\(\int \frac {\sqrt {c+d \sin (e+f x)}}{a+b \sin (e+f x)} \, dx\) [754]

Optimal result

Integrand size = 27, antiderivative size = 153 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{a+b \sin (e+f x)} \, dx=\frac {2 d \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{b f \sqrt {c+d \sin (e+f x)}}+\frac {2 (b c-a d) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{b (a+b) f \sqrt {c+d \sin (e+f x)}} \] Output:

2*d*InverseJacobiAM(1/2*e-1/4*Pi+1/2*f*x,2^(1/2)*(d/(c+d))^(1/2))*((c+d*si 
n(f*x+e))/(c+d))^(1/2)/b/f/(c+d*sin(f*x+e))^(1/2)-2*(-a*d+b*c)*EllipticPi( 
cos(1/2*e+1/4*Pi+1/2*f*x),2*b/(a+b),2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x 
+e))/(c+d))^(1/2)/b/(a+b)/f/(c+d*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 5.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{a+b \sin (e+f x)} \, dx=-\frac {2 \left ((a+b) d \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )+(b c-a d) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{b (a+b) f \sqrt {c+d \sin (e+f x)}} \] Input:

Integrate[Sqrt[c + d*Sin[e + f*x]]/(a + b*Sin[e + f*x]),x]
 

Output:

(-2*((a + b)*d*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] + (b*c - a* 
d)*EllipticPi[(2*b)/(a + b), (-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*Sqrt[( 
c + d*Sin[e + f*x])/(c + d)])/(b*(a + b)*f*Sqrt[c + d*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3282, 3042, 3142, 3042, 3140, 3286, 3042, 3284}

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.

\(\displaystyle \int \frac {\sqrt {c+d \sin (e+f x)}}{a+b \sin (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {c+d \sin (e+f x)}}{a+b \sin (e+f x)}dx\)

\(\Big \downarrow \) 3282

\(\displaystyle \frac {(b c-a d) \int \frac {1}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{b}+\frac {d \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(b c-a d) \int \frac {1}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{b}+\frac {d \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{b}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {(b c-a d) \int \frac {1}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{b}+\frac {d \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{b \sqrt {c+d \sin (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(b c-a d) \int \frac {1}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{b}+\frac {d \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{b \sqrt {c+d \sin (e+f x)}}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {(b c-a d) \int \frac {1}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{b}+\frac {2 d \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{b f \sqrt {c+d \sin (e+f x)}}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {(b c-a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{(a+b \sin (e+f x)) \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{b \sqrt {c+d \sin (e+f x)}}+\frac {2 d \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{b f \sqrt {c+d \sin (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(b c-a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{(a+b \sin (e+f x)) \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{b \sqrt {c+d \sin (e+f x)}}+\frac {2 d \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{b f \sqrt {c+d \sin (e+f x)}}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {2 (b c-a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{b f (a+b) \sqrt {c+d \sin (e+f x)}}+\frac {2 d \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{b f \sqrt {c+d \sin (e+f x)}}\)

Input:

Int[Sqrt[c + d*Sin[e + f*x]]/(a + b*Sin[e + f*x]),x]
 

Output:

(2*d*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x] 
)/(c + d)])/(b*f*Sqrt[c + d*Sin[e + f*x]]) + (2*(b*c - a*d)*EllipticPi[(2* 
b)/(a + b), (e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/( 
c + d)])/(b*(a + b)*f*Sqrt[c + d*Sin[e + f*x]])
 

Defintions of rubi rules used

rule 3042

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

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 3282

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

rule 3284

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

rule 3286

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

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {2 \left (\operatorname {EllipticF}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )-\operatorname {EllipticPi}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, -\frac {\left (c -d \right ) b}{a d -b c}, \sqrt {\frac {c -d}{c +d}}\right )\right ) \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \sqrt {-\frac {\left (\sin \left (f x +e \right )-1\right ) d}{c +d}}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \left (c -d \right )}{b \cos \left (f x +e \right ) \sqrt {c +d \sin \left (f x +e \right )}\, f}\)	\(181\)

Input:

int((c+d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x,method=_RETURNVERBOSE)
 

Output:

2*(EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-EllipticP 
i(((c+d*sin(f*x+e))/(c-d))^(1/2),-(c-d)*b/(a*d-b*c),((c-d)/(c+d))^(1/2)))/ 
b*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*((c+d*si 
n(f*x+e))/(c-d))^(1/2)*(c-d)/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{a+b \sin (e+f x)} \, dx=\text {Timed out} \] Input:

integrate((c+d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate((c+d*sin(f*x+e))**(1/2)/(a+b*sin(f*x+e)),x)
 

Output:

Integral(sqrt(c + d*sin(e + f*x))/(a + b*sin(e + f*x)), x)
 

Maxima [F]

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

integrate((c+d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")
 

Output:

integrate(sqrt(d*sin(f*x + e) + c)/(b*sin(f*x + e) + a), x)
                                                                                    
                                                                                    
 

Giac [F]

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

integrate((c+d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="giac")
 

Output:

integrate(sqrt(d*sin(f*x + e) + c)/(b*sin(f*x + e) + a), x)
 

Mupad [F(-1)]

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

int((c + d*sin(e + f*x))^(1/2)/(a + b*sin(e + f*x)),x)
 

Output:

int((c + d*sin(e + f*x))^(1/2)/(a + b*sin(e + f*x)), x)
 

Reduce [F]

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

int((c+d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x)
 

Output:

int(sqrt(sin(e + f*x)*d + c)/(sin(e + f*x)*b + a),x)