3.1.24 \(\int \sqrt [3]{b \coth ^2(c+d x)} \, dx\) [24]

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

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 264, normalized size of antiderivative = 1.00, 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 = {3739, 3557, 335, 302, 648, 632, 210, 642, 212} \begin {gather*} \frac {\sqrt {3} \text {ArcTan}\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} \text {ArcTan}\left (\frac {2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt {3}}\right ) \sqrt [3]{b \coth ^2(c+d x)}}{2 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]{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)} \tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac {2}{3}}(c+d x)} \end {gather*}

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 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 302

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)/(a*n*s^m))*Int[1/(r^2 - s^2*x^2), x] + 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 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*} \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)} \text {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 ) \text {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)} \text {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)} \text {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)} \text {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)} \text {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)} \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.07, size = 151, normalized size = 0.57 \begin {gather*} \frac {\sqrt [3]{b \coth ^2(c+d x)} \left (2 \sqrt {3} \text {ArcTan}\left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right )-2 \sqrt {3} \text {ArcTan}\left (\frac {1+2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right )+4 \tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right )-\log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )+\log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )\right )}{4 d \coth ^{\frac {2}{3}}(c+d x)} \end {gather*}

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 = 1.76, size = 0, normalized size = 0.00 \[\int \left (b \left (\coth ^{2}\left (d x +c \right )\right )\right )^{\frac {1}{3}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 1618 vs. \(2 (216) = 432\).
time = 0.40, size = 1618, normalized size = 6.13 \begin {gather*} \text {Too large to display} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [3]{b \coth ^{2}{\left (c + d x \right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^2\right )}^{1/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(c + d*x)^2)^(1/3),x)

[Out]

int((b*coth(c + d*x)^2)^(1/3), x)

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