3.162 \(\int \frac{\text{csch}^3(x)}{a+a \cosh (x)} \, dx\)

Optimal. Leaf size=49 \[ -\frac{a}{8 (a \cosh (x)+a)^2}+\frac{1}{8 (a-a \cosh (x))}-\frac{1}{4 (a \cosh (x)+a)}+\frac{3 \tanh ^{-1}(\cosh (x))}{8 a} \]

[Out]

(3*ArcTanh[Cosh[x]])/(8*a) + 1/(8*(a - a*Cosh[x])) - a/(8*(a + a*Cosh[x])^2) - 1/(4*(a + a*Cosh[x]))

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Rubi [A]  time = 0.0811718, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2667, 44, 206} \[ -\frac{a}{8 (a \cosh (x)+a)^2}+\frac{1}{8 (a-a \cosh (x))}-\frac{1}{4 (a \cosh (x)+a)}+\frac{3 \tanh ^{-1}(\cosh (x))}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(a + a*Cosh[x]),x]

[Out]

(3*ArcTanh[Cosh[x]])/(8*a) + 1/(8*(a - a*Cosh[x])) - a/(8*(a + a*Cosh[x])^2) - 1/(4*(a + a*Cosh[x]))

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\text{csch}^3(x)}{a+a \cosh (x)} \, dx &=a^3 \operatorname{Subst}\left (\int \frac{1}{(a-x)^2 (a+x)^3} \, dx,x,a \cosh (x)\right )\\ &=a^3 \operatorname{Subst}\left (\int \left (\frac{1}{8 a^3 (a-x)^2}+\frac{1}{4 a^2 (a+x)^3}+\frac{1}{4 a^3 (a+x)^2}+\frac{3}{8 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,a \cosh (x)\right )\\ &=\frac{1}{8 (a-a \cosh (x))}-\frac{a}{8 (a+a \cosh (x))^2}-\frac{1}{4 (a+a \cosh (x))}+\frac{3}{8} \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \cosh (x)\right )\\ &=\frac{3 \tanh ^{-1}(\cosh (x))}{8 a}+\frac{1}{8 (a-a \cosh (x))}-\frac{a}{8 (a+a \cosh (x))^2}-\frac{1}{4 (a+a \cosh (x))}\\ \end{align*}

Mathematica [A]  time = 0.129217, size = 60, normalized size = 1.22 \[ -\frac{2 \coth ^2\left (\frac{x}{2}\right )+\text{sech}^2\left (\frac{x}{2}\right )-12 \cosh ^2\left (\frac{x}{2}\right ) \left (\log \left (\cosh \left (\frac{x}{2}\right )\right )-\log \left (\sinh \left (\frac{x}{2}\right )\right )\right )+4}{16 a (\cosh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^3/(a + a*Cosh[x]),x]

[Out]

-(4 + 2*Coth[x/2]^2 - 12*Cosh[x/2]^2*(Log[Cosh[x/2]] - Log[Sinh[x/2]]) + Sech[x/2]^2)/(16*a*(1 + Cosh[x]))

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Maple [A]  time = 0.02, size = 45, normalized size = 0.9 \begin{align*} -{\frac{1}{32\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{4}}+{\frac{3}{16\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{3}{8\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{1}{16\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^3/(a+a*cosh(x)),x)

[Out]

-1/32/a*tanh(1/2*x)^4+3/16/a*tanh(1/2*x)^2-3/8/a*ln(tanh(1/2*x))-1/16/a/tanh(1/2*x)^2

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Maxima [B]  time = 1.06089, size = 139, normalized size = 2.84 \begin{align*} -\frac{3 \, e^{\left (-x\right )} + 6 \, e^{\left (-2 \, x\right )} - 2 \, e^{\left (-3 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 3 \, e^{\left (-5 \, x\right )}}{4 \,{\left (2 \, a e^{\left (-x\right )} - a e^{\left (-2 \, x\right )} - 4 \, a e^{\left (-3 \, x\right )} - a e^{\left (-4 \, x\right )} + 2 \, a e^{\left (-5 \, x\right )} + a e^{\left (-6 \, x\right )} + a\right )}} + \frac{3 \, \log \left (e^{\left (-x\right )} + 1\right )}{8 \, a} - \frac{3 \, \log \left (e^{\left (-x\right )} - 1\right )}{8 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^3/(a+a*cosh(x)),x, algorithm="maxima")

[Out]

-1/4*(3*e^(-x) + 6*e^(-2*x) - 2*e^(-3*x) + 6*e^(-4*x) + 3*e^(-5*x))/(2*a*e^(-x) - a*e^(-2*x) - 4*a*e^(-3*x) -
a*e^(-4*x) + 2*a*e^(-5*x) + a*e^(-6*x) + a) + 3/8*log(e^(-x) + 1)/a - 3/8*log(e^(-x) - 1)/a

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Fricas [B]  time = 1.95053, size = 2066, normalized size = 42.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^3/(a+a*cosh(x)),x, algorithm="fricas")

[Out]

-1/8*(6*cosh(x)^5 + 6*(5*cosh(x) + 2)*sinh(x)^4 + 6*sinh(x)^5 + 12*cosh(x)^4 + 4*(15*cosh(x)^2 + 12*cosh(x) -
1)*sinh(x)^3 - 4*cosh(x)^3 + 12*(5*cosh(x)^3 + 6*cosh(x)^2 - cosh(x) + 1)*sinh(x)^2 + 12*cosh(x)^2 - 3*(cosh(x
)^6 + 2*(3*cosh(x) + 1)*sinh(x)^5 + sinh(x)^6 + 2*cosh(x)^5 + (15*cosh(x)^2 + 10*cosh(x) - 1)*sinh(x)^4 - cosh
(x)^4 + 4*(5*cosh(x)^3 + 5*cosh(x)^2 - cosh(x) - 1)*sinh(x)^3 - 4*cosh(x)^3 + (15*cosh(x)^4 + 20*cosh(x)^3 - 6
*cosh(x)^2 - 12*cosh(x) - 1)*sinh(x)^2 - cosh(x)^2 + 2*(3*cosh(x)^5 + 5*cosh(x)^4 - 2*cosh(x)^3 - 6*cosh(x)^2
- cosh(x) + 1)*sinh(x) + 2*cosh(x) + 1)*log(cosh(x) + sinh(x) + 1) + 3*(cosh(x)^6 + 2*(3*cosh(x) + 1)*sinh(x)^
5 + sinh(x)^6 + 2*cosh(x)^5 + (15*cosh(x)^2 + 10*cosh(x) - 1)*sinh(x)^4 - cosh(x)^4 + 4*(5*cosh(x)^3 + 5*cosh(
x)^2 - cosh(x) - 1)*sinh(x)^3 - 4*cosh(x)^3 + (15*cosh(x)^4 + 20*cosh(x)^3 - 6*cosh(x)^2 - 12*cosh(x) - 1)*sin
h(x)^2 - cosh(x)^2 + 2*(3*cosh(x)^5 + 5*cosh(x)^4 - 2*cosh(x)^3 - 6*cosh(x)^2 - cosh(x) + 1)*sinh(x) + 2*cosh(
x) + 1)*log(cosh(x) + sinh(x) - 1) + 6*(5*cosh(x)^4 + 8*cosh(x)^3 - 2*cosh(x)^2 + 4*cosh(x) + 1)*sinh(x) + 6*c
osh(x))/(a*cosh(x)^6 + a*sinh(x)^6 + 2*a*cosh(x)^5 + 2*(3*a*cosh(x) + a)*sinh(x)^5 - a*cosh(x)^4 + (15*a*cosh(
x)^2 + 10*a*cosh(x) - a)*sinh(x)^4 - 4*a*cosh(x)^3 + 4*(5*a*cosh(x)^3 + 5*a*cosh(x)^2 - a*cosh(x) - a)*sinh(x)
^3 - a*cosh(x)^2 + (15*a*cosh(x)^4 + 20*a*cosh(x)^3 - 6*a*cosh(x)^2 - 12*a*cosh(x) - a)*sinh(x)^2 + 2*a*cosh(x
) + 2*(3*a*cosh(x)^5 + 5*a*cosh(x)^4 - 2*a*cosh(x)^3 - 6*a*cosh(x)^2 - a*cosh(x) + a)*sinh(x) + a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)**3/(a+a*cosh(x)),x)

[Out]

Integral(csch(x)**3/(cosh(x) + 1), x)/a

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Giac [B]  time = 1.18777, size = 127, normalized size = 2.59 \begin{align*} \frac{3 \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{16 \, a} - \frac{3 \, \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{16 \, a} + \frac{3 \, e^{\left (-x\right )} + 3 \, e^{x} - 10}{16 \, a{\left (e^{\left (-x\right )} + e^{x} - 2\right )}} - \frac{9 \,{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 52 \, e^{\left (-x\right )} + 52 \, e^{x} + 84}{32 \, a{\left (e^{\left (-x\right )} + e^{x} + 2\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^3/(a+a*cosh(x)),x, algorithm="giac")

[Out]

3/16*log(e^(-x) + e^x + 2)/a - 3/16*log(e^(-x) + e^x - 2)/a + 1/16*(3*e^(-x) + 3*e^x - 10)/(a*(e^(-x) + e^x -
2)) - 1/32*(9*(e^(-x) + e^x)^2 + 52*e^(-x) + 52*e^x + 84)/(a*(e^(-x) + e^x + 2)^2)