\(\int \frac {1}{(b \coth ^4(c+d x))^{2/3}} \, dx\) [48]

Optimal result

Integrand size = 14, antiderivative size = 291 \[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{2/3}} \, dx=-\frac {3 \coth (c+d x)}{5 d \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \coth ^{\frac {8}{3}}(c+d x)}{2 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \coth ^{\frac {8}{3}}(c+d x)}{2 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \coth ^{\frac {8}{3}}(c+d x)}{d \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {\coth ^{\frac {8}{3}}(c+d x) \log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\coth ^{\frac {8}{3}}(c+d x) \log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \left (b \coth ^4(c+d x)\right )^{2/3}} \]

[Out]

-3/5*coth(d*x+c)/d/(b*coth(d*x+c)^4)^(2/3)+arctanh(coth(d*x+c)^(1/3))*coth(d*x+c)^(8/3)/d/(b*coth(d*x+c)^4)^(2
/3)-1/4*coth(d*x+c)^(8/3)*ln(1-coth(d*x+c)^(1/3)+coth(d*x+c)^(2/3))/d/(b*coth(d*x+c)^4)^(2/3)+1/4*coth(d*x+c)^
(8/3)*ln(1+coth(d*x+c)^(1/3)+coth(d*x+c)^(2/3))/d/(b*coth(d*x+c)^4)^(2/3)-1/2*arctan(1/3*(1-2*coth(d*x+c)^(1/3
))*3^(1/2))*coth(d*x+c)^(8/3)*3^(1/2)/d/(b*coth(d*x+c)^4)^(2/3)+1/2*arctan(1/3*(1+2*coth(d*x+c)^(1/3))*3^(1/2)
)*coth(d*x+c)^(8/3)*3^(1/2)/d/(b*coth(d*x+c)^4)^(2/3)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 291, 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, 3555, 3557, 335, 216, 648, 632, 210, 642, 212} \[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{2/3}} \, dx=-\frac {\sqrt {3} \coth ^{\frac {8}{3}}(c+d x) \arctan \left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right )}{2 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\sqrt {3} \coth ^{\frac {8}{3}}(c+d x) \arctan \left (\frac {2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt {3}}\right )}{2 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\coth ^{\frac {8}{3}}(c+d x) \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )}{d \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {3 \coth (c+d x)}{5 d \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {\coth ^{\frac {8}{3}}(c+d x) \log \left (\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\coth ^{\frac {8}{3}}(c+d x) \log \left (\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \left (b \coth ^4(c+d x)\right )^{2/3}} \]

[In]

Int[(b*Coth[c + d*x]^4)^(-2/3),x]

[Out]

(-3*Coth[c + d*x])/(5*d*(b*Coth[c + d*x]^4)^(2/3)) - (Sqrt[3]*ArcTan[(1 - 2*Coth[c + d*x]^(1/3))/Sqrt[3]]*Coth
[c + d*x]^(8/3))/(2*d*(b*Coth[c + d*x]^4)^(2/3)) + (Sqrt[3]*ArcTan[(1 + 2*Coth[c + d*x]^(1/3))/Sqrt[3]]*Coth[c
+ d*x]^(8/3))/(2*d*(b*Coth[c + d*x]^4)^(2/3)) + (ArcTanh[Coth[c + d*x]^(1/3)]*Coth[c + d*x]^(8/3))/(d*(b*Coth
[c + d*x]^4)^(2/3)) - (Coth[c + d*x]^(8/3)*Log[1 - Coth[c + d*x]^(1/3) + Coth[c + d*x]^(2/3)])/(4*d*(b*Coth[c
+ d*x]^4)^(2/3)) + (Coth[c + d*x]^(8/3)*Log[1 + Coth[c + d*x]^(1/3) + Coth[c + d*x]^(2/3)])/(4*d*(b*Coth[c + d
*x]^4)^(2/3))

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 3555

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 {\coth ^{\frac {8}{3}}(c+d x) \int \frac {1}{\coth ^{\frac {8}{3}}(c+d x)} \, dx}{\left (b \coth ^4(c+d x)\right )^{2/3}} \\ & = -\frac {3 \coth (c+d x)}{5 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\coth ^{\frac {8}{3}}(c+d x) \int \frac {1}{\coth ^{\frac {2}{3}}(c+d x)} \, dx}{\left (b \coth ^4(c+d x)\right )^{2/3}} \\ & = -\frac {3 \coth (c+d x)}{5 d \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {\coth ^{\frac {8}{3}}(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 \left (b \coth ^4(c+d x)\right )^{2/3}} \\ & = -\frac {3 \coth (c+d x)}{5 d \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {\left (3 \coth ^{\frac {8}{3}}(c+d x)\right ) \text {Subst}\left (\int \frac {1}{-1+x^6} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{d \left (b \coth ^4(c+d x)\right )^{2/3}} \\ & = -\frac {3 \coth (c+d x)}{5 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\coth ^{\frac {8}{3}}(c+d x) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\coth ^{\frac {8}{3}}(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 \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\coth ^{\frac {8}{3}}(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 \left (b \coth ^4(c+d x)\right )^{2/3}} \\ & = -\frac {3 \coth (c+d x)}{5 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \coth ^{\frac {8}{3}}(c+d x)}{d \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {\coth ^{\frac {8}{3}}(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 \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\coth ^{\frac {8}{3}}(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 \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\left (3 \coth ^{\frac {8}{3}}(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 \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\left (3 \coth ^{\frac {8}{3}}(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 \left (b \coth ^4(c+d x)\right )^{2/3}} \\ & = -\frac {3 \coth (c+d x)}{5 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \coth ^{\frac {8}{3}}(c+d x)}{d \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {\coth ^{\frac {8}{3}}(c+d x) \log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\coth ^{\frac {8}{3}}(c+d x) \log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {\left (3 \coth ^{\frac {8}{3}}(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 \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {\left (3 \coth ^{\frac {8}{3}}(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 \left (b \coth ^4(c+d x)\right )^{2/3}} \\ & = -\frac {3 \coth (c+d x)}{5 d \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \coth ^{\frac {8}{3}}(c+d x)}{2 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \coth ^{\frac {8}{3}}(c+d x)}{2 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \coth ^{\frac {8}{3}}(c+d x)}{d \left (b \coth ^4(c+d x)\right )^{2/3}}-\frac {\coth ^{\frac {8}{3}}(c+d x) \log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \left (b \coth ^4(c+d x)\right )^{2/3}}+\frac {\coth ^{\frac {8}{3}}(c+d x) \log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \left (b \coth ^4(c+d x)\right )^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{2/3}} \, dx=-\frac {\coth (c+d x) \left (6+5 \coth ^2(c+d x)^{5/6} \log \left (1-\sqrt [6]{\coth ^2(c+d x)}\right )-5 \coth ^2(c+d x)^{5/6} \log \left (1+\sqrt [6]{\coth ^2(c+d x)}\right )-5 (-1)^{2/3} \coth ^2(c+d x)^{5/6} \log \left (1-\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)}\right )+5 (-1)^{2/3} \coth ^2(c+d x)^{5/6} \log \left (1+\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)}\right )-5 \sqrt [3]{-1} \coth ^2(c+d x)^{5/6} \log \left (1-(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)}\right )+5 \sqrt [3]{-1} \coth ^2(c+d x)^{5/6} \log \left (1+(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)}\right )\right )}{10 d \left (b \coth ^4(c+d x)\right )^{2/3}} \]

[In]

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

[Out]

-1/10*(Coth[c + d*x]*(6 + 5*(Coth[c + d*x]^2)^(5/6)*Log[1 - (Coth[c + d*x]^2)^(1/6)] - 5*(Coth[c + d*x]^2)^(5/
6)*Log[1 + (Coth[c + d*x]^2)^(1/6)] - 5*(-1)^(2/3)*(Coth[c + d*x]^2)^(5/6)*Log[1 - (-1)^(1/3)*(Coth[c + d*x]^2
)^(1/6)] + 5*(-1)^(2/3)*(Coth[c + d*x]^2)^(5/6)*Log[1 + (-1)^(1/3)*(Coth[c + d*x]^2)^(1/6)] - 5*(-1)^(1/3)*(Co
th[c + d*x]^2)^(5/6)*Log[1 - (-1)^(2/3)*(Coth[c + d*x]^2)^(1/6)] + 5*(-1)^(1/3)*(Coth[c + d*x]^2)^(5/6)*Log[1
+ (-1)^(2/3)*(Coth[c + d*x]^2)^(1/6)]))/(d*(b*Coth[c + d*x]^4)^(2/3))

Maple [F]

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

[In]

int(1/(b*coth(d*x+c)^4)^(2/3),x)

[Out]

int(1/(b*coth(d*x+c)^4)^(2/3),x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1159 vs. \(2 (239) = 478\).

Time = 0.28 (sec) , antiderivative size = 1159, normalized size of antiderivative = 3.98 \[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{2/3}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(b*coth(d*x+c)^4)^(2/3),x, algorithm="fricas")

[Out]

1/20*(10*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)
*sqrt(-(-b^2)^(1/3))*arctan(-1/3*(sqrt(3)*(-b^2)^(1/3)*b*sqrt(-(-b^2)^(1/3)) - 2*sqrt(3)*(-b^2)^(2/3)*(b*cosh(
d*x + c)/sinh(d*x + c))^(1/3)*sqrt(-(-b^2)^(1/3)))/b^2) + 10*sqrt(3)*(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*si
nh(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*c
osh(d*x + c)^3 + b*cosh(d*x + c))*sinh(d*x + c) + b)*(b^2)^(1/6)*arctan(-1/3*sqrt(3)*(b^2)^(1/6)*((b^2)^(1/3)*
b - 2*(b^2)^(2/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))/b^2) + 5*(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)^(2/3)*log(b*(b*cosh(d*x + c)/sinh(d*x + c))^(2/3) - (-b^2)^(1/3)*b
+ (-b^2)^(2/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3)) - 5*(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)^(2/3)*log(b*(b*cosh(d*x + c)/sinh(d*x + c))^(2/3) + (b^2)^(1/3)*b - (b^2)^(2
/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3)) - 10*(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))*s
inh(d*x + c) + 1)*(-b^2)^(2/3)*log(b*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3) - (-b^2)^(2/3)) + 10*(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)^(2/3)*log(b*(b*cosh(d*x + c)/sinh(d*x
+ c))^(1/3) + (b^2)^(2/3)) - 12*(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))*sin
h(d*x + c) + b)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))/(b^2*d*cosh(d*x + c)^4 + 4*b^2*d*cosh(d*x + c)*sinh(d*x
+ c)^3 + b^2*d*sinh(d*x + c)^4 + 2*b^2*d*cosh(d*x + c)^2 + b^2*d + 2*(3*b^2*d*cosh(d*x + c)^2 + b^2*d)*sinh(d
*x + c)^2 + 4*(b^2*d*cosh(d*x + c)^3 + b^2*d*cosh(d*x + c))*sinh(d*x + c))

Sympy [F]

\[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{2/3}} \, dx=\int \frac {1}{\left (b \coth ^{4}{\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]

[In]

integrate(1/(b*coth(d*x+c)**4)**(2/3),x)

[Out]

Integral((b*coth(c + d*x)**4)**(-2/3), x)

Maxima [F]

\[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{2/3}} \, dx=\int { \frac {1}{\left (b \coth \left (d x + c\right )^{4}\right )^{\frac {2}{3}}} \,d x } \]

[In]

integrate(1/(b*coth(d*x+c)^4)^(2/3),x, algorithm="maxima")

[Out]

integrate((b*coth(d*x + c)^4)^(-2/3), x)

Giac [F]

\[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{2/3}} \, dx=\int { \frac {1}{\left (b \coth \left (d x + c\right )^{4}\right )^{\frac {2}{3}}} \,d x } \]

[In]

integrate(1/(b*coth(d*x+c)^4)^(2/3),x, algorithm="giac")

[Out]

integrate((b*coth(d*x + c)^4)^(-2/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{2/3}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^4\right )}^{2/3}} \,d x \]

[In]

int(1/(b*coth(c + d*x)^4)^(2/3),x)

[Out]

int(1/(b*coth(c + d*x)^4)^(2/3), x)