3.187 \(\int \frac{\tanh ^6(x)}{a+a \cosh (x)} \, dx\)

Optimal. Leaf size=46 \[ -\frac{\tanh ^5(x)}{5 a}+\frac{3 \tan ^{-1}(\sinh (x))}{8 a}-\frac{\tanh ^3(x) \text{sech}(x)}{4 a}-\frac{3 \tanh (x) \text{sech}(x)}{8 a} \]

[Out]

(3*ArcTan[Sinh[x]])/(8*a) - (3*Sech[x]*Tanh[x])/(8*a) - (Sech[x]*Tanh[x]^3)/(4*a) - Tanh[x]^5/(5*a)

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Rubi [A]  time = 0.0928183, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2706, 2607, 30, 2611, 3770} \[ -\frac{\tanh ^5(x)}{5 a}+\frac{3 \tan ^{-1}(\sinh (x))}{8 a}-\frac{\tanh ^3(x) \text{sech}(x)}{4 a}-\frac{3 \tanh (x) \text{sech}(x)}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^6/(a + a*Cosh[x]),x]

[Out]

(3*ArcTan[Sinh[x]])/(8*a) - (3*Sech[x]*Tanh[x])/(8*a) - (Sech[x]*Tanh[x]^3)/(4*a) - Tanh[x]^5/(5*a)

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -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 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 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{\tanh ^6(x)}{a+a \cosh (x)} \, dx &=\frac{\int \text{sech}(x) \tanh ^4(x) \, dx}{a}-\frac{\int \text{sech}^2(x) \tanh ^4(x) \, dx}{a}\\ &=-\frac{\text{sech}(x) \tanh ^3(x)}{4 a}+\frac{i \operatorname{Subst}\left (\int x^4 \, dx,x,i \tanh (x)\right )}{a}+\frac{3 \int \text{sech}(x) \tanh ^2(x) \, dx}{4 a}\\ &=-\frac{3 \text{sech}(x) \tanh (x)}{8 a}-\frac{\text{sech}(x) \tanh ^3(x)}{4 a}-\frac{\tanh ^5(x)}{5 a}+\frac{3 \int \text{sech}(x) \, dx}{8 a}\\ &=\frac{3 \tan ^{-1}(\sinh (x))}{8 a}-\frac{3 \text{sech}(x) \tanh (x)}{8 a}-\frac{\text{sech}(x) \tanh ^3(x)}{4 a}-\frac{\tanh ^5(x)}{5 a}\\ \end{align*}

Mathematica [A]  time = 0.0845774, size = 58, normalized size = 1.26 \[ \frac{\cosh ^2\left (\frac{x}{2}\right ) \left (30 \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+\tanh (x) \left (-8 \text{sech}^4(x)+10 \text{sech}^3(x)+16 \text{sech}^2(x)-25 \text{sech}(x)-8\right )\right )}{20 a (\cosh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^6/(a + a*Cosh[x]),x]

[Out]

(Cosh[x/2]^2*(30*ArcTan[Tanh[x/2]] + (-8 - 25*Sech[x] + 16*Sech[x]^2 + 10*Sech[x]^3 - 8*Sech[x]^4)*Tanh[x]))/(
20*a*(1 + Cosh[x]))

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Maple [B]  time = 0.06, size = 115, normalized size = 2.5 \begin{align*}{\frac{3}{4\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{9} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-5}}+{\frac{7}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{7} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-5}}-{\frac{32}{5\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{5} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-5}}-{\frac{7}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-5}}-{\frac{3}{4\,a}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-5}}+{\frac{3}{4\,a}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^6/(a+a*cosh(x)),x)

[Out]

3/4/a/(tanh(1/2*x)^2+1)^5*tanh(1/2*x)^9+7/2/a/(tanh(1/2*x)^2+1)^5*tanh(1/2*x)^7-32/5/a/(tanh(1/2*x)^2+1)^5*tan
h(1/2*x)^5-7/2/a/(tanh(1/2*x)^2+1)^5*tanh(1/2*x)^3-3/4/a/(tanh(1/2*x)^2+1)^5*tanh(1/2*x)+3/4/a*arctan(tanh(1/2
*x))

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Maxima [B]  time = 1.55973, size = 120, normalized size = 2.61 \begin{align*} -\frac{25 \, e^{\left (-x\right )} + 10 \, e^{\left (-3 \, x\right )} + 80 \, e^{\left (-4 \, x\right )} - 10 \, e^{\left (-7 \, x\right )} + 40 \, e^{\left (-8 \, x\right )} - 25 \, e^{\left (-9 \, x\right )} + 8}{20 \,{\left (5 \, a e^{\left (-2 \, x\right )} + 10 \, a e^{\left (-4 \, x\right )} + 10 \, a e^{\left (-6 \, x\right )} + 5 \, a e^{\left (-8 \, x\right )} + a e^{\left (-10 \, x\right )} + a\right )}} - \frac{3 \, \arctan \left (e^{\left (-x\right )}\right )}{4 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^6/(a+a*cosh(x)),x, algorithm="maxima")

[Out]

-1/20*(25*e^(-x) + 10*e^(-3*x) + 80*e^(-4*x) - 10*e^(-7*x) + 40*e^(-8*x) - 25*e^(-9*x) + 8)/(5*a*e^(-2*x) + 10
*a*e^(-4*x) + 10*a*e^(-6*x) + 5*a*e^(-8*x) + a*e^(-10*x) + a) - 3/4*arctan(e^(-x))/a

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Fricas [B]  time = 1.88848, size = 2522, normalized size = 54.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^6/(a+a*cosh(x)),x, algorithm="fricas")

[Out]

-1/20*(25*cosh(x)^9 + 5*(45*cosh(x) - 8)*sinh(x)^8 + 25*sinh(x)^9 - 40*cosh(x)^8 + 10*(90*cosh(x)^2 - 32*cosh(
x) + 1)*sinh(x)^7 + 10*cosh(x)^7 + 70*(30*cosh(x)^3 - 16*cosh(x)^2 + cosh(x))*sinh(x)^6 + 70*(45*cosh(x)^4 - 3
2*cosh(x)^3 + 3*cosh(x)^2)*sinh(x)^5 + 10*(315*cosh(x)^5 - 280*cosh(x)^4 + 35*cosh(x)^3 - 8)*sinh(x)^4 - 80*co
sh(x)^4 + 10*(210*cosh(x)^6 - 224*cosh(x)^5 + 35*cosh(x)^4 - 32*cosh(x) - 1)*sinh(x)^3 - 10*cosh(x)^3 + 10*(90
*cosh(x)^7 - 112*cosh(x)^6 + 21*cosh(x)^5 - 48*cosh(x)^2 - 3*cosh(x))*sinh(x)^2 - 15*(cosh(x)^10 + 10*cosh(x)*
sinh(x)^9 + sinh(x)^10 + 5*(9*cosh(x)^2 + 1)*sinh(x)^8 + 5*cosh(x)^8 + 40*(3*cosh(x)^3 + cosh(x))*sinh(x)^7 +
10*(21*cosh(x)^4 + 14*cosh(x)^2 + 1)*sinh(x)^6 + 10*cosh(x)^6 + 4*(63*cosh(x)^5 + 70*cosh(x)^3 + 15*cosh(x))*s
inh(x)^5 + 10*(21*cosh(x)^6 + 35*cosh(x)^4 + 15*cosh(x)^2 + 1)*sinh(x)^4 + 10*cosh(x)^4 + 40*(3*cosh(x)^7 + 7*
cosh(x)^5 + 5*cosh(x)^3 + cosh(x))*sinh(x)^3 + 5*(9*cosh(x)^8 + 28*cosh(x)^6 + 30*cosh(x)^4 + 12*cosh(x)^2 + 1
)*sinh(x)^2 + 5*cosh(x)^2 + 10*(cosh(x)^9 + 4*cosh(x)^7 + 6*cosh(x)^5 + 4*cosh(x)^3 + cosh(x))*sinh(x) + 1)*ar
ctan(cosh(x) + sinh(x)) + 5*(45*cosh(x)^8 - 64*cosh(x)^7 + 14*cosh(x)^6 - 64*cosh(x)^3 - 6*cosh(x)^2 - 5)*sinh
(x) - 25*cosh(x) - 8)/(a*cosh(x)^10 + 10*a*cosh(x)*sinh(x)^9 + a*sinh(x)^10 + 5*a*cosh(x)^8 + 5*(9*a*cosh(x)^2
+ a)*sinh(x)^8 + 40*(3*a*cosh(x)^3 + a*cosh(x))*sinh(x)^7 + 10*a*cosh(x)^6 + 10*(21*a*cosh(x)^4 + 14*a*cosh(x
)^2 + a)*sinh(x)^6 + 4*(63*a*cosh(x)^5 + 70*a*cosh(x)^3 + 15*a*cosh(x))*sinh(x)^5 + 10*a*cosh(x)^4 + 10*(21*a*
cosh(x)^6 + 35*a*cosh(x)^4 + 15*a*cosh(x)^2 + a)*sinh(x)^4 + 40*(3*a*cosh(x)^7 + 7*a*cosh(x)^5 + 5*a*cosh(x)^3
+ a*cosh(x))*sinh(x)^3 + 5*a*cosh(x)^2 + 5*(9*a*cosh(x)^8 + 28*a*cosh(x)^6 + 30*a*cosh(x)^4 + 12*a*cosh(x)^2
+ a)*sinh(x)^2 + 10*(a*cosh(x)^9 + 4*a*cosh(x)^7 + 6*a*cosh(x)^5 + 4*a*cosh(x)^3 + a*cosh(x))*sinh(x) + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tanh ^{6}{\left (x \right )}}{\cosh{\left (x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**6/(a+a*cosh(x)),x)

[Out]

Integral(tanh(x)**6/(cosh(x) + 1), x)/a

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Giac [A]  time = 1.18083, size = 78, normalized size = 1.7 \begin{align*} \frac{3 \, \arctan \left (e^{x}\right )}{4 \, a} - \frac{25 \, e^{\left (9 \, x\right )} - 40 \, e^{\left (8 \, x\right )} + 10 \, e^{\left (7 \, x\right )} - 80 \, e^{\left (4 \, x\right )} - 10 \, e^{\left (3 \, x\right )} - 25 \, e^{x} - 8}{20 \, a{\left (e^{\left (2 \, x\right )} + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^6/(a+a*cosh(x)),x, algorithm="giac")

[Out]

3/4*arctan(e^x)/a - 1/20*(25*e^(9*x) - 40*e^(8*x) + 10*e^(7*x) - 80*e^(4*x) - 10*e^(3*x) - 25*e^x - 8)/(a*(e^(
2*x) + 1)^5)