\(\int \frac {\coth ^3(x)}{1+\tanh (x)} \, dx\) [123]

Optimal result

Integrand size = 11, antiderivative size = 37 \[ \int \frac {\coth ^3(x)}{1+\tanh (x)} \, dx=-\frac {3 x}{2}+\frac {3 \coth (x)}{2}-\coth ^2(x)+2 \log (\sinh (x))+\frac {\coth ^2(x)}{2 (1+\tanh (x))} \]

[Out]

-3/2*x+3/2*coth(x)-coth(x)^2+2*ln(sinh(x))+1/2*coth(x)^2/(1+tanh(x))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3633, 3610, 3612, 3556} \[ \int \frac {\coth ^3(x)}{1+\tanh (x)} \, dx=-\frac {3 x}{2}-\coth ^2(x)+\frac {3 \coth (x)}{2}+2 \log (\sinh (x))+\frac {\coth ^2(x)}{2 (\tanh (x)+1)} \]

[In]

Int[Coth[x]^3/(1 + Tanh[x]),x]

[Out]

(-3*x)/2 + (3*Coth[x])/2 - Coth[x]^2 + 2*Log[Sinh[x]] + Coth[x]^2/(2*(1 + Tanh[x]))

Rule 3556

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

Rule 3610

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b
*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3612

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c +
b*d)*(x/(a^2 + b^2)), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3633

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-a
)*((c + d*Tan[e + f*x])^(n + 1)/(2*f*(b*c - a*d)*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a*(b*c - a*d)), Int[(c
+ d*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x
] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\coth ^2(x)}{2 (1+\tanh (x))}-\frac {1}{2} \int \coth ^3(x) (-4+3 \tanh (x)) \, dx \\ & = -\coth ^2(x)+\frac {\coth ^2(x)}{2 (1+\tanh (x))}-\frac {1}{2} i \int \coth ^2(x) (-3 i+4 i \tanh (x)) \, dx \\ & = \frac {3 \coth (x)}{2}-\coth ^2(x)+\frac {\coth ^2(x)}{2 (1+\tanh (x))}+\frac {1}{2} \int \coth (x) (4-3 \tanh (x)) \, dx \\ & = -\frac {3 x}{2}+\frac {3 \coth (x)}{2}-\coth ^2(x)+\frac {\coth ^2(x)}{2 (1+\tanh (x))}+2 \int \coth (x) \, dx \\ & = -\frac {3 x}{2}+\frac {3 \coth (x)}{2}-\coth ^2(x)+2 \log (\sinh (x))+\frac {\coth ^2(x)}{2 (1+\tanh (x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.46 \[ \int \frac {\coth ^3(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \left (-2 \coth ^2(x)-\coth ^4(x)+\frac {\coth ^5(x)}{1+\coth (x)}+\coth ^3(x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\tanh ^2(x)\right )+4 (\log (\cosh (x))+\log (\tanh (x)))\right ) \]

[In]

Integrate[Coth[x]^3/(1 + Tanh[x]),x]

[Out]

(-2*Coth[x]^2 - Coth[x]^4 + Coth[x]^5/(1 + Coth[x]) + Coth[x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Tanh[x]^2] +
4*(Log[Cosh[x]] + Log[Tanh[x]]))/2

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81

method	result	size
risch	\(-\frac {7 x}{2}+\frac {{\mathrm e}^{-2 x}}{4}-\frac {2}{\left ({\mathrm e}^{2 x}-1\right )^{2}}+2 \ln \left ({\mathrm e}^{2 x}-1\right )\)	\(30\)
parallelrisch	\(\frac {\left (-4 \tanh \left (x \right )-4\right ) \ln \left (1-\tanh \left (x \right )\right )+\left (4 \tanh \left (x \right )+4\right ) \ln \left (\tanh \left (x \right )\right )-7 \tanh \left (x \right ) x -\coth \left (x \right )^{2}-7 x +\coth \left (x \right )+3}{2+2 \tanh \left (x \right )}\)	\(52\)
default	\(\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {1}{\tanh \left (\frac {x}{2}\right )+1}-\frac {7 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}-\frac {\tanh \left (\frac {x}{2}\right )^{2}}{8}+\frac {\tanh \left (\frac {x}{2}\right )}{2}-\frac {1}{8 \tanh \left (\frac {x}{2}\right )^{2}}+\frac {1}{2 \tanh \left (\frac {x}{2}\right )}+2 \ln \left (\tanh \left (\frac {x}{2}\right )\right )-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}\)	\(75\)

[In]

int(coth(x)^3/(1+tanh(x)),x,method=_RETURNVERBOSE)

[Out]

-7/2*x+1/4*exp(-2*x)-2/(exp(2*x)-1)^2+2*ln(exp(2*x)-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 357 vs. \(2 (31) = 62\).

Time = 0.25 (sec) , antiderivative size = 357, normalized size of antiderivative = 9.65 \[ \int \frac {\coth ^3(x)}{1+\tanh (x)} \, dx=-\frac {14 \, x \cosh \left (x\right )^{6} + 84 \, x \cosh \left (x\right ) \sinh \left (x\right )^{5} + 14 \, x \sinh \left (x\right )^{6} - {\left (28 \, x + 1\right )} \cosh \left (x\right )^{4} + {\left (210 \, x \cosh \left (x\right )^{2} - 28 \, x - 1\right )} \sinh \left (x\right )^{4} + 4 \, {\left (70 \, x \cosh \left (x\right )^{3} - {\left (28 \, x + 1\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 2 \, {\left (7 \, x + 5\right )} \cosh \left (x\right )^{2} + 2 \, {\left (105 \, x \cosh \left (x\right )^{4} - 3 \, {\left (28 \, x + 1\right )} \cosh \left (x\right )^{2} + 7 \, x + 5\right )} \sinh \left (x\right )^{2} - 8 \, {\left (\cosh \left (x\right )^{6} + 6 \, \cosh \left (x\right ) \sinh \left (x\right )^{5} + \sinh \left (x\right )^{6} + {\left (15 \, \cosh \left (x\right )^{2} - 2\right )} \sinh \left (x\right )^{4} - 2 \, \cosh \left (x\right )^{4} + 4 \, {\left (5 \, \cosh \left (x\right )^{3} - 2 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + {\left (15 \, \cosh \left (x\right )^{4} - 12 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + \cosh \left (x\right )^{2} + 2 \, {\left (3 \, \cosh \left (x\right )^{5} - 4 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac {2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 4 \, {\left (21 \, x \cosh \left (x\right )^{5} - {\left (28 \, x + 1\right )} \cosh \left (x\right )^{3} + {\left (7 \, x + 5\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) - 1}{4 \, {\left (\cosh \left (x\right )^{6} + 6 \, \cosh \left (x\right ) \sinh \left (x\right )^{5} + \sinh \left (x\right )^{6} + {\left (15 \, \cosh \left (x\right )^{2} - 2\right )} \sinh \left (x\right )^{4} - 2 \, \cosh \left (x\right )^{4} + 4 \, {\left (5 \, \cosh \left (x\right )^{3} - 2 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + {\left (15 \, \cosh \left (x\right )^{4} - 12 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + \cosh \left (x\right )^{2} + 2 \, {\left (3 \, \cosh \left (x\right )^{5} - 4 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right )\right )}} \]

[In]

integrate(coth(x)^3/(1+tanh(x)),x, algorithm="fricas")

[Out]

-1/4*(14*x*cosh(x)^6 + 84*x*cosh(x)*sinh(x)^5 + 14*x*sinh(x)^6 - (28*x + 1)*cosh(x)^4 + (210*x*cosh(x)^2 - 28*
x - 1)*sinh(x)^4 + 4*(70*x*cosh(x)^3 - (28*x + 1)*cosh(x))*sinh(x)^3 + 2*(7*x + 5)*cosh(x)^2 + 2*(105*x*cosh(x
)^4 - 3*(28*x + 1)*cosh(x)^2 + 7*x + 5)*sinh(x)^2 - 8*(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + (15*cosh(
x)^2 - 2)*sinh(x)^4 - 2*cosh(x)^4 + 4*(5*cosh(x)^3 - 2*cosh(x))*sinh(x)^3 + (15*cosh(x)^4 - 12*cosh(x)^2 + 1)*
sinh(x)^2 + cosh(x)^2 + 2*(3*cosh(x)^5 - 4*cosh(x)^3 + cosh(x))*sinh(x))*log(2*sinh(x)/(cosh(x) - sinh(x))) +
4*(21*x*cosh(x)^5 - (28*x + 1)*cosh(x)^3 + (7*x + 5)*cosh(x))*sinh(x) - 1)/(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 +
sinh(x)^6 + (15*cosh(x)^2 - 2)*sinh(x)^4 - 2*cosh(x)^4 + 4*(5*cosh(x)^3 - 2*cosh(x))*sinh(x)^3 + (15*cosh(x)^4
- 12*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(3*cosh(x)^5 - 4*cosh(x)^3 + cosh(x))*sinh(x))

Sympy [F]

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

[In]

integrate(coth(x)**3/(1+tanh(x)),x)

[Out]

Integral(coth(x)**3/(tanh(x) + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.46 \[ \int \frac {\coth ^3(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \, x + \frac {2 \, {\left (2 \, e^{\left (-2 \, x\right )} - 1\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} + \frac {1}{4} \, e^{\left (-2 \, x\right )} + 2 \, \log \left (e^{\left (-x\right )} + 1\right ) + 2 \, \log \left (e^{\left (-x\right )} - 1\right ) \]

[In]

integrate(coth(x)^3/(1+tanh(x)),x, algorithm="maxima")

[Out]

1/2*x + 2*(2*e^(-2*x) - 1)/(2*e^(-2*x) - e^(-4*x) - 1) + 1/4*e^(-2*x) + 2*log(e^(-x) + 1) + 2*log(e^(-x) - 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.08 \[ \int \frac {\coth ^3(x)}{1+\tanh (x)} \, dx=-\frac {7}{2} \, x + \frac {{\left (e^{\left (4 \, x\right )} - 10 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-2 \, x\right )}}{4 \, {\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} + 2 \, \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \]

[In]

integrate(coth(x)^3/(1+tanh(x)),x, algorithm="giac")

[Out]

-7/2*x + 1/4*(e^(4*x) - 10*e^(2*x) + 1)*e^(-2*x)/(e^(2*x) - 1)^2 + 2*log(abs(e^(2*x) - 1))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {\coth ^3(x)}{1+\tanh (x)} \, dx=2\,\ln \left ({\mathrm {e}}^{2\,x}-1\right )-\frac {7\,x}{2}+\frac {{\mathrm {e}}^{-2\,x}}{4}-\frac {2}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1} \]

[In]

int(coth(x)^3/(tanh(x) + 1),x)

[Out]

2*log(exp(2*x) - 1) - (7*x)/2 + exp(-2*x)/4 - 2/(exp(4*x) - 2*exp(2*x) + 1)