3.190 \(\int \frac{\tanh ^3(x)}{a+a \cosh (x)} \, dx\)

Optimal. Leaf size=19 \[ \frac{\text{sech}^2(x)}{2 a}-\frac{\text{sech}(x)}{a} \]

[Out]

-(Sech[x]/a) + Sech[x]^2/(2*a)

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Rubi [A]  time = 0.0659567, antiderivative size = 19, normalized size of antiderivative = 1., 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, 2606, 30, 8} \[ \frac{\text{sech}^2(x)}{2 a}-\frac{\text{sech}(x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(a + a*Cosh[x]),x]

[Out]

-(Sech[x]/a) + Sech[x]^2/(2*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 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.0213752, size = 17, normalized size = 0.89 \[ \frac{2 \sinh ^4\left (\frac{x}{2}\right ) \text{sech}^2(x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sech[x]^2*Sinh[x/2]^4)/a

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Maple [A]  time = 0.026, size = 18, normalized size = 1. \begin{align*}{\frac{1}{a} \left ( - \left ( \cosh \left ( x \right ) \right ) ^{-1}+{\frac{1}{2\, \left ( \cosh \left ( x \right ) \right ) ^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.07107, size = 95, normalized size = 5. \begin{align*} -\frac{2 \, e^{\left (-x\right )}}{2 \, a e^{\left (-2 \, x\right )} + a e^{\left (-4 \, x\right )} + a} + \frac{2 \, e^{\left (-2 \, x\right )}}{2 \, a e^{\left (-2 \, x\right )} + a e^{\left (-4 \, x\right )} + a} - \frac{2 \, e^{\left (-3 \, x\right )}}{2 \, a e^{\left (-2 \, x\right )} + a e^{\left (-4 \, x\right )} + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.82161, size = 221, normalized size = 11.63 \begin{align*} -\frac{2 \,{\left (\cosh \left (x\right )^{2} +{\left (2 \, \cosh \left (x\right ) - 1\right )} \sinh \left (x\right ) + \sinh \left (x\right )^{2} - \cosh \left (x\right ) + 1\right )}}{a \cosh \left (x\right )^{3} + 3 \, a \cosh \left (x\right ) \sinh \left (x\right )^{2} + a \sinh \left (x\right )^{3} + 3 \, a \cosh \left (x\right ) +{\left (3 \, a \cosh \left (x\right )^{2} + a\right )} \sinh \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(cosh(x)^2 + (2*cosh(x) - 1)*sinh(x) + sinh(x)^2 - cosh(x) + 1)/(a*cosh(x)^3 + 3*a*cosh(x)*sinh(x)^2 + a*si
nh(x)^3 + 3*a*cosh(x) + (3*a*cosh(x)^2 + a)*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tanh ^{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(tanh(x)**3/(a+a*cosh(x)),x)

[Out]

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

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Giac [A]  time = 1.15578, size = 30, normalized size = 1.58 \begin{align*} -\frac{2 \,{\left (e^{\left (-x\right )} + e^{x} - 1\right )}}{a{\left (e^{\left (-x\right )} + e^{x}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(e^(-x) + e^x - 1)/(a*(e^(-x) + e^x)^2)