\(\int \sqrt [3]{b \coth (c+d x)} \, dx\) [11]

Optimal result

Integrand size = 12, antiderivative size = 132 \[ \int \sqrt [3]{b \coth (c+d x)} \, dx=-\frac {\sqrt {3} \sqrt [3]{b} \arctan \left (\frac {b^{2/3}+2 (b \coth (c+d x))^{2/3}}{\sqrt {3} b^{2/3}}\right )}{2 d}-\frac {\sqrt [3]{b} \log \left (b^{2/3}-(b \coth (c+d x))^{2/3}\right )}{2 d}+\frac {\sqrt [3]{b} \log \left (b^{4/3}+b^{2/3} (b \coth (c+d x))^{2/3}+(b \coth (c+d x))^{4/3}\right )}{4 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.82 \[ \int \sqrt [3]{b \coth (c+d x)} \, dx=-\frac {(b \coth (c+d x))^{4/3} \left (\log \left (1-\sqrt [3]{\coth ^2(c+d x)}\right )-\sqrt [3]{-1} \log \left (1+\sqrt [3]{-1} \sqrt [3]{\coth ^2(c+d x)}\right )+(-1)^{2/3} \log \left (1-(-1)^{2/3} \sqrt [3]{\coth ^2(c+d x)}\right )\right )}{2 b d \coth ^2(c+d x)^{2/3}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.84, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3042, 3957, 25, 266, 807, 821, 16, 1142, 1082, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 821

\(\displaystyle \frac {3 b \left (\frac {\int \frac {1}{b^{2/3}-b^2 \coth ^2(c+d x)}d\left (b^2 \coth ^2(c+d x)\right )}{3 b^{2/3}}-\frac {\int \frac {b^{2/3}-b^2 \coth ^2(c+d x)}{b^2 \coth ^2(c+d x)+b^{5/3} \coth (c+d x)+b^{4/3}}d\left (b^2 \coth ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {3 b \left (-\frac {\int \frac {b^{2/3}-b^2 \coth ^2(c+d x)}{b^2 \coth ^2(c+d x)+b^{5/3} \coth (c+d x)+b^{4/3}}d\left (b^2 \coth ^2(c+d x)\right )}{3 b^{2/3}}-\frac {\log \left (b^{2/3}-b^2 \coth ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {3 b \left (-\frac {\frac {3}{2} b^{2/3} \int \frac {1}{b^2 \coth ^2(c+d x)+b^{5/3} \coth (c+d x)+b^{4/3}}d\left (b^2 \coth ^2(c+d x)\right )-\frac {1}{2} \int \frac {2 b^2 \coth ^2(c+d x)+b^{2/3}}{b^2 \coth ^2(c+d x)+b^{5/3} \coth (c+d x)+b^{4/3}}d\left (b^2 \coth ^2(c+d x)\right )}{3 b^{2/3}}-\frac {\log \left (b^{2/3}-b^2 \coth ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {3 b \left (-\frac {-\frac {1}{2} \int \frac {2 b^2 \coth ^2(c+d x)+b^{2/3}}{b^2 \coth ^2(c+d x)+b^{5/3} \coth (c+d x)+b^{4/3}}d\left (b^2 \coth ^2(c+d x)\right )-3 \int \frac {1}{-2 \sqrt [3]{b} \coth (c+d x)-4}d\left (2 \sqrt [3]{b} \coth (c+d x)+1\right )}{3 b^{2/3}}-\frac {\log \left (b^{2/3}-b^2 \coth ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 b \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{b} \coth (c+d x)+1}{\sqrt {3}}\right )-\frac {1}{2} \int \frac {2 b^2 \coth ^2(c+d x)+b^{2/3}}{b^2 \coth ^2(c+d x)+b^{5/3} \coth (c+d x)+b^{4/3}}d\left (b^2 \coth ^2(c+d x)\right )}{3 b^{2/3}}-\frac {\log \left (b^{2/3}-b^2 \coth ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3 b \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{b} \coth (c+d x)+1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (b^{5/3} \coth (c+d x)+b^{4/3}+b^2 \coth ^2(c+d x)\right )}{3 b^{2/3}}-\frac {\log \left (b^{2/3}-b^2 \coth ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

Input:

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

Output:

(3*b*(-1/3*Log[b^(2/3) - b^2*Coth[c + d*x]^2]/b^(2/3) - (Sqrt[3]*ArcTan[(1 
 + 2*b^(1/3)*Coth[c + d*x])/Sqrt[3]] - Log[b^(4/3) + b^(5/3)*Coth[c + d*x] 
 + b^2*Coth[c + d*x]^2]/2)/(3*b^(2/3))))/(2*d)
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 821

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) 
 Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 
*x^2), x], x] /; FreeQ[{a, b}, x]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{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 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]
 

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {3 b \left (\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}-\left (b^{2}\right )^{\frac {1}{3}}\right )}{6 \left (b^{2}\right )^{\frac {1}{3}}}-\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {4}{3}}+\left (b^{2}\right )^{\frac {1}{3}} \left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}+\left (b^{2}\right )^{\frac {2}{3}}\right )}{12 \left (b^{2}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}}{\left (b^{2}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{6 \left (b^{2}\right )^{\frac {1}{3}}}\right )}{d}\)	\(109\)
default	\(-\frac {3 b \left (\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}-\left (b^{2}\right )^{\frac {1}{3}}\right )}{6 \left (b^{2}\right )^{\frac {1}{3}}}-\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {4}{3}}+\left (b^{2}\right )^{\frac {1}{3}} \left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}+\left (b^{2}\right )^{\frac {2}{3}}\right )}{12 \left (b^{2}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}}{\left (b^{2}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{6 \left (b^{2}\right )^{\frac {1}{3}}}\right )}{d}\)	\(109\)

Input:

int((b*coth(d*x+c))^(1/3),x,method=_RETURNVERBOSE)
 

Output:

-3/d*b*(1/6/(b^2)^(1/3)*ln((b*coth(d*x+c))^(2/3)-(b^2)^(1/3))-1/12/(b^2)^( 
1/3)*ln((b*coth(d*x+c))^(4/3)+(b^2)^(1/3)*(b*coth(d*x+c))^(2/3)+(b^2)^(2/3 
))+1/6*3^(1/2)/(b^2)^(1/3)*arctan(1/3*3^(1/2)*(2/(b^2)^(1/3)*(b*coth(d*x+c 
))^(2/3)+1)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (99) = 198\).

Time = 0.15 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.20 \[ \int \sqrt [3]{b \coth (c+d x)} \, dx=-\frac {2 \, \sqrt {3} \left (-b\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} b - 2 \, \sqrt {3} \left (-b\right )^{\frac {1}{3}} \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {2}{3}}}{3 \, b}\right ) - 2 \, \left (-b\right )^{\frac {1}{3}} \log \left (-\left (-b\right )^{\frac {2}{3}} + \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {2}{3}}\right ) + \left (-b\right )^{\frac {1}{3}} \log \left (\frac {{\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \left (-b\right )^{\frac {2}{3}} \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {2}{3}} - {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b\right )} \left (-b\right )^{\frac {1}{3}} + {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + b\right )} \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {1}{3}}}{\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1}\right )}{4 \, d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (99) = 198\).

Time = 0.21 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.64 \[ \int \sqrt [3]{b \coth (c+d x)} \, dx=-\frac {b {\left (\frac {2 \, \sqrt {3} {\left | b \right |}^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {2}{3}} + {\left | b \right |}^{\frac {2}{3}}\right )}}{3 \, {\left | b \right |}^{\frac {2}{3}}}\right )}{b^{2}} - \frac {{\left | b \right |}^{\frac {4}{3}} \log \left (\left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {2}{3}} {\left | b \right |}^{\frac {2}{3}} + {\left | b \right |}^{\frac {4}{3}} + \frac {{\left (b e^{\left (2 \, d x + 2 \, c\right )} + b\right )} \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {1}{3}}}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )}{b^{2}} + \frac {2 \, {\left | b \right |}^{\frac {4}{3}} \log \left ({\left | \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {2}{3}} - {\left | b \right |}^{\frac {2}{3}} \right |}\right )}{b^{2}}\right )}}{4 \, d} \] Input:

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

Output:

-1/4*b*(2*sqrt(3)*abs(b)^(4/3)*arctan(1/3*sqrt(3)*(2*((b*e^(2*d*x + 2*c) + 
 b)/(e^(2*d*x + 2*c) - 1))^(2/3) + abs(b)^(2/3))/abs(b)^(2/3))/b^2 - abs(b 
)^(4/3)*log(((b*e^(2*d*x + 2*c) + b)/(e^(2*d*x + 2*c) - 1))^(2/3)*abs(b)^( 
2/3) + abs(b)^(4/3) + (b*e^(2*d*x + 2*c) + b)*((b*e^(2*d*x + 2*c) + b)/(e^ 
(2*d*x + 2*c) - 1))^(1/3)/(e^(2*d*x + 2*c) - 1))/b^2 + 2*abs(b)^(4/3)*log( 
abs(((b*e^(2*d*x + 2*c) + b)/(e^(2*d*x + 2*c) - 1))^(2/3) - abs(b)^(2/3))) 
/b^2)/d
 

Mupad [B] (verification not implemented)

Time = 2.92 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.11 \[ \int \sqrt [3]{b \coth (c+d x)} \, dx=\frac {{\left (-b\right )}^{1/3}\,\ln \left (81\,{\left (-b\right )}^{16/3}\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{2/3}-81\,b^6\right )}{2\,d}-\frac {{\left (-b\right )}^{1/3}\,\ln \left (-\frac {81\,b^6}{d^4}-\frac {81\,{\left (-b\right )}^{16/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{2/3}}{d^4}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,d}+\frac {{\left (-b\right )}^{1/3}\,\ln \left (-\frac {81\,b^6}{d^4}+\frac {162\,{\left (-b\right )}^{16/3}\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{2/3}}{d^4}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{d} \] Input:

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

Output:

((-b)^(1/3)*log(81*(-b)^(16/3)*(b*coth(c + d*x))^(2/3) - 81*b^6))/(2*d) - 
((-b)^(1/3)*log(- (81*b^6)/d^4 - (81*(-b)^(16/3)*((3^(1/2)*1i)/2 + 1/2)*(b 
*coth(c + d*x))^(2/3))/d^4)*((3^(1/2)*1i)/2 + 1/2))/(2*d) + ((-b)^(1/3)*lo 
g((162*(-b)^(16/3)*((3^(1/2)*1i)/4 - 1/4)*(b*coth(c + d*x))^(2/3))/d^4 - ( 
81*b^6)/d^4)*((3^(1/2)*1i)/4 - 1/4))/d
 

Reduce [F]

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

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

Output:

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