3.25 \(\int (i \sinh (c+d x))^{3/2} \, dx\)

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

[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)*(I*sinh(d*x+c))^(1/2)/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, 2641} \[ \frac {2 i \sqrt {i \sinh (c+d x)} \cosh (c+d x)}{3 d}-\frac {2 i F\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(I*Sinh[c + d*x])^(3/2),x]

[Out]

(((-2*I)/3)*EllipticF[(I*c - Pi/2 + I*d*x)/2, 2])/d + (((2*I)/3)*Cosh[c + d*x]*Sqrt[I*Sinh[c + d*x]])/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 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 (i \sinh (c+d x))^{3/2} \, dx &=\frac {2 i \cosh (c+d x) \sqrt {i \sinh (c+d x)}}{3 d}+\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) \sqrt {i \sinh (c+d x)}}{3 d}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 94, normalized size = 1.52 \[ -\frac {2 i \sqrt {i \sinh (c+d x)} \left (\text {csch}(c+d x) \sqrt {-\sinh (2 c+2 d x)-\cosh (2 c+2 d x)+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\cosh (2 (c+d x))+\sinh (2 (c+d x))\right )-\cosh (c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(I*Sinh[c + d*x])^(3/2),x]

[Out]

(((-2*I)/3)*Sqrt[I*Sinh[c + d*x]]*(-Cosh[c + d*x] + Csch[c + d*x]*Hypergeometric2F1[1/4, 1/2, 5/4, Cosh[2*(c +
d*x)] + Sinh[2*(c + d*x)]]*Sqrt[1 - Cosh[2*c + 2*d*x] - Sinh[2*c + 2*d*x]]))/d

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ \frac {{\left (\sqrt {\frac {1}{2}} {\left (i \, e^{\left (2 \, d x + 2 \, c\right )} + i\right )} \sqrt {i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )} + 3 \, d e^{\left (d x + c\right )} {\rm integral}\left (-\frac {2 i \, \sqrt {\frac {1}{2}} \sqrt {i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )}}{3 \, {\left (d e^{\left (2 \, d x + 2 \, c\right )} - d\right )}}, x\right )\right )} e^{\left (-d x - c\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(1/2)*(I*e^(2*d*x + 2*c) + I)*sqrt(I*e^(2*d*x + 2*c) - I)*e^(-1/2*d*x - 1/2*c) + 3*d*e^(d*x + c)*inte
gral(-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))*e^(-d*x -
c)/d

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i \, \sinh \left (d x + c\right )\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((I*sinh(d*x + c))^(3/2), x)

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maple [A]  time = 0.07, size = 104, normalized size = 1.68 \[ \frac {i \left (\sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+2 i \sinh \left (d x +c \right ) \left (\cosh ^{2}\left (d x +c \right )\right )\right )}{3 \cosh \left (d x +c \right ) \sqrt {i \sinh \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*I*((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*I*sinh(d*x+c)*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 \[ \int \left (i \, \sinh \left (d x + c\right )\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 \[ \int {\left (\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sinh(c + d*x)*1i)^(3/2),x)

[Out]

int((sinh(c + d*x)*1i)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*sinh(c + d*x))**(3/2), x)

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