3.50 \(\int \frac {1}{(a \text {sech}^4(x))^{3/2}} \, dx\)

Optimal. Leaf size=86 \[ \frac {5 x \text {sech}^2(x)}{16 a \sqrt {a \text {sech}^4(x)}}+\frac {5 \tanh (x)}{16 a \sqrt {a \text {sech}^4(x)}}+\frac {\sinh (x) \cosh ^3(x)}{6 a \sqrt {a \text {sech}^4(x)}}+\frac {5 \sinh (x) \cosh (x)}{24 a \sqrt {a \text {sech}^4(x)}} \]

[Out]

5/16*x*sech(x)^2/a/(a*sech(x)^4)^(1/2)+5/24*cosh(x)*sinh(x)/a/(a*sech(x)^4)^(1/2)+1/6*cosh(x)^3*sinh(x)/a/(a*s
ech(x)^4)^(1/2)+5/16*tanh(x)/a/(a*sech(x)^4)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4123, 2635, 8} \[ \frac {5 x \text {sech}^2(x)}{16 a \sqrt {a \text {sech}^4(x)}}+\frac {5 \tanh (x)}{16 a \sqrt {a \text {sech}^4(x)}}+\frac {\sinh (x) \cosh ^3(x)}{6 a \sqrt {a \text {sech}^4(x)}}+\frac {5 \sinh (x) \cosh (x)}{24 a \sqrt {a \text {sech}^4(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x*Sech[x]^2)/(16*a*Sqrt[a*Sech[x]^4]) + (5*Cosh[x]*Sinh[x])/(24*a*Sqrt[a*Sech[x]^4]) + (Cosh[x]^3*Sinh[x])/
(6*a*Sqrt[a*Sech[x]^4]) + (5*Tanh[x])/(16*a*Sqrt[a*Sech[x]^4])

Rule 8

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a \text {sech}^4(x)\right )^{3/2}} \, dx &=\frac {\text {sech}^2(x) \int \cosh ^6(x) \, dx}{a \sqrt {a \text {sech}^4(x)}}\\ &=\frac {\cosh ^3(x) \sinh (x)}{6 a \sqrt {a \text {sech}^4(x)}}+\frac {\left (5 \text {sech}^2(x)\right ) \int \cosh ^4(x) \, dx}{6 a \sqrt {a \text {sech}^4(x)}}\\ &=\frac {5 \cosh (x) \sinh (x)}{24 a \sqrt {a \text {sech}^4(x)}}+\frac {\cosh ^3(x) \sinh (x)}{6 a \sqrt {a \text {sech}^4(x)}}+\frac {\left (5 \text {sech}^2(x)\right ) \int \cosh ^2(x) \, dx}{8 a \sqrt {a \text {sech}^4(x)}}\\ &=\frac {5 \cosh (x) \sinh (x)}{24 a \sqrt {a \text {sech}^4(x)}}+\frac {\cosh ^3(x) \sinh (x)}{6 a \sqrt {a \text {sech}^4(x)}}+\frac {5 \tanh (x)}{16 a \sqrt {a \text {sech}^4(x)}}+\frac {\left (5 \text {sech}^2(x)\right ) \int 1 \, dx}{16 a \sqrt {a \text {sech}^4(x)}}\\ &=\frac {5 x \text {sech}^2(x)}{16 a \sqrt {a \text {sech}^4(x)}}+\frac {5 \cosh (x) \sinh (x)}{24 a \sqrt {a \text {sech}^4(x)}}+\frac {\cosh ^3(x) \sinh (x)}{6 a \sqrt {a \text {sech}^4(x)}}+\frac {5 \tanh (x)}{16 a \sqrt {a \text {sech}^4(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 38, normalized size = 0.44 \[ \frac {(60 x+45 \sinh (2 x)+9 \sinh (4 x)+\sinh (6 x)) \text {sech}^6(x)}{192 \left (a \text {sech}^4(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sech[x]^6*(60*x + 45*Sinh[2*x] + 9*Sinh[4*x] + Sinh[6*x]))/(192*(a*Sech[x]^4)^(3/2))

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fricas [B]  time = 0.45, size = 1141, normalized size = 13.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)^4)^(3/2),x, algorithm="fricas")

[Out]

1/384*((e^(4*x) + 2*e^(2*x) + 1)*sinh(x)^12 + cosh(x)^12 + 12*(cosh(x)*e^(4*x) + 2*cosh(x)*e^(2*x) + cosh(x))*
sinh(x)^11 + 3*(22*cosh(x)^2 + (22*cosh(x)^2 + 3)*e^(4*x) + 2*(22*cosh(x)^2 + 3)*e^(2*x) + 3)*sinh(x)^10 + 9*c
osh(x)^10 + 10*(22*cosh(x)^3 + (22*cosh(x)^3 + 9*cosh(x))*e^(4*x) + 2*(22*cosh(x)^3 + 9*cosh(x))*e^(2*x) + 9*c
osh(x))*sinh(x)^9 + 45*(11*cosh(x)^4 + 9*cosh(x)^2 + (11*cosh(x)^4 + 9*cosh(x)^2 + 1)*e^(4*x) + 2*(11*cosh(x)^
4 + 9*cosh(x)^2 + 1)*e^(2*x) + 1)*sinh(x)^8 + 45*cosh(x)^8 + 72*(11*cosh(x)^5 + 15*cosh(x)^3 + (11*cosh(x)^5 +
15*cosh(x)^3 + 5*cosh(x))*e^(4*x) + 2*(11*cosh(x)^5 + 15*cosh(x)^3 + 5*cosh(x))*e^(2*x) + 5*cosh(x))*sinh(x)^
7 + 120*x*cosh(x)^6 + 6*(154*cosh(x)^6 + 315*cosh(x)^4 + 210*cosh(x)^2 + (154*cosh(x)^6 + 315*cosh(x)^4 + 210*
cosh(x)^2 + 20*x)*e^(4*x) + 2*(154*cosh(x)^6 + 315*cosh(x)^4 + 210*cosh(x)^2 + 20*x)*e^(2*x) + 20*x)*sinh(x)^6
+ 36*(22*cosh(x)^7 + 63*cosh(x)^5 + 70*cosh(x)^3 + 20*x*cosh(x) + (22*cosh(x)^7 + 63*cosh(x)^5 + 70*cosh(x)^3
+ 20*x*cosh(x))*e^(4*x) + 2*(22*cosh(x)^7 + 63*cosh(x)^5 + 70*cosh(x)^3 + 20*x*cosh(x))*e^(2*x))*sinh(x)^5 +
45*(11*cosh(x)^8 + 42*cosh(x)^6 + 70*cosh(x)^4 + 40*x*cosh(x)^2 + (11*cosh(x)^8 + 42*cosh(x)^6 + 70*cosh(x)^4
+ 40*x*cosh(x)^2 - 1)*e^(4*x) + 2*(11*cosh(x)^8 + 42*cosh(x)^6 + 70*cosh(x)^4 + 40*x*cosh(x)^2 - 1)*e^(2*x) -
1)*sinh(x)^4 - 45*cosh(x)^4 + 20*(11*cosh(x)^9 + 54*cosh(x)^7 + 126*cosh(x)^5 + 120*x*cosh(x)^3 + (11*cosh(x)^
9 + 54*cosh(x)^7 + 126*cosh(x)^5 + 120*x*cosh(x)^3 - 9*cosh(x))*e^(4*x) + 2*(11*cosh(x)^9 + 54*cosh(x)^7 + 126
*cosh(x)^5 + 120*x*cosh(x)^3 - 9*cosh(x))*e^(2*x) - 9*cosh(x))*sinh(x)^3 + 3*(22*cosh(x)^10 + 135*cosh(x)^8 +
420*cosh(x)^6 + 600*x*cosh(x)^4 - 90*cosh(x)^2 + (22*cosh(x)^10 + 135*cosh(x)^8 + 420*cosh(x)^6 + 600*x*cosh(x
)^4 - 90*cosh(x)^2 - 3)*e^(4*x) + 2*(22*cosh(x)^10 + 135*cosh(x)^8 + 420*cosh(x)^6 + 600*x*cosh(x)^4 - 90*cosh
(x)^2 - 3)*e^(2*x) - 3)*sinh(x)^2 - 9*cosh(x)^2 + (cosh(x)^12 + 9*cosh(x)^10 + 45*cosh(x)^8 + 120*x*cosh(x)^6
- 45*cosh(x)^4 - 9*cosh(x)^2 - 1)*e^(4*x) + 2*(cosh(x)^12 + 9*cosh(x)^10 + 45*cosh(x)^8 + 120*x*cosh(x)^6 - 45
*cosh(x)^4 - 9*cosh(x)^2 - 1)*e^(2*x) + 6*(2*cosh(x)^11 + 15*cosh(x)^9 + 60*cosh(x)^7 + 120*x*cosh(x)^5 - 30*c
osh(x)^3 + (2*cosh(x)^11 + 15*cosh(x)^9 + 60*cosh(x)^7 + 120*x*cosh(x)^5 - 30*cosh(x)^3 - 3*cosh(x))*e^(4*x) +
2*(2*cosh(x)^11 + 15*cosh(x)^9 + 60*cosh(x)^7 + 120*x*cosh(x)^5 - 30*cosh(x)^3 - 3*cosh(x))*e^(2*x) - 3*cosh(
x))*sinh(x) - 1)*sqrt(a/(e^(8*x) + 4*e^(6*x) + 6*e^(4*x) + 4*e^(2*x) + 1))*e^(2*x)/(a^2*cosh(x)^6*e^(2*x) + 6*
a^2*cosh(x)^5*e^(2*x)*sinh(x) + 15*a^2*cosh(x)^4*e^(2*x)*sinh(x)^2 + 20*a^2*cosh(x)^3*e^(2*x)*sinh(x)^3 + 15*a
^2*cosh(x)^2*e^(2*x)*sinh(x)^4 + 6*a^2*cosh(x)*e^(2*x)*sinh(x)^5 + a^2*e^(2*x)*sinh(x)^6)

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giac [A]  time = 0.13, size = 52, normalized size = 0.60 \[ -\frac {{\left (110 \, e^{\left (6 \, x\right )} + 45 \, e^{\left (4 \, x\right )} + 9 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-6 \, x\right )} - 120 \, x - e^{\left (6 \, x\right )} - 9 \, e^{\left (4 \, x\right )} - 45 \, e^{\left (2 \, x\right )}}{384 \, a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)^4)^(3/2),x, algorithm="giac")

[Out]

-1/384*((110*e^(6*x) + 45*e^(4*x) + 9*e^(2*x) + 1)*e^(-6*x) - 120*x - e^(6*x) - 9*e^(4*x) - 45*e^(2*x))/a^(3/2
)

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maple [B]  time = 0.21, size = 230, normalized size = 2.67 \[ \frac {5 \,{\mathrm e}^{2 x} x}{16 a \left (1+{\mathrm e}^{2 x}\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}}+\frac {{\mathrm e}^{8 x}}{384 a \left (1+{\mathrm e}^{2 x}\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}}+\frac {3 \,{\mathrm e}^{6 x}}{128 a \left (1+{\mathrm e}^{2 x}\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}}+\frac {15 \,{\mathrm e}^{4 x}}{128 a \left (1+{\mathrm e}^{2 x}\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}}-\frac {15}{128 \sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}\, \left (1+{\mathrm e}^{2 x}\right )^{2} a}-\frac {3 \,{\mathrm e}^{-2 x}}{128 a \left (1+{\mathrm e}^{2 x}\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}}-\frac {{\mathrm e}^{-4 x}}{384 a \left (1+{\mathrm e}^{2 x}\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sech(x)^4)^(3/2),x)

[Out]

5/16/a*exp(2*x)/(1+exp(2*x))^2/(a*exp(4*x)/(1+exp(2*x))^4)^(1/2)*x+1/384/a*exp(8*x)/(1+exp(2*x))^2/(a*exp(4*x)
/(1+exp(2*x))^4)^(1/2)+3/128/a*exp(6*x)/(1+exp(2*x))^2/(a*exp(4*x)/(1+exp(2*x))^4)^(1/2)+15/128/a*exp(4*x)/(1+
exp(2*x))^2/(a*exp(4*x)/(1+exp(2*x))^4)^(1/2)-15/128/(a*exp(4*x)/(1+exp(2*x))^4)^(1/2)/(1+exp(2*x))^2/a-3/128/
a*exp(-2*x)/(1+exp(2*x))^2/(a*exp(4*x)/(1+exp(2*x))^4)^(1/2)-1/384/a*exp(-4*x)/(1+exp(2*x))^2/(a*exp(4*x)/(1+e
xp(2*x))^4)^(1/2)

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maxima [A]  time = 0.47, size = 65, normalized size = 0.76 \[ \frac {{\left (9 \, \sqrt {a} e^{\left (-2 \, x\right )} + 45 \, \sqrt {a} e^{\left (-4 \, x\right )} - 45 \, \sqrt {a} e^{\left (-8 \, x\right )} - 9 \, \sqrt {a} e^{\left (-10 \, x\right )} - \sqrt {a} e^{\left (-12 \, x\right )} + \sqrt {a}\right )} e^{\left (6 \, x\right )}}{384 \, a^{2}} + \frac {5 \, x}{16 \, a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)^4)^(3/2),x, algorithm="maxima")

[Out]

1/384*(9*sqrt(a)*e^(-2*x) + 45*sqrt(a)*e^(-4*x) - 45*sqrt(a)*e^(-8*x) - 9*sqrt(a)*e^(-10*x) - sqrt(a)*e^(-12*x
) + sqrt(a))*e^(6*x)/a^2 + 5/16*x/a^(3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (\frac {a}{{\mathrm {cosh}\relax (x)}^4}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a/cosh(x)^4)^(3/2),x)

[Out]

int(1/(a/cosh(x)^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \operatorname {sech}^{4}{\relax (x )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)**4)**(3/2),x)

[Out]

Integral((a*sech(x)**4)**(-3/2), x)

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