3.59 \(\int \frac{\text{csch}^4(x)}{a+a \text{sech}(x)} \, dx\)

Optimal. Leaf size=34 \[ -\frac{\coth ^5(x)}{5 a}+\frac{\coth ^3(x)}{3 a}+\frac{\text{csch}^5(x)}{5 a} \]

[Out]

Coth[x]^3/(3*a) - Coth[x]^5/(5*a) + Csch[x]^5/(5*a)

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Rubi [A]  time = 0.146252, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3872, 2839, 2606, 30, 2607, 14} \[ -\frac{\coth ^5(x)}{5 a}+\frac{\coth ^3(x)}{3 a}+\frac{\text{csch}^5(x)}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^4/(a + a*Sech[x]),x]

[Out]

Coth[x]^3/(3*a) - Coth[x]^5/(5*a) + Csch[x]^5/(5*a)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 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 2607

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{\text{csch}^4(x)}{a+a \text{sech}(x)} \, dx &=-\int \frac{\coth (x) \text{csch}^3(x)}{-a-a \cosh (x)} \, dx\\ &=\frac{\int \coth ^2(x) \text{csch}^4(x) \, dx}{a}-\frac{\int \coth (x) \text{csch}^5(x) \, dx}{a}\\ &=\frac{i \operatorname{Subst}\left (\int x^4 \, dx,x,-i \text{csch}(x)\right )}{a}+\frac{i \operatorname{Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,i \coth (x)\right )}{a}\\ &=\frac{\text{csch}^5(x)}{5 a}+\frac{i \operatorname{Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,i \coth (x)\right )}{a}\\ &=\frac{\coth ^3(x)}{3 a}-\frac{\coth ^5(x)}{5 a}+\frac{\text{csch}^5(x)}{5 a}\\ \end{align*}

Mathematica [A]  time = 0.0599826, size = 39, normalized size = 1.15 \[ \frac{(-6 \cosh (x)-2 \cosh (2 x)+2 \cosh (3 x)+\cosh (4 x)-15) \text{csch}^3(x)}{60 a (\cosh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^4/(a + a*Sech[x]),x]

[Out]

((-15 - 6*Cosh[x] - 2*Cosh[2*x] + 2*Cosh[3*x] + Cosh[4*x])*Csch[x]^3)/(60*a*(1 + Cosh[x]))

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Maple [A]  time = 0.026, size = 39, normalized size = 1.2 \begin{align*}{\frac{1}{16\,a} \left ( -{\frac{1}{5} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{5}}+{\frac{2}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}+2\, \left ( \tanh \left ( x/2 \right ) \right ) ^{-1}-{\frac{1}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^4/(a+a*sech(x)),x)

[Out]

1/16/a*(-1/5*tanh(1/2*x)^5+2/3*tanh(1/2*x)^3+2/tanh(1/2*x)-1/3/tanh(1/2*x)^3)

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Maxima [B]  time = 1.17394, size = 394, normalized size = 11.59 \begin{align*} \frac{8 \, e^{\left (-x\right )}}{15 \,{\left (2 \, a e^{\left (-x\right )} - 2 \, a e^{\left (-2 \, x\right )} - 6 \, a e^{\left (-3 \, x\right )} + 6 \, a e^{\left (-5 \, x\right )} + 2 \, a e^{\left (-6 \, x\right )} - 2 \, a e^{\left (-7 \, x\right )} - a e^{\left (-8 \, x\right )} + a\right )}} - \frac{8 \, e^{\left (-2 \, x\right )}}{15 \,{\left (2 \, a e^{\left (-x\right )} - 2 \, a e^{\left (-2 \, x\right )} - 6 \, a e^{\left (-3 \, x\right )} + 6 \, a e^{\left (-5 \, x\right )} + 2 \, a e^{\left (-6 \, x\right )} - 2 \, a e^{\left (-7 \, x\right )} - a e^{\left (-8 \, x\right )} + a\right )}} - \frac{8 \, e^{\left (-3 \, x\right )}}{5 \,{\left (2 \, a e^{\left (-x\right )} - 2 \, a e^{\left (-2 \, x\right )} - 6 \, a e^{\left (-3 \, x\right )} + 6 \, a e^{\left (-5 \, x\right )} + 2 \, a e^{\left (-6 \, x\right )} - 2 \, a e^{\left (-7 \, x\right )} - a e^{\left (-8 \, x\right )} + a\right )}} - \frac{4 \, e^{\left (-4 \, x\right )}}{2 \, a e^{\left (-x\right )} - 2 \, a e^{\left (-2 \, x\right )} - 6 \, a e^{\left (-3 \, x\right )} + 6 \, a e^{\left (-5 \, x\right )} + 2 \, a e^{\left (-6 \, x\right )} - 2 \, a e^{\left (-7 \, x\right )} - a e^{\left (-8 \, x\right )} + a} + \frac{4}{15 \,{\left (2 \, a e^{\left (-x\right )} - 2 \, a e^{\left (-2 \, x\right )} - 6 \, a e^{\left (-3 \, x\right )} + 6 \, a e^{\left (-5 \, x\right )} + 2 \, a e^{\left (-6 \, x\right )} - 2 \, a e^{\left (-7 \, x\right )} - a e^{\left (-8 \, x\right )} + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+a*sech(x)),x, algorithm="maxima")

[Out]

8/15*e^(-x)/(2*a*e^(-x) - 2*a*e^(-2*x) - 6*a*e^(-3*x) + 6*a*e^(-5*x) + 2*a*e^(-6*x) - 2*a*e^(-7*x) - a*e^(-8*x
) + a) - 8/15*e^(-2*x)/(2*a*e^(-x) - 2*a*e^(-2*x) - 6*a*e^(-3*x) + 6*a*e^(-5*x) + 2*a*e^(-6*x) - 2*a*e^(-7*x)
- a*e^(-8*x) + a) - 8/5*e^(-3*x)/(2*a*e^(-x) - 2*a*e^(-2*x) - 6*a*e^(-3*x) + 6*a*e^(-5*x) + 2*a*e^(-6*x) - 2*a
*e^(-7*x) - a*e^(-8*x) + a) - 4*e^(-4*x)/(2*a*e^(-x) - 2*a*e^(-2*x) - 6*a*e^(-3*x) + 6*a*e^(-5*x) + 2*a*e^(-6*
x) - 2*a*e^(-7*x) - a*e^(-8*x) + a) + 4/15/(2*a*e^(-x) - 2*a*e^(-2*x) - 6*a*e^(-3*x) + 6*a*e^(-5*x) + 2*a*e^(-
6*x) - 2*a*e^(-7*x) - a*e^(-8*x) + a)

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Fricas [B]  time = 2.36489, size = 695, normalized size = 20.44 \begin{align*} -\frac{8 \,{\left (7 \, \cosh \left (x\right )^{2} + 4 \,{\left (4 \, \cosh \left (x\right ) + 1\right )} \sinh \left (x\right ) + 7 \, \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) + 1\right )}}{15 \,{\left (a \cosh \left (x\right )^{6} + a \sinh \left (x\right )^{6} + 2 \, a \cosh \left (x\right )^{5} + 2 \,{\left (3 \, a \cosh \left (x\right ) + a\right )} \sinh \left (x\right )^{5} - 2 \, a \cosh \left (x\right )^{4} +{\left (15 \, a \cosh \left (x\right )^{2} + 10 \, a \cosh \left (x\right ) - 2 \, a\right )} \sinh \left (x\right )^{4} - 6 \, a \cosh \left (x\right )^{3} + 2 \,{\left (10 \, a \cosh \left (x\right )^{3} + 10 \, a \cosh \left (x\right )^{2} - 4 \, a \cosh \left (x\right ) - 3 \, a\right )} \sinh \left (x\right )^{3} - a \cosh \left (x\right )^{2} +{\left (15 \, a \cosh \left (x\right )^{4} + 20 \, a \cosh \left (x\right )^{3} - 12 \, a \cosh \left (x\right )^{2} - 18 \, a \cosh \left (x\right ) - a\right )} \sinh \left (x\right )^{2} + 4 \, a \cosh \left (x\right ) + 2 \,{\left (3 \, a \cosh \left (x\right )^{5} + 5 \, a \cosh \left (x\right )^{4} - 4 \, a \cosh \left (x\right )^{3} - 9 \, a \cosh \left (x\right )^{2} + a \cosh \left (x\right ) + 4 \, a\right )} \sinh \left (x\right ) + 2 \, a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+a*sech(x)),x, algorithm="fricas")

[Out]

-8/15*(7*cosh(x)^2 + 4*(4*cosh(x) + 1)*sinh(x) + 7*sinh(x)^2 + 2*cosh(x) + 1)/(a*cosh(x)^6 + a*sinh(x)^6 + 2*a
*cosh(x)^5 + 2*(3*a*cosh(x) + a)*sinh(x)^5 - 2*a*cosh(x)^4 + (15*a*cosh(x)^2 + 10*a*cosh(x) - 2*a)*sinh(x)^4 -
6*a*cosh(x)^3 + 2*(10*a*cosh(x)^3 + 10*a*cosh(x)^2 - 4*a*cosh(x) - 3*a)*sinh(x)^3 - a*cosh(x)^2 + (15*a*cosh(
x)^4 + 20*a*cosh(x)^3 - 12*a*cosh(x)^2 - 18*a*cosh(x) - a)*sinh(x)^2 + 4*a*cosh(x) + 2*(3*a*cosh(x)^5 + 5*a*co
sh(x)^4 - 4*a*cosh(x)^3 - 9*a*cosh(x)^2 + a*cosh(x) + 4*a)*sinh(x) + 2*a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{csch}^{4}{\left (x \right )}}{\operatorname{sech}{\left (x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)**4/(a+a*sech(x)),x)

[Out]

Integral(csch(x)**4/(sech(x) + 1), x)/a

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Giac [B]  time = 1.182, size = 80, normalized size = 2.35 \begin{align*} \frac{3 \, e^{\left (2 \, x\right )} - 12 \, e^{x} + 5}{24 \, a{\left (e^{x} - 1\right )}^{3}} - \frac{15 \, e^{\left (4 \, x\right )} + 60 \, e^{\left (3 \, x\right )} + 10 \, e^{\left (2 \, x\right )} + 20 \, e^{x} + 7}{120 \, a{\left (e^{x} + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+a*sech(x)),x, algorithm="giac")

[Out]

1/24*(3*e^(2*x) - 12*e^x + 5)/(a*(e^x - 1)^3) - 1/120*(15*e^(4*x) + 60*e^(3*x) + 10*e^(2*x) + 20*e^x + 7)/(a*(
e^x + 1)^5)