3.46 \(\int \sqrt [3]{b \coth ^4(c+d x)} \, dx\)

Optimal. Leaf size=289 \[ -\frac{\sqrt [3]{b \coth ^4(c+d x)} \log \left (\coth ^{\frac{2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \coth ^{\frac{4}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^4(c+d x)} \log \left (\coth ^{\frac{2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \coth ^{\frac{4}{3}}(c+d x)}-\frac{\sqrt{3} \sqrt [3]{b \coth ^4(c+d x)} \tan ^{-1}\left (\frac{1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt{3}}\right )}{2 d \coth ^{\frac{4}{3}}(c+d x)}+\frac{\sqrt{3} \sqrt [3]{b \coth ^4(c+d x)} \tan ^{-1}\left (\frac{2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt{3}}\right )}{2 d \coth ^{\frac{4}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^4(c+d x)} \tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac{4}{3}}(c+d x)}-\frac{3 \tanh (c+d x) \sqrt [3]{b \coth ^4(c+d x)}}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.175738, antiderivative size = 289, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {3658, 3473, 3476, 329, 210, 634, 618, 204, 628, 206} \[ -\frac{\sqrt [3]{b \coth ^4(c+d x)} \log \left (\coth ^{\frac{2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \coth ^{\frac{4}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^4(c+d x)} \log \left (\coth ^{\frac{2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )}{4 d \coth ^{\frac{4}{3}}(c+d x)}-\frac{\sqrt{3} \sqrt [3]{b \coth ^4(c+d x)} \tan ^{-1}\left (\frac{1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt{3}}\right )}{2 d \coth ^{\frac{4}{3}}(c+d x)}+\frac{\sqrt{3} \sqrt [3]{b \coth ^4(c+d x)} \tan ^{-1}\left (\frac{2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt{3}}\right )}{2 d \coth ^{\frac{4}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^4(c+d x)} \tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac{4}{3}}(c+d x)}-\frac{3 \tanh (c+d x) \sqrt [3]{b \coth ^4(c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x]^4)^(1/3),x]

[Out]

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

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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

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 210

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*Cos[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; (2*r^2*Int[1/(r^2 - s^2*x^2), x])/(a*n) +
Dist[(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 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 ^4(c+d x)} \, dx &=\frac{\sqrt [3]{b \coth ^4(c+d x)} \int \coth ^{\frac{4}{3}}(c+d x) \, dx}{\coth ^{\frac{4}{3}}(c+d x)}\\ &=-\frac{3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}+\frac{\sqrt [3]{b \coth ^4(c+d x)} \int \frac{1}{\coth ^{\frac{2}{3}}(c+d x)} \, dx}{\coth ^{\frac{4}{3}}(c+d x)}\\ &=-\frac{3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}-\frac{\sqrt [3]{b \coth ^4(c+d x)} \operatorname{Subst}\left (\int \frac{1}{x^{2/3} \left (-1+x^2\right )} \, dx,x,\coth (c+d x)\right )}{d \coth ^{\frac{4}{3}}(c+d x)}\\ &=-\frac{3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}-\frac{\left (3 \sqrt [3]{b \coth ^4(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^6} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac{4}{3}}(c+d x)}\\ &=-\frac{3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}+\frac{\sqrt [3]{b \coth ^4(c+d x)} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac{4}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^4(c+d x)} \operatorname{Subst}\left (\int \frac{1-\frac{x}{2}}{1-x+x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac{4}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^4(c+d x)} \operatorname{Subst}\left (\int \frac{1+\frac{x}{2}}{1+x+x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac{4}{3}}(c+d x)}\\ &=\frac{\tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{d \coth ^{\frac{4}{3}}(c+d x)}-\frac{3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}-\frac{\sqrt [3]{b \coth ^4(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{4}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^4(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{4}{3}}(c+d x)}+\frac{\left (3 \sqrt [3]{b \coth ^4(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{4}{3}}(c+d x)}+\frac{\left (3 \sqrt [3]{b \coth ^4(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{4}{3}}(c+d x)}\\ &=\frac{\tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{d \coth ^{\frac{4}{3}}(c+d x)}-\frac{\sqrt [3]{b \coth ^4(c+d x)} \log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac{2}{3}}(c+d x)\right )}{4 d \coth ^{\frac{4}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^4(c+d x)} \log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac{2}{3}}(c+d x)\right )}{4 d \coth ^{\frac{4}{3}}(c+d x)}-\frac{3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}-\frac{\left (3 \sqrt [3]{b \coth ^4(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{4}{3}}(c+d x)}-\frac{\left (3 \sqrt [3]{b \coth ^4(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{4}{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 ^4(c+d x)}}{2 d \coth ^{\frac{4}{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 ^4(c+d x)}}{2 d \coth ^{\frac{4}{3}}(c+d x)}+\frac{\tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right ) \sqrt [3]{b \coth ^4(c+d x)}}{d \coth ^{\frac{4}{3}}(c+d x)}-\frac{\sqrt [3]{b \coth ^4(c+d x)} \log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac{2}{3}}(c+d x)\right )}{4 d \coth ^{\frac{4}{3}}(c+d x)}+\frac{\sqrt [3]{b \coth ^4(c+d x)} \log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac{2}{3}}(c+d x)\right )}{4 d \coth ^{\frac{4}{3}}(c+d x)}-\frac{3 \sqrt [3]{b \coth ^4(c+d x)} \tanh (c+d x)}{d}\\ \end{align*}

Mathematica [C]  time = 0.0356893, size = 43, normalized size = 0.15 \[ \frac{3 \tanh (c+d x) \sqrt [3]{b \coth ^4(c+d x)} \left (\, _2F_1\left (\frac{1}{6},1;\frac{7}{6};\coth ^2(c+d x)\right )-1\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Coth[c + d*x]^4)^(1/3),x]

[Out]

(3*(b*Coth[c + d*x]^4)^(1/3)*(-1 + Hypergeometric2F1[1/6, 1, 7/6, Coth[c + d*x]^2])*Tanh[c + d*x])/d

________________________________________________________________________________________

Maple [F]  time = 0.099, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{b \left ({\rm coth} \left (dx+c\right ) \right ) ^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \coth \left (d x + c\right )^{4}\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)^4)^(1/3),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.26736, size = 855, normalized size = 2.96 \begin{align*} -\frac{2 \, \sqrt{3} \left (-b\right )^{\frac{1}{3}} \arctan \left (\frac{\sqrt{3} b + 2 \, \sqrt{3} \left (-b\right )^{\frac{2}{3}} \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}}}{3 \, b}\right ) - 2 \, \sqrt{3} b^{\frac{1}{3}} \arctan \left (-\frac{\sqrt{3} b - 2 \, \sqrt{3} b^{\frac{2}{3}} \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}}}{3 \, b}\right ) + \left (-b\right )^{\frac{1}{3}} \log \left (\left (-b\right )^{\frac{2}{3}} - \left (-b\right )^{\frac{1}{3}} \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}} + \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{2}{3}}\right ) + b^{\frac{1}{3}} \log \left (b^{\frac{2}{3}} - b^{\frac{1}{3}} \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}} + \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{2}{3}}\right ) - 2 \, \left (-b\right )^{\frac{1}{3}} \log \left (\left (-b\right )^{\frac{1}{3}} + \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}}\right ) - 2 \, b^{\frac{1}{3}} \log \left (b^{\frac{1}{3}} + \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}}\right ) + 12 \, \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(3)*(-b)^(1/3)*arctan(1/3*(sqrt(3)*b + 2*sqrt(3)*(-b)^(2/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))
/b) - 2*sqrt(3)*b^(1/3)*arctan(-1/3*(sqrt(3)*b - 2*sqrt(3)*b^(2/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))/b) +
(-b)^(1/3)*log((-b)^(2/3) - (-b)^(1/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3) + (b*cosh(d*x + c)/sinh(d*x + c)
)^(2/3)) + b^(1/3)*log(b^(2/3) - b^(1/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3) + (b*cosh(d*x + c)/sinh(d*x + c
))^(2/3)) - 2*(-b)^(1/3)*log((-b)^(1/3) + (b*cosh(d*x + c)/sinh(d*x + c))^(1/3)) - 2*b^(1/3)*log(b^(1/3) + (b*
cosh(d*x + c)/sinh(d*x + c))^(1/3)) + 12*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))/d

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{b \coth ^{4}{\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)**4)**(1/3),x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \coth \left (d x + c\right )^{4}\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)^4)^(1/3),x, algorithm="giac")

[Out]

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