3.1.24 \(\int (i \sinh (c+d x))^{5/2} \, dx\) [24]

Optimal. Leaf size=62 \[ -\frac {6 i E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right )}{5 d}+\frac {2 i \cosh (c+d x) (i \sinh (c+d x))^{3/2}}{5 d} \]

[Out]

6/5*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+2/5*I*cosh(d*x+c)*(I*sinh(d*x+c))^(3/2)/d

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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 = {2715, 2719} \begin {gather*} \frac {2 i (i \sinh (c+d x))^{3/2} \cosh (c+d x)}{5 d}-\frac {6 i E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(I*Sinh[c + d*x])^(5/2),x]

[Out]

(((-6*I)/5)*EllipticE[(I*c - Pi/2 + I*d*x)/2, 2])/d + (((2*I)/5)*Cosh[c + d*x]*(I*Sinh[c + d*x])^(3/2))/d

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[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 (i \sinh (c+d x))^{5/2} \, dx &=\frac {2 i \cosh (c+d x) (i \sinh (c+d x))^{3/2}}{5 d}+\frac {3}{5} \int \sqrt {i \sinh (c+d x)} \, dx\\ &=-\frac {6 i E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right )}{5 d}+\frac {2 i \cosh (c+d x) (i \sinh (c+d x))^{3/2}}{5 d}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 55, normalized size = 0.89 \begin {gather*} \frac {6 i E\left (\left .\frac {1}{4} (-2 i c+\pi -2 i d x)\right |2\right )-\sqrt {i \sinh (c+d x)} \sinh (2 (c+d x))}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(I*Sinh[c + d*x])^(5/2),x]

[Out]

((6*I)*EllipticE[((-2*I)*c + Pi - (2*I)*d*x)/4, 2] - Sqrt[I*Sinh[c + d*x]]*Sinh[2*(c + d*x)])/(5*d)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (82 ) = 164\).
time = 0.69, size = 169, normalized size = 2.73

method	result	size
default	\(-\frac {i \left (3 \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-6 \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticE \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+2 \left (\cosh ^{4}\left (d x +c \right )\right )-2 \left (\cosh ^{2}\left (d x +c \right )\right )\right )}{5 \cosh \left (d x +c \right ) \sqrt {i \sinh \left (d x +c \right )}\, d}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((I*sinh(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/5*I*(3*(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))-6*(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)*Ellip
ticE((1-I*sinh(d*x+c))^(1/2),1/2*2^(1/2))+2*cosh(d*x+c)^4-2*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((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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 94, normalized size = 1.52 \begin {gather*} -\frac {{\left (\sqrt {\frac {1}{2}} {\left (e^{\left (4 \, d x + 4 \, c\right )} + 12 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} \sqrt {i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )} + 12 \, \sqrt {2} \sqrt {i} e^{\left (2 \, d x + 2 \, c\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, e^{\left (d x + c\right )}\right )\right )\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{10 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(sqrt(1/2)*(e^(4*d*x + 4*c) + 12*e^(2*d*x + 2*c) - 1)*sqrt(I*e^(2*d*x + 2*c) - I)*e^(-1/2*d*x - 1/2*c) +
12*sqrt(2)*sqrt(I)*e^(2*d*x + 2*c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, e^(d*x + c))))*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 \left (i \sinh {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*sinh(c + d*x))**(5/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sinh(c + d*x)*1i)^(5/2),x)

[Out]

int((sinh(c + d*x)*1i)^(5/2), x)

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