\(\int \sqrt {\sinh (a+b x)} \, dx\) [10]

Optimal result

Integrand size = 10, antiderivative size = 54 \[ \int \sqrt {\sinh (a+b x)} \, dx=-\frac {2 i E\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {\sinh (a+b x)}}{b \sqrt {i \sinh (a+b x)}} \]

[Out]

2*I*(sin(1/2*I*a+1/4*Pi+1/2*I*b*x)^2)^(1/2)/sin(1/2*I*a+1/4*Pi+1/2*I*b*x)*EllipticE(cos(1/2*I*a+1/4*Pi+1/2*I*b
*x),2^(1/2))*sinh(b*x+a)^(1/2)/b/(I*sinh(b*x+a))^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2721, 2719} \[ \int \sqrt {\sinh (a+b x)} \, dx=-\frac {2 i \sqrt {\sinh (a+b x)} E\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{b \sqrt {i \sinh (a+b x)}} \]

[In]

Int[Sqrt[Sinh[a + b*x]],x]

[Out]

((-2*I)*EllipticE[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[Sinh[a + b*x]])/(b*Sqrt[I*Sinh[a + b*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 {\sinh (a+b x)} \int \sqrt {i \sinh (a+b x)} \, dx}{\sqrt {i \sinh (a+b x)}} \\ & = -\frac {2 i E\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {\sinh (a+b x)}}{b \sqrt {i \sinh (a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \sqrt {\sinh (a+b x)} \, dx=\frac {2 E\left (\left .\frac {1}{2} \left (\frac {\pi }{2}-i (a+b x)\right )\right |2\right ) \sqrt {i \sinh (a+b x)}}{b \sqrt {\sinh (a+b x)}} \]

[In]

Integrate[Sqrt[Sinh[a + b*x]],x]

[Out]

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

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.00

method	result	size
default	\(\frac {\sqrt {-i \left (\sinh \left (b x +a \right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (-\sinh \left (b x +a \right )+i\right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )\right )}{\cosh \left (b x +a \right ) \sqrt {\sinh \left (b x +a \right )}\, b}\)	\(108\)
risch	\(\frac {\sqrt {2}\, \sqrt {\left ({\mathrm e}^{2 b x +2 a}-1\right ) {\mathrm e}^{-b x -a}}}{b}-\frac {\left (\frac {2 \,{\mathrm e}^{2 b x +2 a}-2}{\sqrt {\left ({\mathrm e}^{2 b x +2 a}-1\right ) {\mathrm e}^{b x +a}}}-\frac {\sqrt {{\mathrm e}^{b x +a}+1}\, \sqrt {-2 \,{\mathrm e}^{b x +a}+2}\, \sqrt {-{\mathrm e}^{b x +a}}\, \left (-2 \operatorname {EllipticE}\left (\sqrt {{\mathrm e}^{b x +a}+1}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticF}\left (\sqrt {{\mathrm e}^{b x +a}+1}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {{\mathrm e}^{3 b x +3 a}-{\mathrm e}^{b x +a}}}\right ) \sqrt {2}\, \sqrt {\left ({\mathrm e}^{2 b x +2 a}-1\right ) {\mathrm e}^{-b x -a}}\, \sqrt {\left ({\mathrm e}^{2 b x +2 a}-1\right ) {\mathrm e}^{b x +a}}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}\)	\(210\)

[In]

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

[Out]

(-I*(sinh(b*x+a)+I))^(1/2)*2^(1/2)*(-I*(-sinh(b*x+a)+I))^(1/2)*(I*sinh(b*x+a))^(1/2)*(2*EllipticE((1-I*sinh(b*
x+a))^(1/2),1/2*2^(1/2))-EllipticF((1-I*sinh(b*x+a))^(1/2),1/2*2^(1/2)))/cosh(b*x+a)/sinh(b*x+a)^(1/2)/b

Fricas [C] (verification not implemented)

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

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.69 \[ \int \sqrt {\sinh (a+b x)} \, dx=-\frac {2 \, {\left (\sqrt {2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )\right ) + \sqrt {\sinh \left (b x + a\right )}\right )}}{b} \]

[In]

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

[Out]

-2*(sqrt(2)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cosh(b*x + a) + sinh(b*x + a))) + sqrt(sinh(b*x +
a)))/b

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(sinh(a + b*x)), x)

Maxima [F]

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

[In]

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

[Out]

integrate(sqrt(sinh(b*x + a)), x)

Giac [F]

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

[In]

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

[Out]

integrate(sqrt(sinh(b*x + a)), x)

Mupad [F(-1)]

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

[In]

int(sinh(a + b*x)^(1/2),x)

[Out]

int(sinh(a + b*x)^(1/2), x)