3.21 \(\int \frac {1}{(a \cosh (x))^{5/2}} \, dx\)

Optimal. Leaf size=50 \[ \frac {2 \sinh (x)}{3 a (a \cosh (x))^{3/2}}-\frac {2 i \sqrt {\cosh (x)} F\left (\left .\frac {i x}{2}\right |2\right )}{3 a^2 \sqrt {a \cosh (x)}} \]

[Out]

2/3*sinh(x)/a/(a*cosh(x))^(3/2)-2/3*I*(cosh(1/2*x)^2)^(1/2)/cosh(1/2*x)*EllipticF(I*sinh(1/2*x),2^(1/2))*cosh(
x)^(1/2)/a^2/(a*cosh(x))^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2636, 2642, 2641} \[ \frac {2 \sinh (x)}{3 a (a \cosh (x))^{3/2}}-\frac {2 i \sqrt {\cosh (x)} F\left (\left .\frac {i x}{2}\right |2\right )}{3 a^2 \sqrt {a \cosh (x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cosh[x])^(-5/2),x]

[Out]

(((-2*I)/3)*Sqrt[Cosh[x]]*EllipticF[(I/2)*x, 2])/(a^2*Sqrt[a*Cosh[x]]) + (2*Sinh[x])/(3*a*(a*Cosh[x])^(3/2))

Rule 2636

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

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

\begin {align*} \int \frac {1}{(a \cosh (x))^{5/2}} \, dx &=\frac {2 \sinh (x)}{3 a (a \cosh (x))^{3/2}}+\frac {\int \frac {1}{\sqrt {a \cosh (x)}} \, dx}{3 a^2}\\ &=\frac {2 \sinh (x)}{3 a (a \cosh (x))^{3/2}}+\frac {\sqrt {\cosh (x)} \int \frac {1}{\sqrt {\cosh (x)}} \, dx}{3 a^2 \sqrt {a \cosh (x)}}\\ &=-\frac {2 i \sqrt {\cosh (x)} F\left (\left .\frac {i x}{2}\right |2\right )}{3 a^2 \sqrt {a \cosh (x)}}+\frac {2 \sinh (x)}{3 a (a \cosh (x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 56, normalized size = 1.12 \[ \frac {2 \left (\sqrt {\sinh (2 x)+\cosh (2 x)+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\cosh (2 x)-\sinh (2 x)\right )+\tanh (x)\right )}{3 a^2 \sqrt {a \cosh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cosh[x])^(-5/2),x]

[Out]

(2*(Hypergeometric2F1[1/4, 1/2, 5/4, -Cosh[2*x] - Sinh[2*x]]*Sqrt[1 + Cosh[2*x] + Sinh[2*x]] + Tanh[x]))/(3*a^
2*Sqrt[a*Cosh[x]])

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \cosh \relax (x)}}{a^{3} \cosh \relax (x)^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*cosh(x))/(a^3*cosh(x)^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \cosh \relax (x)\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x))^(5/2),x, algorithm="giac")

[Out]

integrate((a*cosh(x))^(-5/2), x)

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maple [B]  time = 0.34, size = 177, normalized size = 3.54 \[ \frac {\left (2 \sqrt {-2 \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )-1}\, \sqrt {-\left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \EllipticF \left (\sqrt {2}\, \cosh \left (\frac {x}{2}\right ), \frac {\sqrt {2}}{2}\right ) \sqrt {2}\, \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )+\sqrt {2}\, \sqrt {-2 \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )-1}\, \sqrt {-\left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \EllipticF \left (\sqrt {2}\, \cosh \left (\frac {x}{2}\right ), \frac {\sqrt {2}}{2}\right )+4 \left (\sinh ^{2}\left (\frac {x}{2}\right )\right ) \cosh \left (\frac {x}{2}\right )\right ) \sqrt {a \left (2 \left (\cosh ^{2}\left (\frac {x}{2}\right )\right )-1\right ) \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}}{3 a^{2} \sqrt {a \left (2 \left (\sinh ^{4}\left (\frac {x}{2}\right )\right )+\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \left (2 \left (\cosh ^{2}\left (\frac {x}{2}\right )\right )-1\right ) \sinh \left (\frac {x}{2}\right ) \sqrt {a \left (2 \left (\cosh ^{2}\left (\frac {x}{2}\right )\right )-1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x))^(5/2),x)

[Out]

1/3*(2*(-2*sinh(1/2*x)^2-1)^(1/2)*(-sinh(1/2*x)^2)^(1/2)*EllipticF(2^(1/2)*cosh(1/2*x),1/2*2^(1/2))*2^(1/2)*si
nh(1/2*x)^2+2^(1/2)*(-2*sinh(1/2*x)^2-1)^(1/2)*(-sinh(1/2*x)^2)^(1/2)*EllipticF(2^(1/2)*cosh(1/2*x),1/2*2^(1/2
))+4*sinh(1/2*x)^2*cosh(1/2*x))/a^2*(a*(2*cosh(1/2*x)^2-1)*sinh(1/2*x)^2)^(1/2)/(a*(2*sinh(1/2*x)^4+sinh(1/2*x
)^2))^(1/2)/(2*cosh(1/2*x)^2-1)/sinh(1/2*x)/(a*(2*cosh(1/2*x)^2-1))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \cosh \relax (x)\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*cosh(x))^(-5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (a\,\mathrm {cosh}\relax (x)\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x))^(5/2),x)

[Out]

int(1/(a*cosh(x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \cosh {\relax (x )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x))**(5/2),x)

[Out]

Integral((a*cosh(x))**(-5/2), x)

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