3.188 \(\int \frac {\tanh ^5(x)}{a+a \cosh (x)} \, dx\)

Optimal. Leaf size=30 \[ -\frac {\tanh ^4(x)}{4 a}+\frac {\text {sech}^3(x)}{3 a}-\frac {\text {sech}(x)}{a} \]

[Out]

-sech(x)/a+1/3*sech(x)^3/a-1/4*tanh(x)^4/a

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Rubi [A]  time = 0.09, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2706, 2607, 30, 2606} \[ -\frac {\tanh ^4(x)}{4 a}+\frac {\text {sech}^3(x)}{3 a}-\frac {\text {sech}(x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^5/(a + a*Cosh[x]),x]

[Out]

-(Sech[x]/a) + Sech[x]^3/(3*a) - Tanh[x]^4/(4*a)

Rule 30

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 25, normalized size = 0.83 \[ \frac {2 \sinh ^6\left (\frac {x}{2}\right ) (5 \cosh (x)+3) \text {sech}^4(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^5/(a + a*Cosh[x]),x]

[Out]

(2*(3 + 5*Cosh[x])*Sech[x]^4*Sinh[x/2]^6)/(3*a)

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fricas [B]  time = 0.46, size = 174, normalized size = 5.80 \[ -\frac {2 \, {\left (3 \, \cosh \relax (x)^{4} + 3 \, {\left (4 \, \cosh \relax (x) - 1\right )} \sinh \relax (x)^{3} + 3 \, \sinh \relax (x)^{4} - 3 \, \cosh \relax (x)^{3} + {\left (18 \, \cosh \relax (x)^{2} - 9 \, \cosh \relax (x) + 8\right )} \sinh \relax (x)^{2} + 8 \, \cosh \relax (x)^{2} + {\left (12 \, \cosh \relax (x)^{3} - 9 \, \cosh \relax (x)^{2} + 4 \, \cosh \relax (x) + 3\right )} \sinh \relax (x) - 3 \, \cosh \relax (x) + 5\right )}}{3 \, {\left (a \cosh \relax (x)^{5} + 5 \, a \cosh \relax (x) \sinh \relax (x)^{4} + a \sinh \relax (x)^{5} + 5 \, a \cosh \relax (x)^{3} + {\left (10 \, a \cosh \relax (x)^{2} + 3 \, a\right )} \sinh \relax (x)^{3} + 5 \, {\left (2 \, a \cosh \relax (x)^{3} + 3 \, a \cosh \relax (x)\right )} \sinh \relax (x)^{2} + 10 \, a \cosh \relax (x) + {\left (5 \, a \cosh \relax (x)^{4} + 9 \, a \cosh \relax (x)^{2} + 2 \, a\right )} \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(a+a*cosh(x)),x, algorithm="fricas")

[Out]

-2/3*(3*cosh(x)^4 + 3*(4*cosh(x) - 1)*sinh(x)^3 + 3*sinh(x)^4 - 3*cosh(x)^3 + (18*cosh(x)^2 - 9*cosh(x) + 8)*s
inh(x)^2 + 8*cosh(x)^2 + (12*cosh(x)^3 - 9*cosh(x)^2 + 4*cosh(x) + 3)*sinh(x) - 3*cosh(x) + 5)/(a*cosh(x)^5 +
5*a*cosh(x)*sinh(x)^4 + a*sinh(x)^5 + 5*a*cosh(x)^3 + (10*a*cosh(x)^2 + 3*a)*sinh(x)^3 + 5*(2*a*cosh(x)^3 + 3*
a*cosh(x))*sinh(x)^2 + 10*a*cosh(x) + (5*a*cosh(x)^4 + 9*a*cosh(x)^2 + 2*a)*sinh(x))

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giac [A]  time = 0.15, size = 48, normalized size = 1.60 \[ -\frac {2 \, {\left (3 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 3 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4 \, e^{\left (-x\right )} - 4 \, e^{x} + 6\right )}}{3 \, a {\left (e^{\left (-x\right )} + e^{x}\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(a+a*cosh(x)),x, algorithm="giac")

[Out]

-2/3*(3*(e^(-x) + e^x)^3 - 3*(e^(-x) + e^x)^2 - 4*e^(-x) - 4*e^x + 6)/(a*(e^(-x) + e^x)^4)

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maple [A]  time = 0.11, size = 30, normalized size = 1.00 \[ \frac {\frac {1}{3 \cosh \relax (x )^{3}}-\frac {1}{4 \cosh \relax (x )^{4}}-\frac {1}{\cosh \relax (x )}+\frac {1}{2 \cosh \relax (x )^{2}}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^5/(a+a*cosh(x)),x)

[Out]

1/a*(1/3/cosh(x)^3-1/4/cosh(x)^4-1/cosh(x)+1/2/cosh(x)^2)

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maxima [B]  time = 0.37, size = 223, normalized size = 7.43 \[ -\frac {2 \, e^{\left (-x\right )}}{4 \, a e^{\left (-2 \, x\right )} + 6 \, a e^{\left (-4 \, x\right )} + 4 \, a e^{\left (-6 \, x\right )} + a e^{\left (-8 \, x\right )} + a} + \frac {2 \, e^{\left (-2 \, x\right )}}{4 \, a e^{\left (-2 \, x\right )} + 6 \, a e^{\left (-4 \, x\right )} + 4 \, a e^{\left (-6 \, x\right )} + a e^{\left (-8 \, x\right )} + a} - \frac {10 \, e^{\left (-3 \, x\right )}}{3 \, {\left (4 \, a e^{\left (-2 \, x\right )} + 6 \, a e^{\left (-4 \, x\right )} + 4 \, a e^{\left (-6 \, x\right )} + a e^{\left (-8 \, x\right )} + a\right )}} - \frac {10 \, e^{\left (-5 \, x\right )}}{3 \, {\left (4 \, a e^{\left (-2 \, x\right )} + 6 \, a e^{\left (-4 \, x\right )} + 4 \, a e^{\left (-6 \, x\right )} + a e^{\left (-8 \, x\right )} + a\right )}} + \frac {2 \, e^{\left (-6 \, x\right )}}{4 \, a e^{\left (-2 \, x\right )} + 6 \, a e^{\left (-4 \, x\right )} + 4 \, a e^{\left (-6 \, x\right )} + a e^{\left (-8 \, x\right )} + a} - \frac {2 \, e^{\left (-7 \, x\right )}}{4 \, a e^{\left (-2 \, x\right )} + 6 \, a e^{\left (-4 \, x\right )} + 4 \, a e^{\left (-6 \, x\right )} + a e^{\left (-8 \, x\right )} + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(a+a*cosh(x)),x, algorithm="maxima")

[Out]

-2*e^(-x)/(4*a*e^(-2*x) + 6*a*e^(-4*x) + 4*a*e^(-6*x) + a*e^(-8*x) + a) + 2*e^(-2*x)/(4*a*e^(-2*x) + 6*a*e^(-4
*x) + 4*a*e^(-6*x) + a*e^(-8*x) + a) - 10/3*e^(-3*x)/(4*a*e^(-2*x) + 6*a*e^(-4*x) + 4*a*e^(-6*x) + a*e^(-8*x)
+ a) - 10/3*e^(-5*x)/(4*a*e^(-2*x) + 6*a*e^(-4*x) + 4*a*e^(-6*x) + a*e^(-8*x) + a) + 2*e^(-6*x)/(4*a*e^(-2*x)
+ 6*a*e^(-4*x) + 4*a*e^(-6*x) + a*e^(-8*x) + a) - 2*e^(-7*x)/(4*a*e^(-2*x) + 6*a*e^(-4*x) + 4*a*e^(-6*x) + a*e
^(-8*x) + a)

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mupad [B]  time = 1.01, size = 117, normalized size = 3.90 \[ \frac {\frac {8}{a}-\frac {8\,{\mathrm {e}}^x}{3\,a}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {\frac {6}{a}-\frac {8\,{\mathrm {e}}^x}{3\,a}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {\frac {2}{a}-\frac {2\,{\mathrm {e}}^x}{a}}{{\mathrm {e}}^{2\,x}+1}-\frac {4}{a\,\left (4\,{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^5/(a + a*cosh(x)),x)

[Out]

(8/a - (8*exp(x))/(3*a))/(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1) - (6/a - (8*exp(x))/(3*a))/(2*exp(2*x) + exp
(4*x) + 1) + (2/a - (2*exp(x))/a)/(exp(2*x) + 1) - 4/(a*(4*exp(2*x) + 6*exp(4*x) + 4*exp(6*x) + exp(8*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tanh ^{5}{\relax (x )}}{\cosh {\relax (x )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**5/(a+a*cosh(x)),x)

[Out]

Integral(tanh(x)**5/(cosh(x) + 1), x)/a

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