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

Optimal. Leaf size=264 \[ -\frac{\sqrt [3]{b \coth ^2(c+d x)} \log \left (\coth ^{\frac{2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \coth ^{\frac{2}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^2(c+d x)} \log \left (\coth ^{\frac{2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \coth ^{\frac{2}{3}}(c+d x)}+\frac{\sqrt{3} \sqrt [3]{b \coth ^2(c+d x)} \tan ^{-1}\left (\frac{1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt{3}}\right )}{2 d \coth ^{\frac{2}{3}}(c+d x)}-\frac{\sqrt{3} \sqrt [3]{b \coth ^2(c+d x)} \tan ^{-1}\left (\frac{2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt{3}}\right )}{2 d \coth ^{\frac{2}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^2(c+d x)} \tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac{2}{3}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.211663, antiderivative size = 264, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {3658, 3476, 329, 296, 634, 618, 204, 628, 206} \[ -\frac{\sqrt [3]{b \coth ^2(c+d x)} \log \left (\coth ^{\frac{2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \coth ^{\frac{2}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^2(c+d x)} \log \left (\coth ^{\frac{2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \coth ^{\frac{2}{3}}(c+d x)}+\frac{\sqrt{3} \sqrt [3]{b \coth ^2(c+d x)} \tan ^{-1}\left (\frac{1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt{3}}\right )}{2 d \coth ^{\frac{2}{3}}(c+d x)}-\frac{\sqrt{3} \sqrt [3]{b \coth ^2(c+d x)} \tan ^{-1}\left (\frac{2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt{3}}\right )}{2 d \coth ^{\frac{2}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^2(c+d x)} \tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac{2}{3}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x]^2)^(1/3),x]

[Out]

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

Rule 3658

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
]])

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]

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 296

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)*Int[1/(r^2 - s^2*x^2), x])/(a*n*s^m) + Dist[(2*r^(m + 1))/(a*n*s^m), Sum[u, {k,
1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && NegQ[a/b]

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 618

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 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 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 206

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

Rubi steps

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

Mathematica [A]  time = 0.101159, size = 151, normalized size = 0.57 \[ \frac{\sqrt [3]{b \coth ^2(c+d x)} \left (-\log \left (\coth ^{\frac{2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )+\log \left (\coth ^{\frac{2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt{3}}\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt{3}}\right )+4 \tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{4 d \coth ^{\frac{2}{3}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.106, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{b \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \coth \left (d x + c\right )^{2}\right )^{\frac{1}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.08247, size = 4625, normalized size = 17.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(3)*(-b)^(1/3)*arctan(1/3*(sqrt(3)*b*cosh(d*x + c)^2 + 2*sqrt(3)*b*cosh(d*x + c)*sinh(d*x + c) + s
qrt(3)*b*sinh(d*x + c)^2 + 2*(sqrt(3)*cosh(d*x + c)^2 + 2*sqrt(3)*cosh(d*x + c)*sinh(d*x + c) + sqrt(3)*sinh(d
*x + c)^2 - sqrt(3))*(-b)^(2/3)*((b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + b)/(cosh(d*x + c)^2 + sinh(d*x + c)^
2 - 1))^(1/3) + sqrt(3)*b)/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)) - 2*
sqrt(3)*b^(1/3)*arctan(-1/3*(sqrt(3)*b*cosh(d*x + c)^2 + 2*sqrt(3)*b*cosh(d*x + c)*sinh(d*x + c) + sqrt(3)*b*s
inh(d*x + c)^2 - 2*(sqrt(3)*cosh(d*x + c)^2 + 2*sqrt(3)*cosh(d*x + c)*sinh(d*x + c) + sqrt(3)*sinh(d*x + c)^2
- sqrt(3))*b^(2/3)*((b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + b)/(cosh(d*x + c)^2 + sinh(d*x + c)^2 - 1))^(1/3)
+ sqrt(3)*b)/(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)*log(
((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) - (cosh(d*x + c)^4
+ 4*cosh(d*x + c)^3*sinh(d*x + c) + 6*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sin
h(d*x + c)^4 - 1)*(-b)^(1/3)*((b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + b)/(cosh(d*x + c)^2 + sinh(d*x + c)^2 -
1))^(1/3) + (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*cosh(d*x + c)
^2 + b*sinh(d*x + c)^2 + b)/(cosh(d*x + c)^2 + sinh(d*x + c)^2 - 1))^(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*(cosh(d
*x + c)^3 + cosh(d*x + c))*sinh(d*x + c) + 1)) + b^(1/3)*log(((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 + co
sh(d*x + c))*sinh(d*x + c) + 1)*b^(2/3) - (cosh(d*x + c)^4 + 4*cosh(d*x + c)^3*sinh(d*x + c) + 6*cosh(d*x + c)
^2*sinh(d*x + c)^2 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 - 1)*b^(1/3)*((b*cosh(d*x + c)^2 + b*si
nh(d*x + c)^2 + b)/(cosh(d*x + c)^2 + sinh(d*x + c)^2 - 1))^(1/3) + (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*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + b)/(cosh(d*x + c)^2 + sinh(d*x
+ c)^2 - 1))^(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*(cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c) + 1)) - 2*(-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)^(1/3) + (cosh(d*x + c
)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*((b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + b)/(cosh(
d*x + c)^2 + sinh(d*x + c)^2 - 1))^(1/3))/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 +
1)) - 2*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^(1/3) + (cosh(
d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*((b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + b)
/(cosh(d*x + c)^2 + sinh(d*x + c)^2 - 1))^(1/3))/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x +
c)^2 + 1)))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{b \coth ^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \coth \left (d x + c\right )^{2}\right )^{\frac{1}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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