3.13 \(\int \frac{1}{\sinh ^{\frac{5}{2}}(a+b x)} \, dx\)

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

[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]])

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Rubi [A]  time = 0.0317031, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {2636, 2642, 2641} \[ -\frac{2 \cosh (a+b x)}{3 b \sinh ^{\frac{3}{2}}(a+b x)}+\frac{2 i \sqrt{i \sinh (a+b x)} F\left (\left .\frac{1}{2} \left (i a+i b x-\frac{\pi }{2}\right )\right |2\right )}{3 b \sqrt{\sinh (a+b x)}} \]

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 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}{\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*}

Mathematica [C]  time = 0.0784565, size = 86, normalized size = 1.08 \[ -\frac{2 \left (\sinh (a+b x) \sqrt{-\sinh (2 a+2 b x)-\cosh (2 a+2 b x)+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\cosh (2 (a+b x))+\sinh (2 (a+b x))\right )+\cosh (a+b x)\right )}{3 b \sinh ^{\frac{3}{2}}(a+b x)} \]

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 = 0.044, size = 101, normalized size = 1.3 \begin{align*} -{\frac{1}{3\,b\cosh \left ( bx+a \right ) } \left ( i\sqrt{1-i\sinh \left ( bx+a \right ) }\sqrt{2}\sqrt{1+i\sinh \left ( bx+a \right ) }\sqrt{i\sinh \left ( bx+a \right ) }{\it EllipticF} \left ( \sqrt{1-i\sinh \left ( bx+a \right ) },{\frac{\sqrt{2}}{2}} \right ) \sinh \left ( bx+a \right ) +2\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sinh(b*x + a)^(-5/2), x)

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

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

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)