\(\int \frac {\sinh (x)}{\sqrt {a+b \sinh (x)}} \, dx\) [111]

Optimal result

Integrand size = 13, antiderivative size = 128 \[ \int \frac {\sinh (x)}{\sqrt {a+b \sinh (x)}} \, dx=\frac {2 i E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right ) \sqrt {a+b \sinh (x)}}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {2 i a \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},\frac {2 b}{i a+b}\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{b \sqrt {a+b \sinh (x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.79 \[ \int \frac {\sinh (x)}{\sqrt {a+b \sinh (x)}} \, dx=\frac {2 \left ((i a+b) E\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right )-i a \operatorname {EllipticF}\left (\frac {1}{4} (\pi -2 i x),-\frac {2 i b}{a-i b}\right )\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{b \sqrt {a+b \sinh (x)}} \] Input:

Integrate[Sinh[x]/Sqrt[a + b*Sinh[x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {3042, 26, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sinh (x)}{\sqrt {a+b \sinh (x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \sin (i x)}{\sqrt {a-i b \sin (i x)}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\sin (i x)}{\sqrt {a-i b \sin (i x)}}dx\)

\(\Big \downarrow \) 3231

\(\displaystyle -i \left (\frac {i \int \sqrt {a+b \sinh (x)}dx}{b}-\frac {i a \int \frac {1}{\sqrt {a+b \sinh (x)}}dx}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -i \left (\frac {i \int \sqrt {a-i b \sin (i x)}dx}{b}-\frac {i a \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}\right )\)

\(\Big \downarrow \) 3134

\(\displaystyle -i \left (\frac {i \sqrt {a+b \sinh (x)} \int \sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}dx}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {i a \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -i \left (\frac {i \sqrt {a+b \sinh (x)} \int \sqrt {\frac {a}{a-i b}-\frac {i b \sin (i x)}{a-i b}}dx}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {i a \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}\right )\)

\(\Big \downarrow \) 3132

\(\displaystyle -i \left (-\frac {i a \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}-\frac {2 \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}\right )\)

\(\Big \downarrow \) 3142

\(\displaystyle -i \left (-\frac {i a \sqrt {\frac {a+b \sinh (x)}{a-i b}} \int \frac {1}{\sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}}dx}{b \sqrt {a+b \sinh (x)}}-\frac {2 \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -i \left (-\frac {i a \sqrt {\frac {a+b \sinh (x)}{a-i b}} \int \frac {1}{\sqrt {\frac {a}{a-i b}-\frac {i b \sin (i x)}{a-i b}}}dx}{b \sqrt {a+b \sinh (x)}}-\frac {2 \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}\right )\)

\(\Big \downarrow \) 3140

\(\displaystyle -i \left (\frac {2 a \sqrt {\frac {a+b \sinh (x)}{a-i b}} \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},\frac {2 b}{i a+b}\right )}{b \sqrt {a+b \sinh (x)}}-\frac {2 \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}\right )\)

Input:

Int[Sinh[x]/Sqrt[a + b*Sinh[x]],x]
 

Output:

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

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/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 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[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]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/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 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[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]
 

rule 3231

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.70

method	result	size
default	\(\frac {2 \left (i b -a \right ) \sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \left (x \right )\right ) b}{i b +a}}\, \sqrt {\frac {b \left (i+\sinh \left (x \right )\right )}{i b -a}}\, \left (i \operatorname {EllipticE}\left (\sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) b -i \operatorname {EllipticF}\left (\sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) b +\operatorname {EllipticE}\left (\sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) a \right )}{b^{2} \cosh \left (x \right ) \sqrt {a +b \sinh \left (x \right )}}\)	\(218\)
risch	\(\frac {\left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right ) \sqrt {2}\, {\mathrm e}^{-x}}{b \sqrt {\left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right ) {\mathrm e}^{-x}}}+\frac {\left (-\frac {4 \left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right )}{b \sqrt {\left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right ) {\mathrm e}^{x}}}+\frac {4 \left (a +\sqrt {a^{2}+b^{2}}\right ) \sqrt {\frac {\left ({\mathrm e}^{x}+\frac {a +\sqrt {a^{2}+b^{2}}}{b}\right ) b}{a +\sqrt {a^{2}+b^{2}}}}\, \sqrt {\frac {{\mathrm e}^{x}-\frac {-a +\sqrt {a^{2}+b^{2}}}{b}}{-\frac {a +\sqrt {a^{2}+b^{2}}}{b}-\frac {-a +\sqrt {a^{2}+b^{2}}}{b}}}\, \sqrt {-\frac {{\mathrm e}^{x} b}{a +\sqrt {a^{2}+b^{2}}}}\, \left (\left (-\frac {a +\sqrt {a^{2}+b^{2}}}{b}-\frac {-a +\sqrt {a^{2}+b^{2}}}{b}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {\left ({\mathrm e}^{x}+\frac {a +\sqrt {a^{2}+b^{2}}}{b}\right ) b}{a +\sqrt {a^{2}+b^{2}}}}, \sqrt {-\frac {a +\sqrt {a^{2}+b^{2}}}{b \left (-\frac {a +\sqrt {a^{2}+b^{2}}}{b}-\frac {-a +\sqrt {a^{2}+b^{2}}}{b}\right )}}\right )+\frac {\left (-a +\sqrt {a^{2}+b^{2}}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {\left ({\mathrm e}^{x}+\frac {a +\sqrt {a^{2}+b^{2}}}{b}\right ) b}{a +\sqrt {a^{2}+b^{2}}}}, \sqrt {-\frac {a +\sqrt {a^{2}+b^{2}}}{b \left (-\frac {a +\sqrt {a^{2}+b^{2}}}{b}-\frac {-a +\sqrt {a^{2}+b^{2}}}{b}\right )}}\right )}{b}\right )}{b \sqrt {{\mathrm e}^{3 x} b +2 a \,{\mathrm e}^{2 x}-{\mathrm e}^{x} b}}\right ) \sqrt {2}\, \sqrt {\left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right ) {\mathrm e}^{x}}\, {\mathrm e}^{-x}}{2 \sqrt {\left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right ) {\mathrm e}^{-x}}}\)	\(531\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.36 \[ \int \frac {\sinh (x)}{\sqrt {a+b \sinh (x)}} \, dx=-\frac {2 \, {\left (4 \, \sqrt {\frac {1}{2}} a \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} + 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cosh \left (x\right ) + 3 \, b \sinh \left (x\right ) + 2 \, a}{3 \, b}\right ) + 6 \, \sqrt {\frac {1}{2}} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} + 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} + 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cosh \left (x\right ) + 3 \, b \sinh \left (x\right ) + 2 \, a}{3 \, b}\right )\right ) + 3 \, \sqrt {b \sinh \left (x\right ) + a} b\right )}}{3 \, b^{2}} \] Input:

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

Output:

-2/3*(4*sqrt(1/2)*a*sqrt(b)*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*sinh(x) + 2*a)/b) + 6*s 
qrt(1/2)*b^(3/2)*weierstrassZeta(4/3*(4*a^2 + 3*b^2)/b^2, -8/27*(8*a^3 + 9 
*a*b^2)/b^3, 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*sinh(x) + 2*a)/b)) + 3*sqrt(b*sinh(x) 
+ a)*b)/b^2
 

Sympy [F]

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

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

Output:

Integral(sinh(x)/sqrt(a + b*sinh(x)), x)
 

Maxima [F]

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

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

Output:

integrate(sinh(x)/sqrt(b*sinh(x) + a), x)
 

Giac [F]

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

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

Output:

integrate(sinh(x)/sqrt(b*sinh(x) + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sinh(x)*b + a)*sinh(x))/(sinh(x)*b + a),x)