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

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

[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]])

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Rubi [A]  time = 0.0379677, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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]

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]

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*}

Mathematica [C]  time = 0.0947008, size = 84, normalized size = 0.93 \[ -\frac{2 \left (\sqrt{2} \sqrt{\sinh ^2(c+d x) (-(\coth (c+d x)+1))} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\cosh (2 (c+d x))+\sinh (2 (c+d x))\right )+\coth (c+d x)\right )}{3 b^2 d \sqrt{b \sinh (c+d x)}} \]

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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Maple [A]  time = 0.043, size = 114, normalized size = 1.3 \begin{align*} -{\frac{1}{3\,{b}^{2}\sinh \left ( dx+c \right ) \cosh \left ( dx+c \right ) d} \left ( i\sqrt{1-i\sinh \left ( dx+c \right ) }\sqrt{2}\sqrt{1+i\sinh \left ( dx+c \right ) }\sqrt{i\sinh \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{1-i\sinh \left ( dx+c \right ) },{\frac{\sqrt{2}}{2}} \right ) \sinh \left ( dx+c \right ) +2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{b\sinh \left ( dx+c \right ) }}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sinh \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sinh \left (d x + c\right )}}{b^{3} \sinh \left (d x + c\right )^{3}}, x\right ) \end{align*}

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

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sinh \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

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)