3.75 \(\int \frac{\text{csch}^3(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=18 \[ \text{csch}(x)-\frac{1}{2} \tanh ^{-1}(\cosh (x))-\frac{1}{2} \coth (x) \text{csch}(x) \]

[Out]

-ArcTanh[Cosh[x]]/2 + Csch[x] - (Coth[x]*Csch[x])/2

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Rubi [A]  time = 0.160599, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {3518, 3108, 3107, 2606, 8, 2611, 3770} \[ \text{csch}(x)-\frac{1}{2} \tanh ^{-1}(\cosh (x))-\frac{1}{2} \coth (x) \text{csch}(x) \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(1 + Tanh[x]),x]

[Out]

-ArcTanh[Cosh[x]]/2 + Csch[x] - (Coth[x]*Csch[x])/2

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3108

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[a^p*b^p, Int[(Cos[c + d*x]^m*Sin[c + d*x]^n)/(b*Cos[c + d*x] + a*Sin
[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[a^2 + b^2, 0] && ILtQ[p, 0]

Rule 3107

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_.), x_Symbol] :> Int[ExpandTrig[cos[c + d*x]^m*sin[c + d*x]^n*(a*cos[c + d*x] + b*sin[c +
d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IGtQ[p, 0]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

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

Rubi steps

\begin{align*} \int \frac{\text{csch}^3(x)}{1+\tanh (x)} \, dx &=\int \frac{\coth (x) \text{csch}^2(x)}{\cosh (x)+\sinh (x)} \, dx\\ &=i \int \coth (x) \text{csch}^2(x) (-i \cosh (x)+i \sinh (x)) \, dx\\ &=\int \left (-\coth (x) \text{csch}(x)+\coth ^2(x) \text{csch}(x)\right ) \, dx\\ &=-\int \coth (x) \text{csch}(x) \, dx+\int \coth ^2(x) \text{csch}(x) \, dx\\ &=-\frac{1}{2} \coth (x) \text{csch}(x)+i \operatorname{Subst}(\int 1 \, dx,x,-i \text{csch}(x))+\frac{1}{2} \int \text{csch}(x) \, dx\\ &=-\frac{1}{2} \tanh ^{-1}(\cosh (x))+\text{csch}(x)-\frac{1}{2} \coth (x) \text{csch}(x)\\ \end{align*}

Mathematica [A]  time = 0.0787912, size = 20, normalized size = 1.11 \[ \frac{1}{2} \left (\log \left (\tanh \left (\frac{x}{2}\right )\right )-(\coth (x)-2) \text{csch}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^3/(1 + Tanh[x]),x]

[Out]

(-((-2 + Coth[x])*Csch[x]) + Log[Tanh[x/2]])/2

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Maple [B]  time = 0.027, size = 39, normalized size = 2.2 \begin{align*}{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{1}{2}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{1}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^3/(1+tanh(x)),x)

[Out]

1/8*tanh(1/2*x)^2-1/2*tanh(1/2*x)-1/8/tanh(1/2*x)^2+1/2/tanh(1/2*x)+1/2*ln(tanh(1/2*x))

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Maxima [B]  time = 1.11479, size = 65, normalized size = 3.61 \begin{align*} -\frac{e^{\left (-x\right )} - 3 \, e^{\left (-3 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} - \frac{1}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac{1}{2} \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(e^(-x) - 3*e^(-3*x))/(2*e^(-2*x) - e^(-4*x) - 1) - 1/2*log(e^(-x) + 1) + 1/2*log(e^(-x) - 1)

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Fricas [B]  time = 2.64399, size = 733, normalized size = 40.72 \begin{align*} \frac{2 \, \cosh \left (x\right )^{3} + 6 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + 2 \, \sinh \left (x\right )^{3} -{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) +{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + 6 \,{\left (\cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right ) - 6 \, \cosh \left (x\right )}{2 \,{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*cosh(x)^3 + 6*cosh(x)*sinh(x)^2 + 2*sinh(x)^3 - (cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cos
h(x)^2 - 1)*sinh(x)^2 - 2*cosh(x)^2 + 4*(cosh(x)^3 - cosh(x))*sinh(x) + 1)*log(cosh(x) + sinh(x) + 1) + (cosh(
x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 - 1)*sinh(x)^2 - 2*cosh(x)^2 + 4*(cosh(x)^3 - cosh(x))
*sinh(x) + 1)*log(cosh(x) + sinh(x) - 1) + 6*(cosh(x)^2 - 1)*sinh(x) - 6*cosh(x))/(cosh(x)^4 + 4*cosh(x)*sinh(
x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 - 1)*sinh(x)^2 - 2*cosh(x)^2 + 4*(cosh(x)^3 - cosh(x))*sinh(x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}^{3}{\left (x \right )}}{\tanh{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csch(x)**3/(tanh(x) + 1), x)

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Giac [B]  time = 1.21324, size = 46, normalized size = 2.56 \begin{align*} \frac{e^{\left (3 \, x\right )} - 3 \, e^{x}}{{\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} - \frac{1}{2} \, \log \left (e^{x} + 1\right ) + \frac{1}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^(3*x) - 3*e^x)/(e^(2*x) - 1)^2 - 1/2*log(e^x + 1) + 1/2*log(abs(e^x - 1))