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

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

[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 = {2635, 2639} \[ \frac {2 i (i \sinh (c+d x))^{3/2} \cosh (c+d x)}{5 d}-\frac {6 i E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{5 d} \]

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 2635

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] && Integer
Q[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, 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 = 0.06, size = 55, normalized size = 0.89 \[ \frac {6 i E\left (\left .\frac {1}{4} (-2 i c-2 i d x+\pi )\right |2\right )-\sqrt {i \sinh (c+d x)} \sinh (2 (c+d x))}{5 d} \]

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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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ -\frac {\sqrt {\frac {1}{2}} {\left (e^{\left (5 \, d x + 5 \, c\right )} - 2 \, e^{\left (4 \, d x + 4 \, c\right )} - 12 \, e^{\left (3 \, d x + 3 \, c\right )} - 24 \, e^{\left (2 \, d x + 2 \, c\right )} - e^{\left (d x + c\right )} + 2\right )} \sqrt {i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )} - 10 \, {\left (d e^{\left (3 \, d x + 3 \, c\right )} - 2 \, d e^{\left (2 \, d x + 2 \, c\right )}\right )} {\rm integral}\left (\frac {6 \, \sqrt {\frac {1}{2}} {\left (2 \, e^{\left (2 \, d x + 2 \, c\right )} + 3 \, e^{\left (d x + c\right )} - 2\right )} \sqrt {i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )}}{5 \, {\left (d e^{\left (4 \, d x + 4 \, c\right )} - 4 \, d e^{\left (3 \, d x + 3 \, c\right )} + 3 \, d e^{\left (2 \, d x + 2 \, c\right )} + 4 \, d e^{\left (d x + c\right )} - 4 \, d\right )}}, x\right )}{10 \, {\left (d e^{\left (3 \, d x + 3 \, c\right )} - 2 \, d e^{\left (2 \, d x + 2 \, c\right )}\right )}} \]

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^(5*d*x + 5*c) - 2*e^(4*d*x + 4*c) - 12*e^(3*d*x + 3*c) - 24*e^(2*d*x + 2*c) - e^(d*x + c)
+ 2)*sqrt(I*e^(2*d*x + 2*c) - I)*e^(-1/2*d*x - 1/2*c) - 10*(d*e^(3*d*x + 3*c) - 2*d*e^(2*d*x + 2*c))*integral(
6/5*sqrt(1/2)*(2*e^(2*d*x + 2*c) + 3*e^(d*x + c) - 2)*sqrt(I*e^(2*d*x + 2*c) - I)*e^(-1/2*d*x - 1/2*c)/(d*e^(4
*d*x + 4*c) - 4*d*e^(3*d*x + 3*c) + 3*d*e^(2*d*x + 2*c) + 4*d*e^(d*x + c) - 4*d), x))/(d*e^(3*d*x + 3*c) - 2*d
*e^(2*d*x + 2*c))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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maple [B]  time = 0.07, size = 169, normalized size = 2.73 \[ \frac {i \left (6 \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticE \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\cosh ^{4}\left (d x +c \right )\right )+2 \left (\cosh ^{2}\left (d x +c \right )\right )\right )}{5 \cosh \left (d x +c \right ) \sqrt {i \sinh \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*I*(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)*EllipticE((1-I*sinh(d*x
+c))^(1/2),1/2*2^(1/2))-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)*Ellipt
icF((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 \[ \int \left (i \, \sinh \left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 \[ \int {\left (\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sinh(c + d*x)*1i)^(5/2),x)

[Out]

int((sinh(c + d*x)*1i)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*sinh(c + d*x))**(5/2), x)

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