\(\int \sqrt [3]{b \coth ^4(c+d x)} \, dx\) [46]

Optimal result

Integrand size = 14, antiderivative size = 289 \[ \int \sqrt [3]{b \coth ^4(c+d x)} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{2 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{2 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{d \coth ^{\frac {4}{3}}(c+d x)}-\frac {\sqrt [3]{b \coth ^4(c+d x)} \log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\sqrt [3]{b \coth ^4(c+d x)} \log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)}-\frac {3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d} \]

[Out]

arctanh(coth(d*x+c)^(1/3))*(b*coth(d*x+c)^4)^(1/3)/d/coth(d*x+c)^(4/3)-1/4*(b*coth(d*x+c)^4)^(1/3)*ln(1-coth(d
*x+c)^(1/3)+coth(d*x+c)^(2/3))/d/coth(d*x+c)^(4/3)+1/4*(b*coth(d*x+c)^4)^(1/3)*ln(1+coth(d*x+c)^(1/3)+coth(d*x
+c)^(2/3))/d/coth(d*x+c)^(4/3)-1/2*arctan(1/3*(1-2*coth(d*x+c)^(1/3))*3^(1/2))*(b*coth(d*x+c)^4)^(1/3)*3^(1/2)
/d/coth(d*x+c)^(4/3)+1/2*arctan(1/3*(1+2*coth(d*x+c)^(1/3))*3^(1/2))*(b*coth(d*x+c)^4)^(1/3)*3^(1/2)/d/coth(d*
x+c)^(4/3)-3*(b*coth(d*x+c)^4)^(1/3)*tanh(d*x+c)/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3739, 3554, 3557, 335, 216, 648, 632, 210, 642, 212} \[ \int \sqrt [3]{b \coth ^4(c+d x)} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{2 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt {3}}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{2 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{d \coth ^{\frac {4}{3}}(c+d x)}-\frac {3 \tanh (c+d x) \sqrt [3]{b \coth ^4(c+d x)}}{d}-\frac {\sqrt [3]{b \coth ^4(c+d x)} \log \left (\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\sqrt [3]{b \coth ^4(c+d x)} \log \left (\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)} \]

[In]

Int[(b*Coth[c + d*x]^4)^(1/3),x]

[Out]

-1/2*(Sqrt[3]*ArcTan[(1 - 2*Coth[c + d*x]^(1/3))/Sqrt[3]]*(b*Coth[c + d*x]^4)^(1/3))/(d*Coth[c + d*x]^(4/3)) +
(Sqrt[3]*ArcTan[(1 + 2*Coth[c + d*x]^(1/3))/Sqrt[3]]*(b*Coth[c + d*x]^4)^(1/3))/(2*d*Coth[c + d*x]^(4/3)) + (
ArcTanh[Coth[c + d*x]^(1/3)]*(b*Coth[c + d*x]^4)^(1/3))/(d*Coth[c + d*x]^(4/3)) - ((b*Coth[c + d*x]^4)^(1/3)*L
og[1 - Coth[c + d*x]^(1/3) + Coth[c + d*x]^(2/3)])/(4*d*Coth[c + d*x]^(4/3)) + ((b*Coth[c + d*x]^4)^(1/3)*Log[
1 + Coth[c + d*x]^(1/3) + Coth[c + d*x]^(2/3)])/(4*d*Coth[c + d*x]^(4/3)) - (3*(b*Coth[c + d*x]^4)^(1/3)*Tanh[
c + d*x])/d

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 216

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-a/b, n]], s = Denominator[Rt[-a/b, n
]], k, u}, Simp[u = Int[(r - s*Cos[(2*k*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r + s*C
os[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; 2*(r^2/(a*n))*Int[1/(r^2 - s^2*x^2), x] + Dis
t[2*(r/(a*n)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{b \coth ^4(c+d x)} \int \coth ^{\frac {4}{3}}(c+d x) \, dx}{\coth ^{\frac {4}{3}}(c+d x)} \\ & = -\frac {3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}+\frac {\sqrt [3]{b \coth ^4(c+d x)} \int \frac {1}{\coth ^{\frac {2}{3}}(c+d x)} \, dx}{\coth ^{\frac {4}{3}}(c+d x)} \\ & = -\frac {3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}-\frac {\sqrt [3]{b \coth ^4(c+d x)} \text {Subst}\left (\int \frac {1}{x^{2/3} \left (-1+x^2\right )} \, dx,x,\coth (c+d x)\right )}{d \coth ^{\frac {4}{3}}(c+d x)} \\ & = -\frac {3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}-\frac {\left (3 \sqrt [3]{b \coth ^4(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-1+x^6} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac {4}{3}}(c+d x)} \\ & = -\frac {3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}+\frac {\sqrt [3]{b \coth ^4(c+d x)} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\sqrt [3]{b \coth ^4(c+d x)} \text {Subst}\left (\int \frac {1-\frac {x}{2}}{1-x+x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\sqrt [3]{b \coth ^4(c+d x)} \text {Subst}\left (\int \frac {1+\frac {x}{2}}{1+x+x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac {4}{3}}(c+d x)} \\ & = \frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{d \coth ^{\frac {4}{3}}(c+d x)}-\frac {3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}-\frac {\sqrt [3]{b \coth ^4(c+d x)} \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\sqrt [3]{b \coth ^4(c+d x)} \text {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\left (3 \sqrt [3]{b \coth ^4(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\left (3 \sqrt [3]{b \coth ^4(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)} \\ & = \frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{d \coth ^{\frac {4}{3}}(c+d x)}-\frac {\sqrt [3]{b \coth ^4(c+d x)} \log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\sqrt [3]{b \coth ^4(c+d x)} \log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)}-\frac {3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}-\frac {\left (3 \sqrt [3]{b \coth ^4(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \sqrt [3]{\coth (c+d x)}\right )}{2 d \coth ^{\frac {4}{3}}(c+d x)}-\frac {\left (3 \sqrt [3]{b \coth ^4(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{\coth (c+d x)}\right )}{2 d \coth ^{\frac {4}{3}}(c+d x)} \\ & = -\frac {\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{2 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{2 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{d \coth ^{\frac {4}{3}}(c+d x)}-\frac {\sqrt [3]{b \coth ^4(c+d x)} \log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)}+\frac {\sqrt [3]{b \coth ^4(c+d x)} \log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \coth ^{\frac {4}{3}}(c+d x)}-\frac {3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.69 \[ \int \sqrt [3]{b \coth ^4(c+d x)} \, dx=-\frac {\sqrt [3]{b \coth ^4(c+d x)} \left (6 \sqrt [6]{\coth ^2(c+d x)}+\log \left (1-\sqrt [6]{\coth ^2(c+d x)}\right )-\log \left (1+\sqrt [6]{\coth ^2(c+d x)}\right )-(-1)^{2/3} \log \left (1-\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)}\right )+(-1)^{2/3} \log \left (1+\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)}\right )-\sqrt [3]{-1} \log \left (1-(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)}\right )+\sqrt [3]{-1} \log \left (1+(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)}\right )\right ) \tanh (c+d x)}{2 d \sqrt [6]{\coth ^2(c+d x)}} \]

[In]

Integrate[(b*Coth[c + d*x]^4)^(1/3),x]

[Out]

-1/2*((b*Coth[c + d*x]^4)^(1/3)*(6*(Coth[c + d*x]^2)^(1/6) + Log[1 - (Coth[c + d*x]^2)^(1/6)] - Log[1 + (Coth[
c + d*x]^2)^(1/6)] - (-1)^(2/3)*Log[1 - (-1)^(1/3)*(Coth[c + d*x]^2)^(1/6)] + (-1)^(2/3)*Log[1 + (-1)^(1/3)*(C
oth[c + d*x]^2)^(1/6)] - (-1)^(1/3)*Log[1 - (-1)^(2/3)*(Coth[c + d*x]^2)^(1/6)] + (-1)^(1/3)*Log[1 + (-1)^(2/3
)*(Coth[c + d*x]^2)^(1/6)])*Tanh[c + d*x])/(d*(Coth[c + d*x]^2)^(1/6))

Maple [F]

\[\int \left (b \coth \left (d x +c \right )^{4}\right )^{\frac {1}{3}}d x\]

[In]

int((b*coth(d*x+c)^4)^(1/3),x)

[Out]

int((b*coth(d*x+c)^4)^(1/3),x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.00 \[ \int \sqrt [3]{b \coth ^4(c+d x)} \, dx=-\frac {2 \, \sqrt {3} \left (-b\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} b + 2 \, \sqrt {3} \left (-b\right )^{\frac {2}{3}} \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {1}{3}}}{3 \, b}\right ) - 2 \, \sqrt {3} b^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} b - 2 \, \sqrt {3} b^{\frac {2}{3}} \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {1}{3}}}{3 \, b}\right ) + \left (-b\right )^{\frac {1}{3}} \log \left (\left (-b\right )^{\frac {2}{3}} - \left (-b\right )^{\frac {1}{3}} \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {1}{3}} + \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {2}{3}}\right ) + b^{\frac {1}{3}} \log \left (b^{\frac {2}{3}} - b^{\frac {1}{3}} \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {1}{3}} + \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-b\right )^{\frac {1}{3}} \log \left (\left (-b\right )^{\frac {1}{3}} + \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {1}{3}}\right ) - 2 \, b^{\frac {1}{3}} \log \left (b^{\frac {1}{3}} + \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {1}{3}}\right ) + 12 \, \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {1}{3}}}{4 \, d} \]

[In]

integrate((b*coth(d*x+c)^4)^(1/3),x, algorithm="fricas")

[Out]

-1/4*(2*sqrt(3)*(-b)^(1/3)*arctan(1/3*(sqrt(3)*b + 2*sqrt(3)*(-b)^(2/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))
/b) - 2*sqrt(3)*b^(1/3)*arctan(-1/3*(sqrt(3)*b - 2*sqrt(3)*b^(2/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))/b) +
(-b)^(1/3)*log((-b)^(2/3) - (-b)^(1/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3) + (b*cosh(d*x + c)/sinh(d*x + c)
)^(2/3)) + b^(1/3)*log(b^(2/3) - b^(1/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3) + (b*cosh(d*x + c)/sinh(d*x + c
))^(2/3)) - 2*(-b)^(1/3)*log((-b)^(1/3) + (b*cosh(d*x + c)/sinh(d*x + c))^(1/3)) - 2*b^(1/3)*log(b^(1/3) + (b*
cosh(d*x + c)/sinh(d*x + c))^(1/3)) + 12*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))/d

Sympy [F]

\[ \int \sqrt [3]{b \coth ^4(c+d x)} \, dx=\int \sqrt [3]{b \coth ^{4}{\left (c + d x \right )}}\, dx \]

[In]

integrate((b*coth(d*x+c)**4)**(1/3),x)

[Out]

Integral((b*coth(c + d*x)**4)**(1/3), x)

Maxima [F]

\[ \int \sqrt [3]{b \coth ^4(c+d x)} \, dx=\int { \left (b \coth \left (d x + c\right )^{4}\right )^{\frac {1}{3}} \,d x } \]

[In]

integrate((b*coth(d*x+c)^4)^(1/3),x, algorithm="maxima")

[Out]

integrate((b*coth(d*x + c)^4)^(1/3), x)

Giac [F]

\[ \int \sqrt [3]{b \coth ^4(c+d x)} \, dx=\int { \left (b \coth \left (d x + c\right )^{4}\right )^{\frac {1}{3}} \,d x } \]

[In]

integrate((b*coth(d*x+c)^4)^(1/3),x, algorithm="giac")

[Out]

integrate((b*coth(d*x + c)^4)^(1/3), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt [3]{b \coth ^4(c+d x)} \, dx=\int {\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^4\right )}^{1/3} \,d x \]

[In]

int((b*coth(c + d*x)^4)^(1/3),x)

[Out]

int((b*coth(c + d*x)^4)^(1/3), x)