3.1.13 \(\int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx\) [13]

Optimal. Leaf size=80 \[ -\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}+\frac {2 i F\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {i \sinh (a+b x)}}{3 b \sqrt {\sinh (a+b x)}} \]

[Out]

-2/3*cosh(b*x+a)/b/sinh(b*x+a)^(3/2)-2/3*I*(sin(1/2*I*a+1/4*Pi+1/2*I*b*x)^2)^(1/2)/sin(1/2*I*a+1/4*Pi+1/2*I*b*
x)*EllipticF(cos(1/2*I*a+1/4*Pi+1/2*I*b*x),2^(1/2))*(I*sinh(b*x+a))^(1/2)/b/sinh(b*x+a)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2716, 2721, 2720} \begin {gather*} -\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}+\frac {2 i \sqrt {i \sinh (a+b x)} F\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{3 b \sqrt {\sinh (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^(-5/2),x]

[Out]

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

Rule 2716

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 2720

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

Rule 2721

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.06, size = 86, normalized size = 1.08 \begin {gather*} -\frac {2 \left (\cosh (a+b x)+\, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\cosh (2 (a+b x))+\sinh (2 (a+b x))\right ) \sinh (a+b x) \sqrt {1-\cosh (2 a+2 b x)-\sinh (2 a+2 b x)}\right )}{3 b \sinh ^{\frac {3}{2}}(a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^(-5/2),x]

[Out]

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

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Maple [A]
time = 0.61, size = 101, normalized size = 1.26

method	result	size
default	\(-\frac {i \sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \sinh \left (b x +a \right )+2 \left (\cosh ^{2}\left (b x +a \right )\right )}{3 \sinh \left (b x +a \right )^{\frac {3}{2}} \cosh \left (b x +a \right ) b}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sinh(b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/sinh(b*x+a)^(3/2)*(I*(1-I*sinh(b*x+a))^(1/2)*2^(1/2)*(1+I*sinh(b*x+a))^(1/2)*(I*sinh(b*x+a))^(1/2)*Ellipt
icF((1-I*sinh(b*x+a))^(1/2),1/2*2^(1/2))*sinh(b*x+a)+2*cosh(b*x+a)^2)/cosh(b*x+a)/b

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x + a)^(-5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 314, normalized size = 3.92 \begin {gather*} -\frac {2 \, {\left ({\left (\sqrt {2} \cosh \left (b x + a\right )^{4} + 4 \, \sqrt {2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sqrt {2} \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \sqrt {2} \cosh \left (b x + a\right )^{2} - \sqrt {2}\right )} \sinh \left (b x + a\right )^{2} - 2 \, \sqrt {2} \cosh \left (b x + a\right )^{2} + 4 \, {\left (\sqrt {2} \cosh \left (b x + a\right )^{3} - \sqrt {2} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 2 \, {\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} + {\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) + \cosh \left (b x + a\right )\right )} \sqrt {\sinh \left (b x + a\right )}\right )}}{3 \, {\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*((sqrt(2)*cosh(b*x + a)^4 + 4*sqrt(2)*cosh(b*x + a)*sinh(b*x + a)^3 + sqrt(2)*sinh(b*x + a)^4 + 2*(3*sqrt
(2)*cosh(b*x + a)^2 - sqrt(2))*sinh(b*x + a)^2 - 2*sqrt(2)*cosh(b*x + a)^2 + 4*(sqrt(2)*cosh(b*x + a)^3 - sqrt
(2)*cosh(b*x + a))*sinh(b*x + a) + sqrt(2))*weierstrassPInverse(4, 0, cosh(b*x + a) + sinh(b*x + a)) + 2*(cosh
(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (3*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + cosh
(b*x + a))*sqrt(sinh(b*x + a)))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4 - 2
*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sin
h(b*x + a) + b)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sinh ^{\frac {5}{2}}{\left (a + b x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sinh(b*x+a)**(5/2),x)

[Out]

Integral(sinh(a + b*x)**(-5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x + a)^(-5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {sinh}\left (a+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sinh(a + b*x)^(5/2),x)

[Out]

int(1/sinh(a + b*x)^(5/2), x)

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