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

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

[Out]

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))/d+2/3*I*cosh(d*x+c)/d/(I*sinh(d*x+c))^(3/2)

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Rubi [A]  time = 0.02, antiderivative size = 62, normalized size of antiderivative = 1.00, 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*}

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Mathematica [C]  time = 0.06, size = 83, normalized size = 1.34 \[ \frac {2 \left (\sqrt {2} \sqrt {-\left (\sinh ^2(c+d x) (\coth (c+d x)+1)\right )} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\cosh (2 (c+d x))+\sinh (2 (c+d x))\right )+\coth (c+d x)\right )}{3 d \sqrt {i \sinh (c+d x)}} \]

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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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ \frac {\sqrt {\frac {1}{2}} {\left (-4 i \, e^{\left (3 \, d x + 3 \, c\right )} - 4 i \, e^{\left (d x + c\right )}\right )} \sqrt {i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )} + 3 \, {\left (d e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d e^{\left (2 \, d x + 2 \, c\right )} + d\right )} {\rm integral}\left (-\frac {2 i \, \sqrt {\frac {1}{2}} \sqrt {i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )}}{3 \, {\left (d e^{\left (2 \, d x + 2 \, c\right )} - d\right )}}, x\right )}{3 \, {\left (d e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d e^{\left (2 \, d x + 2 \, c\right )} + d\right )}} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 113, normalized size = 1.82 \[ -\frac {i \left (-\sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) \sinh \left (d x +c \right )+2 i \left (\cosh ^{2}\left (d x +c \right )\right )\right )}{3 \sinh \left (d x +c \right ) \cosh \left (d x +c \right ) \sqrt {i \sinh \left (d x +c \right )}\, d} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)*1i)^(5/2),x)

[Out]

int(1/(sinh(c + d*x)*1i)^(5/2), x)

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

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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