3.1.37 \(\int \frac {1}{(b \coth ^3(c+d x))^{2/3}} \, dx\) [37]

Optimal. Leaf size=50 \[ -\frac {\coth (c+d x)}{d \left (b \coth ^3(c+d x)\right )^{2/3}}+\frac {x \coth ^2(c+d x)}{\left (b \coth ^3(c+d x)\right )^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 8} \begin {gather*} \frac {x \coth ^2(c+d x)}{\left (b \coth ^3(c+d x)\right )^{2/3}}-\frac {\coth (c+d x)}{d \left (b \coth ^3(c+d x)\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x]^3)^(-2/3),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3554

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 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 \frac {1}{\left (b \coth ^3(c+d x)\right )^{2/3}} \, dx &=\frac {\coth ^2(c+d x) \int \tanh ^2(c+d x) \, dx}{\left (b \coth ^3(c+d x)\right )^{2/3}}\\ &=-\frac {\coth (c+d x)}{d \left (b \coth ^3(c+d x)\right )^{2/3}}+\frac {\coth ^2(c+d x) \int 1 \, dx}{\left (b \coth ^3(c+d x)\right )^{2/3}}\\ &=-\frac {\coth (c+d x)}{d \left (b \coth ^3(c+d x)\right )^{2/3}}+\frac {x \coth ^2(c+d x)}{\left (b \coth ^3(c+d x)\right )^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 40, normalized size = 0.80 \begin {gather*} \frac {\coth (c+d x) \left (-1+\tanh ^{-1}(\tanh (c+d x)) \coth (c+d x)\right )}{d \left (b \coth ^3(c+d x)\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(118\) vs. \(2(46)=92\).
time = 2.59, size = 119, normalized size = 2.38

method	result	size
risch	\(\frac {\left (1+{\mathrm e}^{2 d x +2 c}\right )^{2} x}{\left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {2}{3}} \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}+\frac {2+2 \,{\mathrm e}^{2 d x +2 c}}{\left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {2}{3}} \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} d}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(b*(1+exp(2*d*x+2*c))^3/(exp(2*d*x+2*c)-1)^3)^(2/3)/(exp(2*d*x+2*c)-1)^2*(1+exp(2*d*x+2*c))^2*x+2/(b*(1+exp(
2*d*x+2*c))^3/(exp(2*d*x+2*c)-1)^3)^(2/3)/(exp(2*d*x+2*c)-1)^2*(1+exp(2*d*x+2*c))/d

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Maxima [A]
time = 0.48, size = 37, normalized size = 0.74 \begin {gather*} \frac {d x + c}{b^{\frac {2}{3}} d} - \frac {2}{{\left (b^{\frac {2}{3}} e^{\left (-2 \, d x - 2 \, c\right )} + b^{\frac {2}{3}}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x + c)/(b^(2/3)*d) - 2/((b^(2/3)*e^(-2*d*x - 2*c) + b^(2/3))*d)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (46) = 92\).
time = 0.35, size = 287, normalized size = 5.74 \begin {gather*} -\frac {{\left (d x \cosh \left (d x + c\right )^{2} - {\left (d x e^{\left (2 \, d x + 2 \, c\right )} - d x\right )} \sinh \left (d x + c\right )^{2} + d x - {\left (d x \cosh \left (d x + c\right )^{2} + d x + 2\right )} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, {\left (d x \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} - d x \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 2\right )} \left (\frac {b e^{\left (6 \, d x + 6 \, c\right )} + 3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (6 \, d x + 6 \, c\right )} - 3 \, e^{\left (4 \, d x + 4 \, c\right )} + 3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {1}{3}}}{b d \cosh \left (d x + c\right )^{2} + {\left (b d e^{\left (2 \, d x + 2 \, c\right )} + b d\right )} \sinh \left (d x + c\right )^{2} + b d + {\left (b d \cosh \left (d x + c\right )^{2} + b d\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (b d \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} + b d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*x*cosh(d*x + c)^2 - (d*x*e^(2*d*x + 2*c) - d*x)*sinh(d*x + c)^2 + d*x - (d*x*cosh(d*x + c)^2 + d*x + 2)*e^
(2*d*x + 2*c) - 2*(d*x*cosh(d*x + c)*e^(2*d*x + 2*c) - d*x*cosh(d*x + c))*sinh(d*x + c) + 2)*((b*e^(6*d*x + 6*
c) + 3*b*e^(4*d*x + 4*c) + 3*b*e^(2*d*x + 2*c) + b)/(e^(6*d*x + 6*c) - 3*e^(4*d*x + 4*c) + 3*e^(2*d*x + 2*c) -
1))^(1/3)/(b*d*cosh(d*x + c)^2 + (b*d*e^(2*d*x + 2*c) + b*d)*sinh(d*x + c)^2 + b*d + (b*d*cosh(d*x + c)^2 + b
*d)*e^(2*d*x + 2*c) + 2*(b*d*cosh(d*x + c)*e^(2*d*x + 2*c) + b*d*cosh(d*x + c))*sinh(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^3\right )}^{2/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*coth(c + d*x)^3)^(2/3),x)

[Out]

int(1/(b*coth(c + d*x)^3)^(2/3), x)

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