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

Optimal. Leaf size=62 \[ \frac{2 i \cosh (c+d x)}{3 d (i \sinh (c+d x))^{3/2}}-\frac{2 i \text{EllipticF}\left (\frac{1}{2} \left (i c+i d x-\frac{\pi }{2}\right ),2\right )}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0201901, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2636, 2641} \[ \frac{2 i \cosh (c+d x)}{3 d (i \sinh (c+d x))^{3/2}}-\frac{2 i F\left (\left .\frac{1}{2} \left (i c+i d x-\frac{\pi }{2}\right )\right |2\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(I*Sinh[c + d*x])^(-5/2),x]

[Out]

(((-2*I)/3)*EllipticF[(I*c - Pi/2 + I*d*x)/2, 2])/d + (((2*I)/3)*Cosh[c + d*x])/(d*(I*Sinh[c + d*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]

Rubi steps

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

Mathematica [C]  time = 0.0586841, size = 83, normalized size = 1.34 \[ \frac{2 \left (\sqrt{2} \sqrt{\sinh ^2(c+d x) (-(\coth (c+d x)+1))} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\cosh (2 (c+d x))+\sinh (2 (c+d x))\right )+\coth (c+d x)\right )}{3 d \sqrt{i \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(I*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*d*Sqrt[I*Sinh[c + d*x]])

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Maple [A]  time = 0.048, size = 113, normalized size = 1.8 \begin{align*}{\frac{-{\frac{i}{3}}}{\sinh \left ( dx+c \right ) \cosh \left ( dx+c \right ) d} \left ( -\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\,i \left ( \cosh \left ( dx+c \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{i\sinh \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I/sinh(d*x+c)*(-(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)*EllipticF((
1-I*sinh(d*x+c))^(1/2),1/2*2^(1/2))*sinh(d*x+c)+2*I*cosh(d*x+c)^2)/cosh(d*x+c)/(I*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 (i \, \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/(I*sinh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{\frac{1}{2}}{\left (-4 i \, e^{\left (3 \, d x + 3 \, c\right )} - 4 i \, e^{\left (d x + c\right )}\right )} \sqrt{i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c\right )} + 3 \,{\left (d e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d e^{\left (2 \, d x + 2 \, c\right )} + d\right )}{\rm integral}\left (-\frac{2 i \, \sqrt{\frac{1}{2}} \sqrt{i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c\right )}}{3 \,{\left (d e^{\left (2 \, d x + 2 \, c\right )} - d\right )}}, x\right )}{3 \,{\left (d e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d e^{\left (2 \, d x + 2 \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(1/2)*(-4*I*e^(3*d*x + 3*c) - 4*I*e^(d*x + c))*sqrt(I*e^(2*d*x + 2*c) - I)*e^(-1/2*d*x - 1/2*c) + 3*(
d*e^(4*d*x + 4*c) - 2*d*e^(2*d*x + 2*c) + d)*integral(-2/3*I*sqrt(1/2)*sqrt(I*e^(2*d*x + 2*c) - I)*e^(-1/2*d*x
- 1/2*c)/(d*e^(2*d*x + 2*c) - d), x))/(d*e^(4*d*x + 4*c) - 2*d*e^(2*d*x + 2*c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (i \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/(I*sinh(d*x+c))**(5/2),x)

[Out]

Integral((I*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 (i \, \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/(I*sinh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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