\(\int \frac {1}{\sqrt [3]{b \coth ^4(c+d x)}} \, dx\) [47]

Optimal result

Integrand size = 14, antiderivative size = 236 \[ \int \frac {1}{\sqrt [3]{b \coth ^4(c+d x)}} \, dx=-\frac {3 \coth (c+d x)}{d \sqrt [3]{b \coth ^4(c+d x)}}+\frac {\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \coth ^{\frac {4}{3}}(c+d x)}{2 d \sqrt [3]{b \coth ^4(c+d x)}}-\frac {\sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \coth ^{\frac {4}{3}}(c+d x)}{2 d \sqrt [3]{b \coth ^4(c+d x)}}+\frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \coth ^{\frac {4}{3}}(c+d x)}{d \sqrt [3]{b \coth ^4(c+d x)}}+\frac {\text {arctanh}\left (\frac {\sqrt [3]{\coth (c+d x)}}{1+\coth ^{\frac {2}{3}}(c+d x)}\right ) \coth ^{\frac {4}{3}}(c+d x)}{2 d \sqrt [3]{b \coth ^4(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\sqrt [3]{b \coth ^4(c+d x)}} \, dx=-\frac {\coth (c+d x) \left (6+\sqrt [6]{\coth ^2(c+d x)} \log \left (1-\sqrt [6]{\coth ^2(c+d x)}\right )-\sqrt [6]{\coth ^2(c+d x)} \log \left (1+\sqrt [6]{\coth ^2(c+d x)}\right )+\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)} \log \left (1-\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)}\right )-\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)} \log \left (1+\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)}\right )+(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)} \log \left (1-(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)}\right )-(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)} \log \left (1+(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)}\right )\right )}{2 d \sqrt [3]{b \coth ^4(c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.78, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.143, Rules used = {3042, 4141, 3042, 3955, 3042, 3957, 25, 266, 825, 27, 219, 1142, 25, 1083, 217, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt [3]{b \tan \left (i c+i d x+\frac {\pi }{2}\right )^4}}dx\)

\(\Big \downarrow \) 4141

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \int \frac {1}{\left (-i \tan \left (i c+i d x+\frac {\pi }{2}\right )\right )^{4/3}}dx}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 3955

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (\int \coth ^{\frac {2}{3}}(c+d x)dx-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (-\frac {3}{d \sqrt [3]{\coth (c+d x)}}+\int \left (-i \tan \left (i c+i d x+\frac {\pi }{2}\right )\right )^{2/3}dx\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (-\frac {\int -\frac {\coth ^{\frac {2}{3}}(c+d x)}{1-\coth ^2(c+d x)}d\coth (c+d x)}{d}-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (\frac {\int \frac {\coth ^{\frac {2}{3}}(c+d x)}{1-\coth ^2(c+d x)}d\coth (c+d x)}{d}-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (\frac {3 \int \frac {\coth ^{\frac {4}{3}}(c+d x)}{1-\coth ^2(c+d x)}d\sqrt [3]{\coth (c+d x)}}{d}-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 825

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (\frac {3 \left (\frac {1}{3} \int \frac {1}{1-\coth ^{\frac {2}{3}}(c+d x)}d\sqrt [3]{\coth (c+d x)}+\frac {1}{3} \int -\frac {\sqrt [3]{\coth (c+d x)}+1}{2 \left (\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )}d\sqrt [3]{\coth (c+d x)}+\frac {1}{3} \int -\frac {1-\sqrt [3]{\coth (c+d x)}}{2 \left (\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )}d\sqrt [3]{\coth (c+d x)}\right )}{d}-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (\frac {3 \left (\frac {1}{3} \int \frac {1}{1-\coth ^{\frac {2}{3}}(c+d x)}d\sqrt [3]{\coth (c+d x)}-\frac {1}{6} \int \frac {\sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {1}{6} \int \frac {1-\sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )}{d}-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (\frac {3 \left (-\frac {1}{6} \int \frac {\sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {1}{6} \int \frac {1-\sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d}-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (\frac {3 \left (\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {1}{2} \int -\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {2 \sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {3}{2} \int \frac {1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d}-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (\frac {3 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 \sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {3}{2} \int \frac {1}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {2 \sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {3}{2} \int \frac {1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d}-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (\frac {3 \left (\frac {1}{6} \left (3 \int \frac {1}{-\coth ^{\frac {2}{3}}(c+d x)-3}d\left (2 \sqrt [3]{\coth (c+d x)}-1\right )+\frac {1}{2} \int \frac {1-2 \sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{6} \left (3 \int \frac {1}{-\coth ^{\frac {2}{3}}(c+d x)-3}d\left (2 \sqrt [3]{\coth (c+d x)}+1\right )+\frac {1}{2} \int \frac {2 \sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d}-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (\frac {3 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 \sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{\coth (c+d x)}-1}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {2 \sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt {3}}\right )\right )+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d}-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\coth ^{\frac {4}{3}}(c+d x) \left (\frac {3 \left (\frac {1}{6} \left (-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{\coth (c+d x)}-1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )\right )+\frac {1}{6} \left (\frac {1}{2} \log \left (\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt {3}}\right )\right )+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d}-\frac {3}{d \sqrt [3]{\coth (c+d x)}}\right )}{\sqrt [3]{b \coth ^4(c+d x)}}\)

Input:

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

Output:

(Coth[c + d*x]^(4/3)*(-3/(d*Coth[c + d*x]^(1/3)) + (3*(ArcTanh[Coth[c + d* 
x]^(1/3)]/3 + (-(Sqrt[3]*ArcTan[(-1 + 2*Coth[c + d*x]^(1/3))/Sqrt[3]]) - L 
og[1 - Coth[c + d*x]^(1/3) + Coth[c + d*x]^(2/3)]/2)/6 + (-(Sqrt[3]*ArcTan 
[(1 + 2*Coth[c + d*x]^(1/3))/Sqrt[3]]) + Log[1 + Coth[c + d*x]^(1/3) + Cot 
h[c + d*x]^(2/3)]/2)/6))/d))/(b*Coth[c + d*x]^4)^(1/3)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 825

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

rule 1083

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

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3955

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

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[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 4141

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff 
= FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ 
n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p]))   Int[ActivateTrig[u]*(Ta 
n[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]])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 585 vs. \(2 (196) = 392\).

Time = 0.18 (sec) , antiderivative size = 3316, normalized size of antiderivative = 14.05 \[ \int \frac {1}{\sqrt [3]{b \coth ^4(c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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