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

Optimal. Leaf size=132 \[ -\frac {\log \left (b^{2/3}-(b \coth (c+d x))^{2/3}\right )}{2 \sqrt [3]{b} d}+\frac {\log \left (b^{2/3} (b \coth (c+d x))^{2/3}+b^{4/3}+(b \coth (c+d x))^{4/3}\right )}{4 \sqrt [3]{b} d}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {b^{2/3}+2 (b \coth (c+d x))^{2/3}}{\sqrt {3} b^{2/3}}\right )}{2 \sqrt [3]{b} d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3476, 329, 275, 200, 31, 634, 617, 204, 628} \[ -\frac {\log \left (b^{2/3}-(b \coth (c+d x))^{2/3}\right )}{2 \sqrt [3]{b} d}+\frac {\log \left (b^{2/3} (b \coth (c+d x))^{2/3}+b^{4/3}+(b \coth (c+d x))^{4/3}\right )}{4 \sqrt [3]{b} d}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {b^{2/3}+2 (b \coth (c+d x))^{2/3}}{\sqrt {3} b^{2/3}}\right )}{2 \sqrt [3]{b} d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x])^(-1/3),x]

[Out]

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

Rule 31

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

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 204

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

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[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 329

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 617

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

Rule 628

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 634

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 3476

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 98, normalized size = 0.74 \[ \frac {\sqrt [3]{\coth (c+d x)} \left (-2 \log \left (1-\coth ^{\frac {2}{3}}(c+d x)\right )+\log \left (\coth ^{\frac {4}{3}}(c+d x)+\coth ^{\frac {2}{3}}(c+d x)+1\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2 \coth ^{\frac {2}{3}}(c+d x)+1}{\sqrt {3}}\right )\right )}{4 d \sqrt [3]{b \coth (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Coth[c + d*x])^(-1/3),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 1598, normalized size = 12.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(3)*b*sqrt((-b)^(1/3)/b)*log((3*b*cosh(d*x + c)^4 + 12*b*cosh(d*x + c)*sinh(d*x + c)^3 + 3*b*sinh(d*
x + c)^4 + 2*b*cosh(d*x + c)^2 + 2*(9*b*cosh(d*x + c)^2 + b)*sinh(d*x + c)^2 + 3*(cosh(d*x + c)^4 + 4*cosh(d*x
+ c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 2*(3*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - 2*cosh(d*x + c)^2 + 4*(c
osh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c) + 1)*(-b)^(1/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(2/3) - sqrt(3)*
((cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 2*(3*cosh(d*x + c)^2 - 1)*sinh(d*x + c
)^2 - 2*cosh(d*x + c)^2 + 4*(cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c) + 1)*(-b)^(2/3)*(b*cosh(d*x + c)/s
inh(d*x + c))^(2/3) - (b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*b*cosh(d*
x + c)^2 + 2*(3*b*cosh(d*x + c)^2 - b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 - b*cosh(d*x + c))*sinh(d*x + c)
+ b)*(-b)^(1/3) - 2*(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x + c) + 6*b*cosh(d*x + c)^2*sinh(d*x + c
)^2 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - b)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))*sqrt((
-b)^(1/3)/b) + 4*(3*b*cosh(d*x + c)^3 + b*cosh(d*x + c))*sinh(d*x + c) + 3*b)/(cosh(d*x + c)^2 + 2*cosh(d*x +
c)*sinh(d*x + c) + sinh(d*x + c)^2)) - 2*(-b)^(2/3)*log(-(-b)^(2/3) + (b*cosh(d*x + c)/sinh(d*x + c))^(2/3)) +
(-b)^(2/3)*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)*sinh(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)))/(b*d), 1/4*(2*sqrt
(3)*b*sqrt(-(-b)^(1/3)/b)*arctan((2*sqrt(3)*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)
^4 + 2*(3*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - 2*cosh(d*x + c)^2 + 4*(cosh(d*x + c)^3 - cosh(d*x + c))*sinh(
d*x + c) + 1)*(-b)^(2/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(2/3)*sqrt(-(-b)^(1/3)/b) + sqrt(3)*(b*cosh(d*x + c)^
4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*b*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 - b)*
sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 - b*cosh(d*x + c))*sinh(d*x + c) + b)*(-b)^(1/3)*sqrt(-(-b)^(1/3)/b) -
4*sqrt(3)*(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x + c) + 6*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b*c
osh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - b)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3)*sqrt(-(-b)^(1/3)/b
))/(9*b*cosh(d*x + c)^4 + 36*b*cosh(d*x + c)*sinh(d*x + c)^3 + 9*b*sinh(d*x + c)^4 + 14*b*cosh(d*x + c)^2 + 2*
(27*b*cosh(d*x + c)^2 + 7*b)*sinh(d*x + c)^2 + 4*(9*b*cosh(d*x + c)^3 + 7*b*cosh(d*x + c))*sinh(d*x + c) + 9*b
)) - 2*(-b)^(2/3)*log(-(-b)^(2/3) + (b*cosh(d*x + c)/sinh(d*x + c))^(2/3)) + (-b)^(2/3)*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*c
osh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(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)))/(b*d)]

________________________________________________________________________________________

giac [B]  time = 0.64, size = 216, normalized size = 1.64 \[ \frac {b {\left (\frac {2 \, \sqrt {3} {\left | b \right |}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {2}{3}} + {\left | b \right |}^{\frac {2}{3}}\right )}}{3 \, {\left | b \right |}^{\frac {2}{3}}}\right )}{b^{2}} + \frac {{\left | b \right |}^{\frac {2}{3}} \log \left (\left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {2}{3}} {\left | b \right |}^{\frac {2}{3}} + {\left | b \right |}^{\frac {4}{3}} + \frac {{\left (b e^{\left (2 \, d x + 2 \, c\right )} + b\right )} \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {1}{3}}}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )}{b^{2}} - \frac {2 \, {\left | b \right |}^{\frac {2}{3}} \log \left ({\left | \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {2}{3}} - {\left | b \right |}^{\frac {2}{3}} \right |}\right )}{b^{2}}\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.09, size = 115, normalized size = 0.87 \[ -\frac {b \ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}-\left (b^{2}\right )^{\frac {1}{3}}\right )}{2 d \left (b^{2}\right )^{\frac {2}{3}}}+\frac {b \ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {4}{3}}+\left (b^{2}\right )^{\frac {1}{3}} \left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}+\left (b^{2}\right )^{\frac {2}{3}}\right )}{4 d \left (b^{2}\right )^{\frac {2}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}}{\left (b^{2}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d \left (b^{2}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \coth \left (d x + c\right )\right )^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.65, size = 147, normalized size = 1.11 \[ \frac {\ln \left (162\,{\left (-b\right )}^{11/3}+162\,b^3\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{2/3}\right )}{2\,{\left (-b\right )}^{1/3}\,d}+\frac {\ln \left (\frac {81\,{\left (-b\right )}^{11/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{d^3}+\frac {162\,b^3\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{2/3}}{d^3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,{\left (-b\right )}^{1/3}\,d}-\frac {\ln \left (\frac {81\,{\left (-b\right )}^{11/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{d^3}-\frac {162\,b^3\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{2/3}}{d^3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,{\left (-b\right )}^{1/3}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*coth(c + d*x))^(1/3),x)

[Out]

log(162*(-b)^(11/3) + 162*b^3*(b*coth(c + d*x))^(2/3))/(2*(-b)^(1/3)*d) + (log((81*(-b)^(11/3)*(3^(1/2)*1i - 1
))/d^3 + (162*b^3*(b*coth(c + d*x))^(2/3))/d^3)*(3^(1/2)*1i - 1))/(4*(-b)^(1/3)*d) - (log((81*(-b)^(11/3)*(3^(
1/2)*1i + 1))/d^3 - (162*b^3*(b*coth(c + d*x))^(2/3))/d^3)*(3^(1/2)*1i + 1))/(4*(-b)^(1/3)*d)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt [3]{b \coth {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________