3.104 \(\int \frac{\tanh ^5(x)}{a+a \text{sech}(x)} \, dx\)

Optimal. Leaf size=36 \[ -\frac{\text{sech}^3(x)}{3 a}+\frac{\text{sech}^2(x)}{2 a}+\frac{\text{sech}(x)}{a}+\frac{\log (\cosh (x))}{a} \]

[Out]

Log[Cosh[x]]/a + Sech[x]/a + Sech[x]^2/(2*a) - Sech[x]^3/(3*a)

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Rubi [A]  time = 0.0591837, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3879, 75} \[ -\frac{\text{sech}^3(x)}{3 a}+\frac{\text{sech}^2(x)}{2 a}+\frac{\text{sech}(x)}{a}+\frac{\log (\cosh (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^5/(a + a*Sech[x]),x]

[Out]

Log[Cosh[x]]/a + Sech[x]/a + Sech[x]^2/(2*a) - Sech[x]^3/(3*a)

Rule 3879

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[((a - b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x^(m + n), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

Mathematica [A]  time = 0.0714647, size = 38, normalized size = 1.06 \[ \frac{\text{sech}^3(x) (6 \cosh (2 x)+3 \cosh (3 x) \log (\cosh (x))+\cosh (x) (9 \log (\cosh (x))+6)+2)}{12 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2 + 6*Cosh[2*x] + 3*Cosh[3*x]*Log[Cosh[x]] + Cosh[x]*(6 + 9*Log[Cosh[x]]))*Sech[x]^3)/(12*a)

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Maple [A]  time = 0.037, size = 34, normalized size = 0.9 \begin{align*} -{\frac{ \left ({\rm sech} \left (x\right ) \right ) ^{3}}{3\,a}}+{\frac{ \left ({\rm sech} \left (x\right ) \right ) ^{2}}{2\,a}}+{\frac{{\rm sech} \left (x\right )}{a}}-{\frac{\ln \left ({\rm sech} \left (x\right ) \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sech(x)^3/a+1/2*sech(x)^2/a+sech(x)/a-1/a*ln(sech(x))

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Maxima [B]  time = 1.50146, size = 100, normalized size = 2.78 \begin{align*} \frac{x}{a} + \frac{2 \,{\left (3 \, e^{\left (-x\right )} + 3 \, e^{\left (-2 \, x\right )} + 2 \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + 3 \, e^{\left (-5 \, x\right )}\right )}}{3 \,{\left (3 \, a e^{\left (-2 \, x\right )} + 3 \, a e^{\left (-4 \, x\right )} + a e^{\left (-6 \, x\right )} + a\right )}} + \frac{\log \left (e^{\left (-2 \, x\right )} + 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.56271, size = 1403, normalized size = 38.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*x*cosh(x)^6 + 3*x*sinh(x)^6 + 6*(3*x*cosh(x) - 1)*sinh(x)^5 + 3*(3*x - 2)*cosh(x)^4 - 6*cosh(x)^5 + 3*
(15*x*cosh(x)^2 + 3*x - 10*cosh(x) - 2)*sinh(x)^4 + 4*(15*x*cosh(x)^3 + 3*(3*x - 2)*cosh(x) - 15*cosh(x)^2 - 1
)*sinh(x)^3 + 3*(3*x - 2)*cosh(x)^2 - 4*cosh(x)^3 + 3*(15*x*cosh(x)^4 + 6*(3*x - 2)*cosh(x)^2 - 20*cosh(x)^3 +
3*x - 4*cosh(x) - 2)*sinh(x)^2 - 3*(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 3*(5*cosh(x)^2 + 1)*sinh(x)
^4 + 3*cosh(x)^4 + 4*(5*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 3*(5*cosh(x)^4 + 6*cosh(x)^2 + 1)*sinh(x)^2 + 3*cos
h(x)^2 + 6*(cosh(x)^5 + 2*cosh(x)^3 + cosh(x))*sinh(x) + 1)*log(2*cosh(x)/(cosh(x) - sinh(x))) + 6*(3*x*cosh(x
)^5 + 2*(3*x - 2)*cosh(x)^3 - 5*cosh(x)^4 + (3*x - 2)*cosh(x) - 2*cosh(x)^2 - 1)*sinh(x) + 3*x - 6*cosh(x))/(a
*cosh(x)^6 + 6*a*cosh(x)*sinh(x)^5 + a*sinh(x)^6 + 3*a*cosh(x)^4 + 3*(5*a*cosh(x)^2 + a)*sinh(x)^4 + 4*(5*a*co
sh(x)^3 + 3*a*cosh(x))*sinh(x)^3 + 3*a*cosh(x)^2 + 3*(5*a*cosh(x)^4 + 6*a*cosh(x)^2 + a)*sinh(x)^2 + 6*(a*cosh
(x)^5 + 2*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 ^{5}{\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(tanh(x)**5/(a+a*sech(x)),x)

[Out]

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

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Giac [A]  time = 1.15256, size = 82, normalized size = 2.28 \begin{align*} \frac{\log \left (e^{\left (-x\right )} + e^{x}\right )}{a} - \frac{11 \,{\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 12 \,{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 12 \, e^{\left (-x\right )} - 12 \, e^{x} + 16}{6 \, a{\left (e^{\left (-x\right )} + e^{x}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e^(-x) + e^x)/a - 1/6*(11*(e^(-x) + e^x)^3 - 12*(e^(-x) + e^x)^2 - 12*e^(-x) - 12*e^x + 16)/(a*(e^(-x) + e
^x)^3)