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*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

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Rubi [A]  time = 0.0198261, 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 = {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*}

Mathematica [A]  time = 0.0588187, 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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Maple [B]  time = 0.05, size = 169, normalized size = 2.7 \begin{align*}{\frac{{\frac{i}{5}}}{d\cosh \left ( dx+c \right ) } \left ( 6\,\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 EllipticE} \left ( \sqrt{1-i\sinh \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) -3\,\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 ) },1/2\,\sqrt{2} \right ) -2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{4}+2\, \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((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., size = 0, normalized size = 0. \begin{align*} \int \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((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 (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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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((I*sinh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((I*sinh(d*x + c))^(5/2), x)