\(\int x^3 \coth ^{-1}(\tanh (a+b x))^3 \, dx\) [148]

Optimal result

Integrand size = 13, antiderivative size = 61 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {1}{140} b^3 x^7+\frac {1}{20} b^2 x^6 \coth ^{-1}(\tanh (a+b x))-\frac {3}{20} b x^5 \coth ^{-1}(\tanh (a+b x))^2+\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^3 \]

[Out]

-1/140*b^3*x^7+1/20*b^2*x^6*arccoth(tanh(b*x+a))-3/20*b*x^5*arccoth(tanh(b*x+a))^2+1/4*x^4*arccoth(tanh(b*x+a)
)^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2199, 30} \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {1}{20} b^2 x^6 \coth ^{-1}(\tanh (a+b x))-\frac {3}{20} b x^5 \coth ^{-1}(\tanh (a+b x))^2+\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^3-\frac {1}{140} b^3 x^7 \]

[In]

Int[x^3*ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

-1/140*(b^3*x^7) + (b^2*x^6*ArcCoth[Tanh[a + b*x]])/20 - (3*b*x^5*ArcCoth[Tanh[a + b*x]]^2)/20 + (x^4*ArcCoth[
Tanh[a + b*x]]^3)/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))^3-\frac {1}{4} (3 b) \int x^4 \coth ^{-1}(\tanh (a+b x))^2 \, dx \\ & = -\frac {3}{20} b x^5 \coth ^{-1}(\tanh (a+b x))^2+\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^3+\frac {1}{10} \left (3 b^2\right ) \int x^5 \coth ^{-1}(\tanh (a+b x)) \, dx \\ & = \frac {1}{20} b^2 x^6 \coth ^{-1}(\tanh (a+b x))-\frac {3}{20} b x^5 \coth ^{-1}(\tanh (a+b x))^2+\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^3-\frac {1}{20} b^3 \int x^6 \, dx \\ & = -\frac {1}{140} b^3 x^7+\frac {1}{20} b^2 x^6 \coth ^{-1}(\tanh (a+b x))-\frac {3}{20} b x^5 \coth ^{-1}(\tanh (a+b x))^2+\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {1}{140} x^4 \left (b^3 x^3-7 b^2 x^2 \coth ^{-1}(\tanh (a+b x))+21 b x \coth ^{-1}(\tanh (a+b x))^2-35 \coth ^{-1}(\tanh (a+b x))^3\right ) \]

[In]

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

[Out]

-1/140*(x^4*(b^3*x^3 - 7*b^2*x^2*ArcCoth[Tanh[a + b*x]] + 21*b*x*ArcCoth[Tanh[a + b*x]]^2 - 35*ArcCoth[Tanh[a
+ b*x]]^3))

Maple [F(-1)]

Timed out.

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {1}{7} \, b^{3} x^{7} + \frac {1}{2} \, a b^{2} x^{6} + \frac {3}{5} \, a^{2} b x^{5} - \frac {1}{32} i \, \pi ^{3} x^{4} + \frac {1}{4} \, a^{3} x^{4} - \frac {3}{80} \, \pi ^{2} {\left (4 \, b x^{5} + 5 \, a x^{4}\right )} + \frac {1}{40} i \, \pi {\left (10 \, b^{2} x^{6} + 24 \, a b x^{5} + 15 \, a^{2} x^{4}\right )} \]

[In]

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

[Out]

1/7*b^3*x^7 + 1/2*a*b^2*x^6 + 3/5*a^2*b*x^5 - 1/32*I*pi^3*x^4 + 1/4*a^3*x^4 - 3/80*pi^2*(4*b*x^5 + 5*a*x^4) +
1/40*I*pi*(10*b^2*x^6 + 24*a*b*x^5 + 15*a^2*x^4)

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.95 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^3 \, dx=- \frac {b^{3} x^{7}}{140} + \frac {b^{2} x^{6} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{20} - \frac {3 b x^{5} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{20} + \frac {x^{4} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{4} \]

[In]

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

[Out]

-b**3*x**7/140 + b**2*x**6*acoth(tanh(a + b*x))/20 - 3*b*x**5*acoth(tanh(a + b*x))**2/20 + x**4*acoth(tanh(a +
b*x))**3/4

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {3}{20} \, b x^{5} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} + \frac {1}{4} \, x^{4} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {1}{140} \, {\left (b^{2} x^{7} - 7 \, b x^{6} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )\right )} b \]

[In]

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

[Out]

-3/20*b*x^5*arccoth(tanh(b*x + a))^2 + 1/4*x^4*arccoth(tanh(b*x + a))^3 - 1/140*(b^2*x^7 - 7*b*x^6*arccoth(tan
h(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 = 1.26 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {1}{7} \, b^{3} x^{7} - \frac {1}{4} \, {\left (-i \, \pi b^{2} - 2 \, a b^{2}\right )} x^{6} - \frac {3}{20} \, {\left (\pi ^{2} b - 4 i \, \pi a b - 4 \, a^{2} b\right )} x^{5} - \frac {1}{32} \, {\left (i \, \pi ^{3} + 6 \, \pi ^{2} a - 12 i \, \pi a^{2} - 8 \, a^{3}\right )} x^{4} \]

[In]

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

[Out]

1/7*b^3*x^7 - 1/4*(-I*pi*b^2 - 2*a*b^2)*x^6 - 3/20*(pi^2*b - 4*I*pi*a*b - 4*a^2*b)*x^5 - 1/32*(I*pi^3 + 6*pi^2
*a - 12*I*pi*a^2 - 8*a^3)*x^4

Mupad [B] (verification not implemented)

Time = 4.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int x^3 \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {b^3\,x^7}{140}+\frac {b^2\,x^6\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{20}-\frac {3\,b\,x^5\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{20}+\frac {x^4\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{4} \]

[In]

int(x^3*acoth(tanh(a + b*x))^3,x)

[Out]

(x^4*acoth(tanh(a + b*x))^3)/4 - (b^3*x^7)/140 - (3*b*x^5*acoth(tanh(a + b*x))^2)/20 + (b^2*x^6*acoth(tanh(a +
b*x)))/20