3.47 \(\int (a \text{sech}^4(x))^{3/2} \, dx\)

Optimal. Leaf size=61 \[ a \sinh (x) \cosh (x) \sqrt{a \text{sech}^4(x)}+\frac{1}{5} a \sinh ^2(x) \tanh ^3(x) \sqrt{a \text{sech}^4(x)}-\frac{2}{3} a \sinh ^2(x) \tanh (x) \sqrt{a \text{sech}^4(x)} \]

[Out]

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

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Rubi [A]  time = 0.0234475, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4123, 3767} \[ a \sinh (x) \cosh (x) \sqrt{a \text{sech}^4(x)}+\frac{1}{5} a \sinh ^2(x) \tanh ^3(x) \sqrt{a \text{sech}^4(x)}-\frac{2}{3} a \sinh ^2(x) \tanh (x) \sqrt{a \text{sech}^4(x)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sech[x]^4)^(3/2),x]

[Out]

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

Rule 4123

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Sec[e + f*x])^n)^
FracPart[p])/(c*Sec[e + f*x])^(n*FracPart[p]), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \left (a \text{sech}^4(x)\right )^{3/2} \, dx &=\left (a \cosh ^2(x) \sqrt{a \text{sech}^4(x)}\right ) \int \text{sech}^6(x) \, dx\\ &=\left (i a \cosh ^2(x) \sqrt{a \text{sech}^4(x)}\right ) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-i \tanh (x)\right )\\ &=a \cosh (x) \sqrt{a \text{sech}^4(x)} \sinh (x)-\frac{2}{3} a \sqrt{a \text{sech}^4(x)} \sinh ^2(x) \tanh (x)+\frac{1}{5} a \sqrt{a \text{sech}^4(x)} \sinh ^2(x) \tanh ^3(x)\\ \end{align*}

Mathematica [A]  time = 0.0543789, size = 30, normalized size = 0.49 \[ \frac{1}{15} \sinh (x) \cosh (x) (6 \cosh (2 x)+\cosh (4 x)+8) \left (a \text{sech}^4(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sech[x]^4)^(3/2),x]

[Out]

(Cosh[x]*(8 + 6*Cosh[2*x] + Cosh[4*x])*(a*Sech[x]^4)^(3/2)*Sinh[x])/15

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Maple [A]  time = 0.06, size = 46, normalized size = 0.8 \begin{align*} -{\frac{16\,a{{\rm e}^{-2\,x}} \left ( 10\,{{\rm e}^{4\,x}}+5\,{{\rm e}^{2\,x}}+1 \right ) }{15\, \left ({{\rm e}^{2\,x}}+1 \right ) ^{3}}\sqrt{{\frac{a{{\rm e}^{4\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{4}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sech(x)^4)^(3/2),x)

[Out]

-16/15*a*exp(-2*x)/(exp(2*x)+1)^3*(a*exp(4*x)/(exp(2*x)+1)^4)^(1/2)*(10*exp(4*x)+5*exp(2*x)+1)

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Maxima [B]  time = 1.72302, size = 162, normalized size = 2.66 \begin{align*} \frac{16 \, a^{\frac{3}{2}} e^{\left (-2 \, x\right )}}{3 \,{\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} + \frac{32 \, a^{\frac{3}{2}} e^{\left (-4 \, x\right )}}{3 \,{\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} + \frac{16 \, a^{\frac{3}{2}}}{15 \,{\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sech(x)^4)^(3/2),x, algorithm="maxima")

[Out]

16/3*a^(3/2)*e^(-2*x)/(5*e^(-2*x) + 10*e^(-4*x) + 10*e^(-6*x) + 5*e^(-8*x) + e^(-10*x) + 1) + 32/3*a^(3/2)*e^(
-4*x)/(5*e^(-2*x) + 10*e^(-4*x) + 10*e^(-6*x) + 5*e^(-8*x) + e^(-10*x) + 1) + 16/15*a^(3/2)/(5*e^(-2*x) + 10*e
^(-4*x) + 10*e^(-6*x) + 5*e^(-8*x) + e^(-10*x) + 1)

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Fricas [B]  time = 2.03966, size = 1623, normalized size = 26.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sech(x)^4)^(3/2),x, algorithm="fricas")

[Out]

-16/15*(10*a*cosh(x)^4 + 10*(a*e^(4*x) + 2*a*e^(2*x) + a)*sinh(x)^4 + 40*(a*cosh(x)*e^(4*x) + 2*a*cosh(x)*e^(2
*x) + a*cosh(x))*sinh(x)^3 + 5*a*cosh(x)^2 + 5*(12*a*cosh(x)^2 + (12*a*cosh(x)^2 + a)*e^(4*x) + 2*(12*a*cosh(x
)^2 + a)*e^(2*x) + a)*sinh(x)^2 + (10*a*cosh(x)^4 + 5*a*cosh(x)^2 + a)*e^(4*x) + 2*(10*a*cosh(x)^4 + 5*a*cosh(
x)^2 + a)*e^(2*x) + 10*(4*a*cosh(x)^3 + a*cosh(x) + (4*a*cosh(x)^3 + a*cosh(x))*e^(4*x) + 2*(4*a*cosh(x)^3 + a
*cosh(x))*e^(2*x))*sinh(x) + a)*sqrt(a/(e^(8*x) + 4*e^(6*x) + 6*e^(4*x) + 4*e^(2*x) + 1))*e^(2*x)/(10*cosh(x)*
e^(2*x)*sinh(x)^9 + e^(2*x)*sinh(x)^10 + 5*(9*cosh(x)^2 + 1)*e^(2*x)*sinh(x)^8 + 40*(3*cosh(x)^3 + cosh(x))*e^
(2*x)*sinh(x)^7 + 10*(21*cosh(x)^4 + 14*cosh(x)^2 + 1)*e^(2*x)*sinh(x)^6 + 4*(63*cosh(x)^5 + 70*cosh(x)^3 + 15
*cosh(x))*e^(2*x)*sinh(x)^5 + 10*(21*cosh(x)^6 + 35*cosh(x)^4 + 15*cosh(x)^2 + 1)*e^(2*x)*sinh(x)^4 + 40*(3*co
sh(x)^7 + 7*cosh(x)^5 + 5*cosh(x)^3 + cosh(x))*e^(2*x)*sinh(x)^3 + 5*(9*cosh(x)^8 + 28*cosh(x)^6 + 30*cosh(x)^
4 + 12*cosh(x)^2 + 1)*e^(2*x)*sinh(x)^2 + 10*(cosh(x)^9 + 4*cosh(x)^7 + 6*cosh(x)^5 + 4*cosh(x)^3 + cosh(x))*e
^(2*x)*sinh(x) + (cosh(x)^10 + 5*cosh(x)^8 + 10*cosh(x)^6 + 10*cosh(x)^4 + 5*cosh(x)^2 + 1)*e^(2*x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \operatorname{sech}^{4}{\left (x \right )}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sech(x)**4)**(3/2),x)

[Out]

Integral((a*sech(x)**4)**(3/2), x)

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Giac [A]  time = 1.17369, size = 36, normalized size = 0.59 \begin{align*} -\frac{16 \, a^{\frac{3}{2}}{\left (10 \, e^{\left (4 \, x\right )} + 5 \, e^{\left (2 \, x\right )} + 1\right )}}{15 \,{\left (e^{\left (2 \, x\right )} + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sech(x)^4)^(3/2),x, algorithm="giac")

[Out]

-16/15*a^(3/2)*(10*e^(4*x) + 5*e^(2*x) + 1)/(e^(2*x) + 1)^5