1}(\tanh (a+b x))^2\right ) \]

[In]

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

[Out]

(x^4*(b^2*x^2 - 6*b*x*ArcCoth[Tanh[a + b*x]] + 15*ArcCoth[Tanh[a + b*x]]^2))/60

Maple [A] (verified)

Time = 28.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(\frac {b^{2} x^{6}}{60}-\frac {b \,x^{5} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{10}+\frac {x^{4} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}{4}\)	\(37\)
risch	\(\text {Expression too large to display}\)	\(3418\)

[In]

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

[Out]

1/60*b^2*x^6-1/10*b*x^5*arccoth(tanh(b*x+a))+1/4*x^4*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 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {1}{6} \, b^{2} x^{6} + \frac {2}{5} \, a b x^{5} - \frac {1}{16} \, \pi ^{2} x^{4} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{20} i \, \pi {\left (4 \, b x^{5} + 5 \, a x^{4}\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {b^{2} x^{6}}{60} - \frac {b x^{5} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{10} + \frac {x^{4} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {1}{60} \, b^{2} x^{6} - \frac {1}{10} \, b x^{5} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) + \frac {1}{4} \, x^{4} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \]

[In]

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

[Out]

1/60*b^2*x^6 - 1/10*b*x^5*arccoth(tanh(b*x + a)) + 1/4*x^4*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 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {1}{6} \, b^{2} x^{6} - \frac {1}{5} \, {\left (-i \, \pi b - 2 \, a b\right )} x^{5} - \frac {1}{16} \, {\left (\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}\right )} x^{4} \]

[In]

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

[Out]

1/6*b^2*x^6 - 1/5*(-I*pi*b - 2*a*b)*x^5 - 1/16*(pi^2 - 4*I*pi*a - 4*a^2)*x^4

Mupad [B] (verification not implemented)

Time = 3.84 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {b^2\,x^6}{60}-\frac {b\,x^5\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{10}+\frac {x^4\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{4} \]

[In]

int(x^3*acoth(tanh(a + b*x))^2,x)

[Out]

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