\(\int \frac {1}{\sqrt [3]{b \coth ^2(c+d x)}} \, dx\) [25]

Optimal result

Integrand size = 14, antiderivative size = 264 \[ \int \frac {1}{\sqrt [3]{b \coth ^2(c+d x)}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \coth ^{\frac {2}{3}}(c+d x)}{2 d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \coth ^{\frac {2}{3}}(c+d x)}{2 d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \coth ^{\frac {2}{3}}(c+d x)}{d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {\coth ^{\frac {2}{3}}(c+d x) \log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(c+d x) \log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \sqrt [3]{b \coth ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3739, 3557, 335, 216, 648, 632, 210, 642, 212} \[ \int \frac {1}{\sqrt [3]{b \coth ^2(c+d x)}} \, dx=-\frac {\sqrt {3} \coth ^{\frac {2}{3}}(c+d x) \arctan \left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right )}{2 d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\sqrt {3} \coth ^{\frac {2}{3}}(c+d x) \arctan \left (\frac {2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt {3}}\right )}{2 d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(c+d x) \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )}{d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {\coth ^{\frac {2}{3}}(c+d x) \log \left (\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(c+d x) \log \left (\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \sqrt [3]{b \coth ^2(c+d x)}} \]

[In]

Int[(b*Coth[c + d*x]^2)^(-1/3),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\sqrt [3]{b \coth ^2(c+d x)}} \, dx=-\frac {\coth (c+d x) \left (\log \left (1-\sqrt [6]{\coth ^2(c+d x)}\right )-\log \left (1+\sqrt [6]{\coth ^2(c+d x)}\right )+\sqrt [3]{-1} \left (-\sqrt [3]{-1} \log \left (1-\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)}\right )+\sqrt [3]{-1} \log \left (1+\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)}\right )-\log \left (1-(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)}\right )+\log \left (1+(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)}\right )\right )\right )}{2 d \sqrt [6]{\coth ^2(c+d x)} \sqrt [3]{b \coth ^2(c+d x)}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1594 vs. \(2 (216) = 432\).

Time = 0.41 (sec) , antiderivative size = 8338, normalized size of antiderivative = 31.58 \[ \int \frac {1}{\sqrt [3]{b \coth ^2(c+d x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

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

[In]

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

[Out]

Integral((b*coth(c + d*x)**2)**(-1/3), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(b*coth(c + d*x)^2)^(1/3),x)

[Out]

int(1/(b*coth(c + d*x)^2)^(1/3), x)