3.1.28 \(\int \frac {1}{(i \sinh (c+d x))^{3/2}} \, dx\) [28]

Optimal. Leaf size=58 \[ \frac {2 i E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right )}{d}+\frac {2 i \cosh (c+d x)}{d \sqrt {i \sinh (c+d x)}} \]

[Out]

-2*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)*EllipticE(cos(1/2*I*c+1/4*Pi+1/2*I*
d*x),2^(1/2))/d+2*I*cosh(d*x+c)/d/(I*sinh(d*x+c))^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 58, 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 = {2716, 2719} \begin {gather*} \frac {2 i E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{d}+\frac {2 i \cosh (c+d x)}{d \sqrt {i \sinh (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(I*Sinh[c + d*x])^(-3/2),x]

[Out]

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

Rule 2716

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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 50, normalized size = 0.86 \begin {gather*} \frac {2 \left (-i E\left (\left .\frac {1}{4} (-2 i c+\pi -2 i d x)\right |2\right )+\coth (c+d x) \sqrt {i \sinh (c+d x)}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(I*Sinh[c + d*x])^(-3/2),x]

[Out]

(2*((-I)*EllipticE[((-2*I)*c + Pi - (2*I)*d*x)/4, 2] + Coth[c + d*x]*Sqrt[I*Sinh[c + d*x]]))/d

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Maple [A]
time = 0.82, size = 159, normalized size = 2.74

method	result	size
default	\(-\frac {i \left (2 \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticE \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\cosh ^{2}\left (d x +c \right )\right )\right )}{\cosh \left (d x +c \right ) \sqrt {i \sinh \left (d x +c \right )}\, d}\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(I*sinh(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-I*(2*(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)*EllipticE((1-I*sinh(d*x+c)
)^(1/2),1/2*2^(1/2))-(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))-2*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(I*sinh(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((I*sinh(d*x + c))^(-3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.08, size = 87, normalized size = 1.50 \begin {gather*} \frac {2 \, {\left (2 \, \sqrt {\frac {1}{2}} \sqrt {i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )} + {\left (\sqrt {2} \sqrt {i} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {2} \sqrt {i}\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, e^{\left (d x + c\right )}\right )\right )\right )}}{d e^{\left (2 \, d x + 2 \, c\right )} - d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(I*sinh(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2*(2*sqrt(1/2)*sqrt(I*e^(2*d*x + 2*c) - I)*e^(3/2*d*x + 3/2*c) + (sqrt(2)*sqrt(I)*e^(2*d*x + 2*c) - sqrt(2)*sq
rt(I))*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, e^(d*x + c))))/(d*e^(2*d*x + 2*c) - d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (i \sinh {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(I*sinh(d*x+c))**(3/2),x)

[Out]

Integral((I*sinh(c + d*x))**(-3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(I*sinh(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((I*sinh(d*x + c))^(-3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)*1i)^(3/2),x)

[Out]

int(1/(sinh(c + d*x)*1i)^(3/2), x)

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