3.33 \(\int (b \coth ^3(c+d x))^{4/3} \, dx\)

Optimal. Leaf size=74 \[ -\frac {b \sqrt [3]{b \coth ^3(c+d x)}}{d}-\frac {b \coth ^2(c+d x) \sqrt [3]{b \coth ^3(c+d x)}}{3 d}+b x \tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3658, 3473, 8} \[ -\frac {b \coth ^2(c+d x) \sqrt [3]{b \coth ^3(c+d x)}}{3 d}-\frac {b \sqrt [3]{b \coth ^3(c+d x)}}{d}+b x \tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x]^3)^(4/3),x]

[Out]

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

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

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Mathematica [C]  time = 0.07, size = 43, normalized size = 0.58 \[ -\frac {\tanh (c+d x) \left (b \coth ^3(c+d x)\right )^{4/3} \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\tanh ^2(c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.45, size = 1046, normalized size = 14.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*b*d*x*cosh(d*x + c)^6 - 3*(b*d*x*e^(2*d*x + 2*c) - b*d*x)*sinh(d*x + c)^6 - 18*(b*d*x*cosh(d*x + c)*e^
(2*d*x + 2*c) - b*d*x*cosh(d*x + c))*sinh(d*x + c)^5 - 3*(3*b*d*x + 4*b)*cosh(d*x + c)^4 + 3*(15*b*d*x*cosh(d*
x + c)^2 - 3*b*d*x - (15*b*d*x*cosh(d*x + c)^2 - 3*b*d*x - 4*b)*e^(2*d*x + 2*c) - 4*b)*sinh(d*x + c)^4 + 12*(5
*b*d*x*cosh(d*x + c)^3 - (3*b*d*x + 4*b)*cosh(d*x + c) - (5*b*d*x*cosh(d*x + c)^3 - (3*b*d*x + 4*b)*cosh(d*x +
c))*e^(2*d*x + 2*c))*sinh(d*x + c)^3 - 3*b*d*x + 3*(3*b*d*x + 4*b)*cosh(d*x + c)^2 + 3*(15*b*d*x*cosh(d*x + c
)^4 + 3*b*d*x - 6*(3*b*d*x + 4*b)*cosh(d*x + c)^2 - (15*b*d*x*cosh(d*x + c)^4 + 3*b*d*x - 6*(3*b*d*x + 4*b)*co
sh(d*x + c)^2 + 4*b)*e^(2*d*x + 2*c) + 4*b)*sinh(d*x + c)^2 - (3*b*d*x*cosh(d*x + c)^6 - 3*(3*b*d*x + 4*b)*cos
h(d*x + c)^4 - 3*b*d*x + 3*(3*b*d*x + 4*b)*cosh(d*x + c)^2 - 8*b)*e^(2*d*x + 2*c) + 6*(3*b*d*x*cosh(d*x + c)^5
- 2*(3*b*d*x + 4*b)*cosh(d*x + c)^3 + (3*b*d*x + 4*b)*cosh(d*x + c) - (3*b*d*x*cosh(d*x + c)^5 - 2*(3*b*d*x +
4*b)*cosh(d*x + c)^3 + (3*b*d*x + 4*b)*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c) - 8*b)*((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)/(d*cosh(d*x + c)^6 + (d*e^(2*d*x + 2*c) + d)*sinh(d*x + c)^6 + 6*(d*cosh(d*x + c)*e^(2*d*x + 2*c) +
d*cosh(d*x + c))*sinh(d*x + c)^5 - 3*d*cosh(d*x + c)^4 + 3*(5*d*cosh(d*x + c)^2 + (5*d*cosh(d*x + c)^2 - d)*e
^(2*d*x + 2*c) - d)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*x + c)^3 - 3*d*cosh(d*x + c) + (5*d*cosh(d*x + c)^3 - 3*d*
cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c)^3 + 3*d*cosh(d*x + c)^2 + 3*(5*d*cosh(d*x + c)^4 - 6*d*cosh(d*x
+ c)^2 + (5*d*cosh(d*x + c)^4 - 6*d*cosh(d*x + c)^2 + d)*e^(2*d*x + 2*c) + d)*sinh(d*x + c)^2 + (d*cosh(d*x +
c)^6 - 3*d*cosh(d*x + c)^4 + 3*d*cosh(d*x + c)^2 - d)*e^(2*d*x + 2*c) + 6*(d*cosh(d*x + c)^5 - 2*d*cosh(d*x +
c)^3 + d*cosh(d*x + c) + (d*cosh(d*x + c)^5 - 2*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x
+ c) - d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.40, size = 145, normalized size = 1.96 \[ \frac {b \left ({\mathrm e}^{2 d x +2 c}-1\right ) \left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {1}{3}} x}{1+{\mathrm e}^{2 d x +2 c}}-\frac {4 b \left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {1}{3}} \left (3 \,{\mathrm e}^{4 d x +4 c}-3 \,{\mathrm e}^{2 d x +2 c}+2\right )}{3 \left (1+{\mathrm e}^{2 d x +2 c}\right ) \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b/(1+exp(2*d*x+2*c))*(exp(2*d*x+2*c)-1)*(b*(1+exp(2*d*x+2*c))^3/(exp(2*d*x+2*c)-1)^3)^(1/3)*x-4/3*b/(1+exp(2*d
*x+2*c))/(exp(2*d*x+2*c)-1)^2*(b*(1+exp(2*d*x+2*c))^3/(exp(2*d*x+2*c)-1)^3)^(1/3)*(3*exp(4*d*x+4*c)-3*exp(2*d*
x+2*c)+2)/d

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maxima [A]  time = 0.42, size = 87, normalized size = 1.18 \[ \frac {{\left (d x + c\right )} b^{\frac {4}{3}}}{d} - \frac {4 \, {\left (3 \, b^{\frac {4}{3}} e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, b^{\frac {4}{3}} e^{\left (-4 \, d x - 4 \, c\right )} - 2 \, b^{\frac {4}{3}}\right )}}{3 \, d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^3\right )}^{4/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(c + d*x)^3)^(4/3),x)

[Out]

int((b*coth(c + d*x)^3)^(4/3), x)

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sympy [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((b*coth(d*x+c)**3)**(4/3),x)

[Out]

Timed out

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