3.1018 \(\int \text{sech}(x) (5-11 \text{sech}^2(x)) \tanh (x) \, dx\)

Optimal. Leaf size=13 \[ \frac{11 \text{sech}^3(x)}{3}-5 \text{sech}(x) \]

[Out]

-5*Sech[x] + (11*Sech[x]^3)/3

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Rubi [A]  time = 0.035788, antiderivative size = 13, 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 = {4339, 14} \[ \frac{11 \text{sech}^3(x)}{3}-5 \text{sech}(x) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]*(5 - 11*Sech[x]^2)*Tanh[x],x]

[Out]

-5*Sech[x] + (11*Sech[x]^3)/3

Rule 4339

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[(b*
c)^(-1), Subst[Int[SubstFor[1/x, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[
c*(a + b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Tan] || EqQ[F, tan])

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 \text{sech}(x) \left (5-11 \text{sech}^2(x)\right ) \tanh (x) \, dx &=\operatorname{Subst}\left (\int \frac{-11+5 x^2}{x^4} \, dx,x,\cosh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{11}{x^4}+\frac{5}{x^2}\right ) \, dx,x,\cosh (x)\right )\\ &=-5 \text{sech}(x)+\frac{11 \text{sech}^3(x)}{3}\\ \end{align*}

Mathematica [A]  time = 0.007138, size = 13, normalized size = 1. \[ \frac{11 \text{sech}^3(x)}{3}-5 \text{sech}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]*(5 - 11*Sech[x]^2)*Tanh[x],x]

[Out]

-5*Sech[x] + (11*Sech[x]^3)/3

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Maple [A]  time = 0.013, size = 12, normalized size = 0.9 \begin{align*} -5\,{\rm sech} \left (x\right )+{\frac{11\, \left ({\rm sech} \left (x\right ) \right ) ^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)*(5-11*sech(x)^2)*tanh(x),x)

[Out]

-5*sech(x)+11/3*sech(x)^3

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Maxima [B]  time = 1.03603, size = 31, normalized size = 2.38 \begin{align*} -\frac{10}{e^{\left (-x\right )} + e^{x}} + \frac{88}{3 \,{\left (e^{\left (-x\right )} + e^{x}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)*(5-11*sech(x)^2)*tanh(x),x, algorithm="maxima")

[Out]

-10/(e^(-x) + e^x) + 88/3/(e^(-x) + e^x)^3

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Fricas [B]  time = 2.02602, size = 309, normalized size = 23.77 \begin{align*} -\frac{2 \,{\left (15 \, \cosh \left (x\right )^{3} + 45 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + 15 \, \sinh \left (x\right )^{3} +{\left (45 \, \cosh \left (x\right )^{2} - 29\right )} \sinh \left (x\right ) + \cosh \left (x\right )\right )}}{3 \,{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} + 2\right )} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)*(5-11*sech(x)^2)*tanh(x),x, algorithm="fricas")

[Out]

-2/3*(15*cosh(x)^3 + 45*cosh(x)*sinh(x)^2 + 15*sinh(x)^3 + (45*cosh(x)^2 - 29)*sinh(x) + cosh(x))/(cosh(x)^4 +
4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 + 2)*sinh(x)^2 + 4*cosh(x)^2 + 4*(cosh(x)^3 + cosh(x))*sinh(
x) + 3)

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Sympy [A]  time = 0.741429, size = 12, normalized size = 0.92 \begin{align*} \frac{11 \operatorname{sech}^{3}{\left (x \right )}}{3} - 5 \operatorname{sech}{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)*(5-11*sech(x)**2)*tanh(x),x)

[Out]

11*sech(x)**3/3 - 5*sech(x)

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Giac [B]  time = 1.11225, size = 32, normalized size = 2.46 \begin{align*} -\frac{2 \,{\left (15 \,{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 44\right )}}{3 \,{\left (e^{\left (-x\right )} + e^{x}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)*(5-11*sech(x)^2)*tanh(x),x, algorithm="giac")

[Out]

-2/3*(15*(e^(-x) + e^x)^2 - 44)/(e^(-x) + e^x)^3