\(\int (i \sinh (c+d x))^{3/2} \, dx\) [25]

Optimal result

Integrand size = 14, antiderivative size = 62 \[ \int (i \sinh (c+d x))^{3/2} \, dx=-\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right )}{3 d}+\frac {2 i \cosh (c+d x) \sqrt {i \sinh (c+d x)}}{3 d} \]

[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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2715, 2720} \[ \int (i \sinh (c+d x))^{3/2} \, dx=\frac {2 i \sqrt {i \sinh (c+d x)} \cosh (c+d x)}{3 d}-\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right ),2\right )}{3 d} \]

[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 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 2720

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

Rubi steps \begin{align*} \text {integral}& = \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 \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right )}{3 d}+\frac {2 i \cosh (c+d x) \sqrt {i \sinh (c+d x)}}{3 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.52 \[ \int (i \sinh (c+d x))^{3/2} \, dx=-\frac {2 i \sqrt {i \sinh (c+d x)} \left (-\cosh (c+d x)+\text {csch}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cosh (2 (c+d x))+\sinh (2 (c+d x))\right ) \sqrt {1-\cosh (2 c+2 d x)-\sinh (2 c+2 d x)}\right )}{3 d} \]

[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

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.68

method	result	size
default	\(\frac {i \left (\sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+2 i \cosh \left (d x +c \right )^{2} \sinh \left (d x +c \right )\right )}{3 \cosh \left (d x +c \right ) \sqrt {i \sinh \left (d x +c \right )}\, d}\)	\(104\)

[In]

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

[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*cosh(d*x+c)^2*sinh(d*x+c))/cosh(d*x+c)/(I*sinh(d*x+c))^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.27 \[ \int (i \sinh (c+d x))^{3/2} \, dx=\frac {{\left (\sqrt {\frac {1}{2}} {\left (i \, e^{\left (2 \, d x + 2 \, c\right )} + i\right )} \sqrt {i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )} - 2 i \, \sqrt {2} \sqrt {i} e^{\left (d x + c\right )} {\rm weierstrassPInverse}\left (4, 0, e^{\left (d x + c\right )}\right )\right )} e^{\left (-d x - c\right )}}{3 \, d} \]

[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) - 2*I*sqrt(2)*sqrt(I)*
e^(d*x + c)*weierstrassPInverse(4, 0, e^(d*x + c)))*e^(-d*x - c)/d

Sympy [F]

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

[In]

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

[Out]

Integral((I*sinh(c + d*x))**(3/2), x)

Maxima [F]

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

[In]

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

[Out]

integrate((I*sinh(d*x + c))^(3/2), x)

Giac [F]

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

[In]

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

[Out]

integrate((I*sinh(d*x + c))^(3/2), x)

Mupad [F(-1)]

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

[In]

int((sinh(c + d*x)*1i)^(3/2),x)

[Out]

int((sinh(c + d*x)*1i)^(3/2), x)