\(\int \sqrt {b \sinh (c+d x)} \, dx\) [18]

Optimal result

Integrand size = 12, antiderivative size = 56 \[ \int \sqrt {b \sinh (c+d x)} \, dx=-\frac {2 i E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{d \sqrt {i \sinh (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2721, 2719} \[ \int \sqrt {b \sinh (c+d x)} \, dx=-\frac {2 i E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{d \sqrt {i \sinh (c+d x)}} \]

[In]

Int[Sqrt[b*Sinh[c + d*x]],x]

[Out]

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

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]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {b \sinh (c+d x)} \int \sqrt {i \sinh (c+d x)} \, dx}{\sqrt {i \sinh (c+d x)}} \\ & = -\frac {2 i E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{d \sqrt {i \sinh (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.93 \[ \int \sqrt {b \sinh (c+d x)} \, dx=\frac {2 i E\left (\left .\frac {1}{4} (-2 i c+\pi -2 i d x)\right |2\right ) \sqrt {b \sinh (c+d x)}}{d \sqrt {i \sinh (c+d x)}} \]

[In]

Integrate[Sqrt[b*Sinh[c + d*x]],x]

[Out]

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

Maple [A] (verified)

Time = 1.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.98

method	result	size
default	\(\frac {b \sqrt {-i \left (\sinh \left (d x +c \right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (i-\sinh \left (d x +c \right )\right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )\right )}{\cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d}\)	\(111\)
risch	\(\frac {\sqrt {2}\, \sqrt {b \left ({\mathrm e}^{2 d x +2 c}-1\right ) {\mathrm e}^{-d x -c}}}{d}-\frac {\left (\frac {2 b \,{\mathrm e}^{2 d x +2 c}-2 b}{b \sqrt {{\mathrm e}^{d x +c} \left (b \,{\mathrm e}^{2 d x +2 c}-b \right )}}-\frac {\sqrt {{\mathrm e}^{d x +c}+1}\, \sqrt {-2 \,{\mathrm e}^{d x +c}+2}\, \sqrt {-{\mathrm e}^{d x +c}}\, \left (-2 \operatorname {EllipticE}\left (\sqrt {{\mathrm e}^{d x +c}+1}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticF}\left (\sqrt {{\mathrm e}^{d x +c}+1}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {b \,{\mathrm e}^{3 d x +3 c}-b \,{\mathrm e}^{d x +c}}}\right ) \sqrt {2}\, \sqrt {b \left ({\mathrm e}^{2 d x +2 c}-1\right ) {\mathrm e}^{-d x -c}}\, \sqrt {b \left ({\mathrm e}^{2 d x +2 c}-1\right ) {\mathrm e}^{d x +c}}}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )}\)	\(227\)

[In]

int((b*sinh(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

b*(-I*(sinh(d*x+c)+I))^(1/2)*2^(1/2)*(-I*(I-sinh(d*x+c)))^(1/2)*(I*sinh(d*x+c))^(1/2)*(2*EllipticE((1-I*sinh(d
*x+c))^(1/2),1/2*2^(1/2))-EllipticF((1-I*sinh(d*x+c))^(1/2),1/2*2^(1/2)))/cosh(d*x+c)/(b*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 = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.75 \[ \int \sqrt {b \sinh (c+d x)} \, dx=-\frac {2 \, {\left (\sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )\right ) + \sqrt {b \sinh \left (d x + c\right )}\right )}}{d} \]

[In]

integrate((b*sinh(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-2*(sqrt(2)*sqrt(b)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cosh(d*x + c) + sinh(d*x + c))) + sqrt(b*s
inh(d*x + c)))/d

Sympy [F]

\[ \int \sqrt {b \sinh (c+d x)} \, dx=\int \sqrt {b \sinh {\left (c + d x \right )}}\, dx \]

[In]

integrate((b*sinh(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(b*sinh(c + d*x)), x)

Maxima [F]

\[ \int \sqrt {b \sinh (c+d x)} \, dx=\int { \sqrt {b \sinh \left (d x + c\right )} \,d x } \]

[In]

integrate((b*sinh(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sinh(d*x + c)), x)

Giac [F]

\[ \int \sqrt {b \sinh (c+d x)} \, dx=\int { \sqrt {b \sinh \left (d x + c\right )} \,d x } \]

[In]

integrate((b*sinh(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sinh(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {b \sinh (c+d x)} \, dx=\int \sqrt {b\,\mathrm {sinh}\left (c+d\,x\right )} \,d x \]

[In]

int((b*sinh(c + d*x))^(1/2),x)

[Out]

int((b*sinh(c + d*x))^(1/2), x)