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

Optimal. Leaf size=309 \[ -\frac {3 \tanh (c+d x)}{5 b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {\coth ^{\frac {2}{3}}(c+d x) \log \left (\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )}{4 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(c+d x) \log \left (\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )}{4 b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {\sqrt {3} \coth ^{\frac {2}{3}}(c+d x) \tan ^{-1}\left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right )}{2 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\sqrt {3} \coth ^{\frac {2}{3}}(c+d x) \tan ^{-1}\left (\frac {2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt {3}}\right )}{2 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(c+d x) \tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right )}{b d \sqrt [3]{b \coth ^2(c+d x)}} \]

[Out]

arctanh(coth(d*x+c)^(1/3))*coth(d*x+c)^(2/3)/b/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))/b/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))/b/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)
/b/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)/b/d/(b*
coth(d*x+c)^2)^(1/3)-3/5*tanh(d*x+c)/b/d/(b*coth(d*x+c)^2)^(1/3)

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Rubi [A]  time = 0.18, antiderivative size = 309, normalized size of antiderivative = 1.00, 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, 3474, 3476, 329, 210, 634, 618, 204, 628, 206} \[ -\frac {\coth ^{\frac {2}{3}}(c+d x) \log \left (\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )}{4 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(c+d x) \log \left (\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )}{4 b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {\sqrt {3} \coth ^{\frac {2}{3}}(c+d x) \tan ^{-1}\left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right )}{2 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\sqrt {3} \coth ^{\frac {2}{3}}(c+d x) \tan ^{-1}\left (\frac {2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt {3}}\right )}{2 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(c+d x) \tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right )}{b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {3 \tanh (c+d x)}{5 b d \sqrt [3]{b \coth ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x]^2)^(-4/3),x]

[Out]

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

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 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])

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 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 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 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 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 3474

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 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 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
]])

Rubi steps

\begin {align*} \int \frac {1}{\left (b \coth ^2(c+d x)\right )^{4/3}} \, dx &=\frac {\coth ^{\frac {2}{3}}(c+d x) \int \frac {1}{\coth ^{\frac {8}{3}}(c+d x)} \, dx}{b \sqrt [3]{b \coth ^2(c+d x)}}\\ &=-\frac {3 \tanh (c+d x)}{5 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(c+d x) \int \frac {1}{\coth ^{\frac {2}{3}}(c+d x)} \, dx}{b \sqrt [3]{b \coth ^2(c+d x)}}\\ &=-\frac {3 \tanh (c+d x)}{5 b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {\coth ^{\frac {2}{3}}(c+d x) \operatorname {Subst}\left (\int \frac {1}{x^{2/3} \left (-1+x^2\right )} \, dx,x,\coth (c+d x)\right )}{b d \sqrt [3]{b \coth ^2(c+d x)}}\\ &=-\frac {3 \tanh (c+d x)}{5 b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {\left (3 \coth ^{\frac {2}{3}}(c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^6} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{b d \sqrt [3]{b \coth ^2(c+d x)}}\\ &=-\frac {3 \tanh (c+d x)}{5 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(c+d x) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(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 )}{b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(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 )}{b d \sqrt [3]{b \coth ^2(c+d x)}}\\ &=\frac {\tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right ) \coth ^{\frac {2}{3}}(c+d x)}{b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {3 \tanh (c+d x)}{5 b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {\coth ^{\frac {2}{3}}(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 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\coth ^{\frac {2}{3}}(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 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\left (3 \coth ^{\frac {2}{3}}(c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{4 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\left (3 \coth ^{\frac {2}{3}}(c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{\coth (c+d x)}\right )}{4 b d \sqrt [3]{b \coth ^2(c+d x)}}\\ &=\frac {\tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right ) \coth ^{\frac {2}{3}}(c+d x)}{b 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 b 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 b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {3 \tanh (c+d x)}{5 b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {\left (3 \coth ^{\frac {2}{3}}(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 b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {\left (3 \coth ^{\frac {2}{3}}(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 b d \sqrt [3]{b \coth ^2(c+d x)}}\\ &=-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \coth ^{\frac {2}{3}}(c+d x)}{2 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \coth ^{\frac {2}{3}}(c+d x)}{2 b d \sqrt [3]{b \coth ^2(c+d x)}}+\frac {\tanh ^{-1}\left (\sqrt [3]{\coth (c+d x)}\right ) \coth ^{\frac {2}{3}}(c+d x)}{b 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 b 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 b d \sqrt [3]{b \coth ^2(c+d x)}}-\frac {3 \tanh (c+d x)}{5 b d \sqrt [3]{b \coth ^2(c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 43, normalized size = 0.14 \[ -\frac {3 \coth (c+d x) \, _2F_1\left (-\frac {5}{6},1;\frac {1}{6};\coth ^2(c+d x)\right )}{5 d \left (b \coth ^2(c+d x)\right )^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Coth[c + d*x]*Hypergeometric2F1[-5/6, 1, 1/6, Coth[c + d*x]^2])/(5*d*(b*Coth[c + d*x]^2)^(4/3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \coth \left (d x + c\right )^{2}\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \left (\coth ^{2}\left (d x +c \right )\right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \coth \left (d x + c\right )^{2}\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^2\right )}^{4/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*coth(c + d*x)^2)^(4/3),x)

[Out]

int(1/(b*coth(c + d*x)^2)^(4/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \coth ^{2}{\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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