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

Optimal. Leaf size=31 \[ \frac {\tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \log (\sinh (c+d x))}{d} \]

[Out]

(b*coth(d*x+c)^3)^(1/3)*ln(sinh(d*x+c))*tanh(d*x+c)/d

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3658, 3475} \[ \frac {\tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \log (\sinh (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x]^3)^(1/3),x]

[Out]

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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

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Mathematica [A]  time = 0.03, size = 39, normalized size = 1.26 \[ \frac {\tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} (\log (\tanh (c+d x))+\log (\cosh (c+d x)))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.57, size = 148, normalized size = 4.77 \[ -\frac {{\left (d x e^{\left (2 \, d x + 2 \, c\right )} - d x - {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )\right )} \left (\frac {b e^{\left (6 \, d x + 6 \, c\right )} + 3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (6 \, d x + 6 \, c\right )} - 3 \, e^{\left (4 \, d x + 4 \, c\right )} + 3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {1}{3}}}{d e^{\left (2 \, d x + 2 \, c\right )} + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*x*e^(2*d*x + 2*c) - d*x - (e^(2*d*x + 2*c) - 1)*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))))*((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*e^(2*d*x + 2*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.38, size = 192, normalized size = 6.19 \[ \frac {\left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {1}{3}} \left ({\mathrm e}^{2 d x +2 c}-1\right ) x}{1+{\mathrm e}^{2 d x +2 c}}-\frac {2 \left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {1}{3}} \left ({\mathrm e}^{2 d x +2 c}-1\right ) \left (d x +c \right )}{\left (1+{\mathrm e}^{2 d x +2 c}\right ) d}+\frac {\left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {1}{3}} \left ({\mathrm e}^{2 d x +2 c}-1\right ) \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{\left (1+{\mathrm e}^{2 d x +2 c}\right ) d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 51, normalized size = 1.65 \[ \frac {{\left (d x + c\right )} b^{\frac {1}{3}}}{d} + \frac {b^{\frac {1}{3}} \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {b^{\frac {1}{3}} \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^3\right )}^{1/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(c + d*x)^3)^(1/3),x)

[Out]

int((b*coth(c + d*x)^3)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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