3.36 \(\int \frac{1}{\sqrt [3]{b \coth ^3(c+d x)}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0202842, antiderivative size = 31, normalized size of antiderivative = 1., 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} \[ \frac{\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt [3]{b \coth ^3(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x]^3)^(-1/3),x]

[Out]

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

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
]])

Rule 3475

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt [3]{b \coth ^3(c+d x)}} \, dx &=\frac{\coth (c+d x) \int \tanh (c+d x) \, dx}{\sqrt [3]{b \coth ^3(c+d x)}}\\ &=\frac{\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt [3]{b \coth ^3(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0284736, size = 31, normalized size = 1. \[ \frac{\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt [3]{b \coth ^3(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.093, size = 192, normalized size = 6.2 \begin{align*}{\frac{ \left ( 1+{{\rm e}^{2\,dx+2\,c}} \right ) x}{{{\rm e}^{2\,dx+2\,c}}-1}{\frac{1}{\sqrt [3]{{\frac{b \left ( 1+{{\rm e}^{2\,dx+2\,c}} \right ) ^{3}}{ \left ({{\rm e}^{2\,dx+2\,c}}-1 \right ) ^{3}}}}}}}-2\,{\frac{ \left ( 1+{{\rm e}^{2\,dx+2\,c}} \right ) \left ( dx+c \right ) }{ \left ({{\rm e}^{2\,dx+2\,c}}-1 \right ) d}{\frac{1}{\sqrt [3]{{\frac{b \left ( 1+{{\rm e}^{2\,dx+2\,c}} \right ) ^{3}}{ \left ({{\rm e}^{2\,dx+2\,c}}-1 \right ) ^{3}}}}}}}+{\frac{ \left ( 1+{{\rm e}^{2\,dx+2\,c}} \right ) \ln \left ( 1+{{\rm e}^{2\,dx+2\,c}} \right ) }{ \left ({{\rm e}^{2\,dx+2\,c}}-1 \right ) d}{\frac{1}{\sqrt [3]{{\frac{b \left ( 1+{{\rm e}^{2\,dx+2\,c}} \right ) ^{3}}{ \left ({{\rm e}^{2\,dx+2\,c}}-1 \right ) ^{3}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.71109, size = 43, normalized size = 1.39 \begin{align*} \frac{d x + c}{b^{\frac{1}{3}} d} + \frac{\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{b^{\frac{1}{3}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.44438, size = 460, normalized size = 14.84 \begin{align*} -\frac{{\left (d x e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d x e^{\left (2 \, d x + 2 \, c\right )} + d x -{\left (e^{\left (4 \, d x + 4 \, c\right )} - 2 \, e^{\left (2 \, d x + 2 \, c\right )} + 1\right )} \log \left (\frac{2 \, \cosh \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{2}{3}}}{b d e^{\left (4 \, d x + 4 \, c\right )} + 2 \, b d e^{\left (2 \, d x + 2 \, c\right )} + b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \coth \left (d x + c\right )^{3}\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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