3.77 \(\int \frac{\text{csch}^5(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=34 \[ \frac{\text{csch}^3(x)}{3}+\frac{1}{8} \tanh ^{-1}(\cosh (x))-\frac{1}{4} \coth (x) \text{csch}^3(x)-\frac{1}{8} \coth (x) \text{csch}(x) \]

[Out]

ArcTanh[Cosh[x]]/8 - (Coth[x]*Csch[x])/8 + Csch[x]^3/3 - (Coth[x]*Csch[x]^3)/4

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Rubi [A]  time = 0.191537, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.727, Rules used = {3518, 3108, 3107, 2606, 30, 2611, 3768, 3770} \[ \frac{\text{csch}^3(x)}{3}+\frac{1}{8} \tanh ^{-1}(\cosh (x))-\frac{1}{4} \coth (x) \text{csch}^3(x)-\frac{1}{8} \coth (x) \text{csch}(x) \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^5/(1 + Tanh[x]),x]

[Out]

ArcTanh[Cosh[x]]/8 - (Coth[x]*Csch[x])/8 + Csch[x]^3/3 - (Coth[x]*Csch[x]^3)/4

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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}^5(x)}{1+\tanh (x)} \, dx &=\int \frac{\coth (x) \text{csch}^4(x)}{\cosh (x)+\sinh (x)} \, dx\\ &=i \int \coth (x) \text{csch}^4(x) (-i \cosh (x)+i \sinh (x)) \, dx\\ &=-\int \left (\coth (x) \text{csch}^3(x)-\coth ^2(x) \text{csch}^3(x)\right ) \, dx\\ &=-\int \coth (x) \text{csch}^3(x) \, dx+\int \coth ^2(x) \text{csch}^3(x) \, dx\\ &=-\frac{1}{4} \coth (x) \text{csch}^3(x)-i \operatorname{Subst}\left (\int x^2 \, dx,x,-i \text{csch}(x)\right )+\frac{1}{4} \int \text{csch}^3(x) \, dx\\ &=-\frac{1}{8} \coth (x) \text{csch}(x)+\frac{\text{csch}^3(x)}{3}-\frac{1}{4} \coth (x) \text{csch}^3(x)-\frac{1}{8} \int \text{csch}(x) \, dx\\ &=\frac{1}{8} \tanh ^{-1}(\cosh (x))-\frac{1}{8} \coth (x) \text{csch}(x)+\frac{\text{csch}^3(x)}{3}-\frac{1}{4} \coth (x) \text{csch}^3(x)\\ \end{align*}

Mathematica [A]  time = 0.118608, size = 49, normalized size = 1.44 \[ -\frac{1}{192} \text{csch}^4(x) \left (42 \cosh (x)+6 \cosh (3 x)+2 \sinh (x) \left (-9 \sinh (x) \log \left (\tanh \left (\frac{x}{2}\right )\right )+3 \sinh (3 x) \log \left (\tanh \left (\frac{x}{2}\right )\right )-32\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^5/(1 + Tanh[x]),x]

[Out]

-(Csch[x]^4*(42*Cosh[x] + 6*Cosh[3*x] + 2*Sinh[x]*(-32 - 9*Log[Tanh[x/2]]*Sinh[x] + 3*Log[Tanh[x/2]]*Sinh[3*x]
)))/192

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Maple [B]  time = 0.033, size = 55, normalized size = 1.6 \begin{align*}{\frac{1}{64} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{4}}-{\frac{1}{24} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}+{\frac{1}{8}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{1}{24} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}-{\frac{1}{8}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{1}{64} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^5/(1+tanh(x)),x)

[Out]

1/64*tanh(1/2*x)^4-1/24*tanh(1/2*x)^3+1/8*tanh(1/2*x)-1/8/tanh(1/2*x)+1/24/tanh(1/2*x)^3-1/8*ln(tanh(1/2*x))-1
/64/tanh(1/2*x)^4

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Maxima [B]  time = 1.07946, size = 100, normalized size = 2.94 \begin{align*} \frac{3 \, e^{\left (-x\right )} - 11 \, e^{\left (-3 \, x\right )} + 53 \, e^{\left (-5 \, x\right )} + 3 \, e^{\left (-7 \, x\right )}}{12 \,{\left (4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1\right )}} + \frac{1}{8} \, \log \left (e^{\left (-x\right )} + 1\right ) - \frac{1}{8} \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^5/(1+tanh(x)),x, algorithm="maxima")

[Out]

1/12*(3*e^(-x) - 11*e^(-3*x) + 53*e^(-5*x) + 3*e^(-7*x))/(4*e^(-2*x) - 6*e^(-4*x) + 4*e^(-6*x) - e^(-8*x) - 1)
+ 1/8*log(e^(-x) + 1) - 1/8*log(e^(-x) - 1)

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Fricas [B]  time = 2.18859, size = 2136, normalized size = 62.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^5/(1+tanh(x)),x, algorithm="fricas")

[Out]

-1/24*(6*cosh(x)^7 + 42*cosh(x)*sinh(x)^6 + 6*sinh(x)^7 + 2*(63*cosh(x)^2 - 11)*sinh(x)^5 - 22*cosh(x)^5 + 10*
(21*cosh(x)^3 - 11*cosh(x))*sinh(x)^4 + 2*(105*cosh(x)^4 - 110*cosh(x)^2 + 53)*sinh(x)^3 + 106*cosh(x)^3 + 2*(
63*cosh(x)^5 - 110*cosh(x)^3 + 159*cosh(x))*sinh(x)^2 - 3*(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*
cosh(x)^2 - 1)*sinh(x)^6 - 4*cosh(x)^6 + 8*(7*cosh(x)^3 - 3*cosh(x))*sinh(x)^5 + 2*(35*cosh(x)^4 - 30*cosh(x)^
2 + 3)*sinh(x)^4 + 6*cosh(x)^4 + 8*(7*cosh(x)^5 - 10*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 - 15*co
sh(x)^4 + 9*cosh(x)^2 - 1)*sinh(x)^2 - 4*cosh(x)^2 + 8*(cosh(x)^7 - 3*cosh(x)^5 + 3*cosh(x)^3 - cosh(x))*sinh(
x) + 1)*log(cosh(x) + sinh(x) + 1) + 3*(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh(x)^2 - 1)*sinh
(x)^6 - 4*cosh(x)^6 + 8*(7*cosh(x)^3 - 3*cosh(x))*sinh(x)^5 + 2*(35*cosh(x)^4 - 30*cosh(x)^2 + 3)*sinh(x)^4 +
6*cosh(x)^4 + 8*(7*cosh(x)^5 - 10*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 - 15*cosh(x)^4 + 9*cosh(x)
^2 - 1)*sinh(x)^2 - 4*cosh(x)^2 + 8*(cosh(x)^7 - 3*cosh(x)^5 + 3*cosh(x)^3 - cosh(x))*sinh(x) + 1)*log(cosh(x)
+ sinh(x) - 1) + 2*(21*cosh(x)^6 - 55*cosh(x)^4 + 159*cosh(x)^2 + 3)*sinh(x) + 6*cosh(x))/(cosh(x)^8 + 8*cosh
(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh(x)^2 - 1)*sinh(x)^6 - 4*cosh(x)^6 + 8*(7*cosh(x)^3 - 3*cosh(x))*sinh(x)^
5 + 2*(35*cosh(x)^4 - 30*cosh(x)^2 + 3)*sinh(x)^4 + 6*cosh(x)^4 + 8*(7*cosh(x)^5 - 10*cosh(x)^3 + 3*cosh(x))*s
inh(x)^3 + 4*(7*cosh(x)^6 - 15*cosh(x)^4 + 9*cosh(x)^2 - 1)*sinh(x)^2 - 4*cosh(x)^2 + 8*(cosh(x)^7 - 3*cosh(x)
^5 + 3*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}^{5}{\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)**5/(1+tanh(x)),x)

[Out]

Integral(csch(x)**5/(tanh(x) + 1), x)

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Giac [A]  time = 1.32025, size = 66, normalized size = 1.94 \begin{align*} -\frac{3 \, e^{\left (7 \, x\right )} - 11 \, e^{\left (5 \, x\right )} + 53 \, e^{\left (3 \, x\right )} + 3 \, e^{x}}{12 \,{\left (e^{\left (2 \, x\right )} - 1\right )}^{4}} + \frac{1}{8} \, \log \left (e^{x} + 1\right ) - \frac{1}{8} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^5/(1+tanh(x)),x, algorithm="giac")

[Out]

-1/12*(3*e^(7*x) - 11*e^(5*x) + 53*e^(3*x) + 3*e^x)/(e^(2*x) - 1)^4 + 1/8*log(e^x + 1) - 1/8*log(abs(e^x - 1))