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

Optimal result

Integrand size = 14, antiderivative size = 31 \[ \int \frac {1}{\sqrt [3]{b \coth ^3(c+d x)}} \, dx=\frac {\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt [3]{b \coth ^3(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3739, 3556} \[ \int \frac {1}{\sqrt [3]{b \coth ^3(c+d x)}} \, dx=\frac {\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt [3]{b \coth ^3(c+d x)}} \]

[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*} \text {integral}& = \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] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt [3]{b \coth ^3(c+d x)}} \, dx=\frac {\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt [3]{b \coth ^3(c+d x)}} \]

[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))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(29)=58\).

Time = 0.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 6.19

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (29) = 58\).

Time = 0.25 (sec) , antiderivative size = 187, normalized size of antiderivative = 6.03 \[ \int \frac {1}{\sqrt [3]{b \coth ^3(c+d x)}} \, dx=-\frac {{\left (d x e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d x e^{\left (2 \, d x + 2 \, c\right )} + d x - {\left (e^{\left (4 \, d x + 4 \, c\right )} - 2 \, e^{\left (2 \, d x + 2 \, c\right )} + 1\right )} \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )\right )} \left (\frac {b e^{\left (6 \, d x + 6 \, c\right )} + 3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (6 \, d x + 6 \, c\right )} - 3 \, e^{\left (4 \, d x + 4 \, c\right )} + 3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {2}{3}}}{b d e^{\left (4 \, d x + 4 \, c\right )} + 2 \, b d e^{\left (2 \, d x + 2 \, c\right )} + b d} \]

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\sqrt [3]{b \coth ^3(c+d x)}} \, dx=\frac {d x + c}{b^{\frac {1}{3}} d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{b^{\frac {1}{3}} d} \]

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [3]{b \coth ^3(c+d x)}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^3\right )}^{1/3}} \,d x \]

[In]

int(1/(b*coth(c + d*x)^3)^(1/3),x)

[Out]

int(1/(b*coth(c + d*x)^3)^(1/3), x)