3.21 \(\int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=90 \[ -\frac {2 \cosh (c+d x)}{3 b d (b \sinh (c+d x))^{3/2}}+\frac {2 i \sqrt {i \sinh (c+d x)} F\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{3 b^2 d \sqrt {b \sinh (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2636, 2642, 2641} \[ -\frac {2 \cosh (c+d x)}{3 b d (b \sinh (c+d x))^{3/2}}+\frac {2 i \sqrt {i \sinh (c+d x)} F\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{3 b^2 d \sqrt {b \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sinh[c + d*x])^(-5/2),x]

[Out]

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

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

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Mathematica [C]  time = 0.10, size = 84, normalized size = 0.93 \[ -\frac {2 \left (\sqrt {2} \sqrt {-\left (\sinh ^2(c+d x) (\coth (c+d x)+1)\right )} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\cosh (2 (c+d x))+\sinh (2 (c+d x))\right )+\coth (c+d x)\right )}{3 b^2 d \sqrt {b \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sinh[c + d*x])^(-5/2),x]

[Out]

(-2*(Coth[c + d*x] + Sqrt[2]*Hypergeometric2F1[1/4, 1/2, 5/4, Cosh[2*(c + d*x)] + Sinh[2*(c + d*x)]]*Sqrt[-((1
+ Coth[c + d*x])*Sinh[c + d*x]^2)]))/(3*b^2*d*Sqrt[b*Sinh[c + d*x]])

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sinh \left (d x + c\right )}}{b^{3} \sinh \left (d x + c\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sinh(d*x + c))/(b^3*sinh(d*x + c)^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(-5/2), x)

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maple [A]  time = 0.07, size = 114, normalized size = 1.27 \[ -\frac {i \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) \sinh \left (d x +c \right )+2 \left (\cosh ^{2}\left (d x +c \right )\right )}{3 b^{2} \sinh \left (d x +c \right ) \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sinh(d*x+c))^(5/2),x)

[Out]

-1/3/b^2/sinh(d*x+c)*(I*(1-I*sinh(d*x+c))^(1/2)*2^(1/2)*(1+I*sinh(d*x+c))^(1/2)*(I*sinh(d*x+c))^(1/2)*Elliptic
F((1-I*sinh(d*x+c))^(1/2),1/2*2^(1/2))*sinh(d*x+c)+2*cosh(d*x+c)^2)/cosh(d*x+c)/(b*sinh(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(-5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sinh(c + d*x))^(5/2),x)

[Out]

int(1/(b*sinh(c + d*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sinh {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*sinh(c + d*x))**(-5/2), x)

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