3.34 \(\int (b \coth ^3(c+d x))^{2/3} \, dx\)

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

[Out]

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

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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 = {3658, 3473, 8} \[ x \tanh ^2(c+d x) \left (b \coth ^3(c+d x)\right )^{2/3}-\frac {\tanh (c+d x) \left (b \coth ^3(c+d x)\right )^{2/3}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x]^3)^(2/3),x]

[Out]

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

Rule 8

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

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

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Mathematica [C]  time = 0.03, size = 41, normalized size = 0.82 \[ -\frac {\tanh (c+d x) \left (b \coth ^3(c+d x)\right )^{2/3} \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\tanh ^2(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.66, size = 392, normalized size = 7.84 \[ \frac {{\left (d x \cosh \left (d x + c\right )^{2} + {\left (d x e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d x e^{\left (2 \, d x + 2 \, c\right )} + d x\right )} \sinh \left (d x + c\right )^{2} - d x + {\left (d x \cosh \left (d x + c\right )^{2} - d x - 2\right )} e^{\left (4 \, d x + 4 \, c\right )} - 2 \, {\left (d x \cosh \left (d x + c\right )^{2} - d x - 2\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (d x \cosh \left (d x + c\right ) e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d x \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} + d x \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 2\right )} \left (\frac {b e^{\left (6 \, d x + 6 \, c\right )} + 3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (6 \, d x + 6 \, c\right )} - 3 \, e^{\left (4 \, d x + 4 \, c\right )} + 3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {2}{3}}}{d \cosh \left (d x + c\right )^{2} + {\left (d e^{\left (4 \, d x + 4 \, c\right )} + 2 \, d e^{\left (2 \, d x + 2 \, c\right )} + d\right )} \sinh \left (d x + c\right )^{2} + {\left (d \cosh \left (d x + c\right )^{2} - d\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (d \cosh \left (d x + c\right )^{2} - d\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (d \cosh \left (d x + c\right ) e^{\left (4 \, d x + 4 \, c\right )} + 2 \, d \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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maple [B]  time = 0.40, size = 119, normalized size = 2.38 \[ \frac {\left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {2}{3}} \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} x}{\left (1+{\mathrm e}^{2 d x +2 c}\right )^{2}}-\frac {2 \left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {2}{3}} \left ({\mathrm e}^{2 d x +2 c}-1\right )}{\left (1+{\mathrm e}^{2 d x +2 c}\right )^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 34, normalized size = 0.68 \[ \frac {{\left (d x + c\right )} b^{\frac {2}{3}}}{d} + \frac {2 \, b^{\frac {2}{3}}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(c + d*x)^3)^(2/3),x)

[Out]

int((b*coth(c + d*x)^3)^(2/3), x)

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sympy [A]  time = 23.76, size = 90, normalized size = 1.80 \[ \begin {cases} \tilde {\infty } b^{\frac {2}{3}} x & \text {for}\: c = \log {\left (- e^{- d x} \right )} \vee c = \log {\left (e^{- d x} \right )} \\x \left (b \coth ^{3}{\relax (c )}\right )^{\frac {2}{3}} & \text {for}\: d = 0 \\b^{\frac {2}{3}} x \left (\frac {1}{\tanh ^{3}{\left (c + d x \right )}}\right )^{\frac {2}{3}} \tanh ^{2}{\left (c + d x \right )} - \frac {b^{\frac {2}{3}} \left (\frac {1}{\tanh ^{3}{\left (c + d x \right )}}\right )^{\frac {2}{3}} \tanh {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*b**(2/3)*x, Eq(c, log(exp(-d*x))) | Eq(c, log(-exp(-d*x)))), (x*(b*coth(c)**3)**(2/3), Eq(d, 0)
), (b**(2/3)*x*(tanh(c + d*x)**(-3))**(2/3)*tanh(c + d*x)**2 - b**(2/3)*(tanh(c + d*x)**(-3))**(2/3)*tanh(c +
d*x)/d, True))

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