\(\int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx\) [108]

Optimal result

Integrand size = 10, antiderivative size = 60 \[ \int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx=\frac {2 i \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},\frac {2 b}{i a+b}\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{\sqrt {a+b \sinh (x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 60, 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 = {2742, 2740} \[ \int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx=\frac {2 i \sqrt {\frac {a+b \sinh (x)}{a-i b}} \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},\frac {2 b}{i a+b}\right )}{\sqrt {a+b \sinh (x)}} \]

[In]

Int[1/Sqrt[a + b*Sinh[x]],x]

[Out]

((2*I)*EllipticF[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[(a + b*Sinh[x])/(a - I*b)])/Sqrt[a + b*Sinh[x]]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\frac {a+b \sinh (x)}{a-i b}} \int \frac {1}{\sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}} \, dx}{\sqrt {a+b \sinh (x)}} \\ & = \frac {2 i \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},\frac {2 b}{i a+b}\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{\sqrt {a+b \sinh (x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx=\frac {2 i \operatorname {EllipticF}\left (\frac {1}{4} (\pi -2 i x),-\frac {2 i b}{a-i b}\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{\sqrt {a+b \sinh (x)}} \]

[In]

Integrate[1/Sqrt[a + b*Sinh[x]],x]

[Out]

((2*I)*EllipticF[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*Sqrt[(a + b*Sinh[x])/(a - I*b)])/Sqrt[a + b*Sinh[x]]

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.08

method	result	size
default	\(-\frac {2 \left (i b -a \right ) \sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \left (x \right )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \left (x \right )\right ) b}{i b -a}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right )}{b \cosh \left (x \right ) \sqrt {a +b \sinh \left (x \right )}}\)	\(125\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx=\frac {2 \, \sqrt {2} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} + 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cosh \left (x\right ) + 3 \, b \sinh \left (x\right ) + 2 \, a}{3 \, b}\right )}{\sqrt {b}} \]

[In]

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

[Out]

2*sqrt(2)*weierstrassPInverse(4/3*(4*a^2 + 3*b^2)/b^2, -8/27*(8*a^3 + 9*a*b^2)/b^3, 1/3*(3*b*cosh(x) + 3*b*sin
h(x) + 2*a)/b)/sqrt(b)

Sympy [F]

\[ \int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx=\int \frac {1}{\sqrt {a + b \sinh {\left (x \right )}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(a + b*sinh(x)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx=\int { \frac {1}{\sqrt {b \sinh \left (x\right ) + a}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(b*sinh(x) + a), x)

Giac [F]

\[ \int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx=\int { \frac {1}{\sqrt {b \sinh \left (x\right ) + a}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(b*sinh(x) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx=\int \frac {1}{\sqrt {a+b\,\mathrm {sinh}\left (x\right )}} \,d x \]

[In]

int(1/(a + b*sinh(x))^(1/2),x)

[Out]

int(1/(a + b*sinh(x))^(1/2), x)