3.1.36 \(\int \frac {1}{\sqrt [3]{b \coth ^3(c+d x)}} \, dx\) [36]

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

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Rubi [A]
time = 0.01, 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 = {3739, 3556} \begin {gather*} \frac {\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt [3]{b \coth ^3(c+d x)}} \end {gather*}

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 3556

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

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}{\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*}

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Mathematica [A]
time = 0.02, size = 31, normalized size = 1.00 \begin {gather*} \frac {\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt [3]{b \coth ^3(c+d x)}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(29)=58\).
time = 2.60, size = 192, normalized size = 6.19

method	result	size
risch	\(\frac {\left (1+{\mathrm e}^{2 d x +2 c}\right ) x}{\left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {1}{3}} \left ({\mathrm e}^{2 d x +2 c}-1\right )}-\frac {2 \left (1+{\mathrm e}^{2 d x +2 c}\right ) \left (d x +c \right )}{\left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {1}{3}} \left ({\mathrm e}^{2 d x +2 c}-1\right ) d}+\frac {\left (1+{\mathrm e}^{2 d x +2 c}\right ) \ln \left (1+{\mathrm e}^{2 d x +2 c}\right )}{\left (\frac {b \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )^{\frac {1}{3}} \left ({\mathrm e}^{2 d x +2 c}-1\right ) d}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.50, size = 32, normalized size = 1.03 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (29) = 58\).
time = 0.38, size = 187, normalized size = 6.03 \begin {gather*} -\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 {gather*}

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

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 = 0.00, 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)^(1/3),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*coth(c + d*x)^3)^(1/3),x)

[Out]

int(1/(b*coth(c + d*x)^3)^(1/3), x)

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