\(\int \frac {\sqrt {a+b \sin (e+f x)}}{\sqrt {g \sin (e+f x)} (c+c \sin (e+f x))} \, dx\) [32]

Optimal result

Integrand size = 39, antiderivative size = 116 \[ \int \frac {\sqrt {a+b \sin (e+f x)}}{\sqrt {g \sin (e+f x)} (c+c \sin (e+f x))} \, dx=-\frac {E\left (\arcsin \left (\frac {\cos (e+f x)}{1+\sin (e+f x)}\right )|-\frac {a-b}{a+b}\right ) \sqrt {\frac {\sin (e+f x)}{1+\sin (e+f x)}} \sqrt {a+b \sin (e+f x)}}{c f \sqrt {g \sin (e+f x)} \sqrt {\frac {a+b \sin (e+f x)}{(a+b) (1+\sin (e+f x))}}} \] Output:

-EllipticE(cos(f*x+e)/(1+sin(f*x+e)),(-(a-b)/(a+b))^(1/2))*(sin(f*x+e)/(1+ 
sin(f*x+e)))^(1/2)*(a+b*sin(f*x+e))^(1/2)/c/f/(g*sin(f*x+e))^(1/2)/((a+b*s 
in(f*x+e))/(a+b)/(1+sin(f*x+e)))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(4679\) vs. \(2(116)=232\).

Time = 33.08 (sec) , antiderivative size = 4679, normalized size of antiderivative = 40.34 \[ \int \frac {\sqrt {a+b \sin (e+f x)}}{\sqrt {g \sin (e+f x)} (c+c \sin (e+f x))} \, dx=\text {Result too large to show} \] Input:

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

Output:

(-2*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sin[e + f*x]*Sq 
rt[a + b*Sin[e + f*x]])/(f*Sqrt[g*Sin[e + f*x]]*(c + c*Sin[e + f*x])) + (( 
Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*Sqrt[Sin[e + f*x]]*((a*Sqrt[Sin[e + 
 f*x]])/(2*Sqrt[a + b*Sin[e + f*x]]) - (b*Sqrt[Sin[e + f*x]])/(2*Sqrt[a + 
b*Sin[e + f*x]]) + (a*Cot[(e + f*x)/2]*Sqrt[Sin[e + f*x]])/(2*Sqrt[a + b*S 
in[e + f*x]]) + (b*Cot[(e + f*x)/2]*Sqrt[Sin[e + f*x]])/(2*Sqrt[a + b*Sin[ 
e + f*x]]) - (b*Cos[(3*(e + f*x))/2]*Csc[(e + f*x)/2]*Sqrt[Sin[e + f*x]])/ 
(2*Sqrt[a + b*Sin[e + f*x]]) + (b*Csc[(e + f*x)/2]*Sqrt[Sin[e + f*x]]*Sin[ 
(3*(e + f*x))/2])/(2*Sqrt[a + b*Sin[e + f*x]]))*Sqrt[(a + 2*b*Tan[(e + f*x 
)/2] + a*Tan[(e + f*x)/2]^2)/(1 + Tan[(e + f*x)/2]^2)]*((Tan[(e + f*x)/2]* 
(1 + Tan[(e + f*x)/2]))/(1 + Tan[(e + f*x)/2]^2) + (Sqrt[-a^2 + b^2]*Sqrt[ 
(a*(a + 2*b*Tan[(e + f*x)/2] + a*Tan[(e + f*x)/2]^2))/(a^2 - b^2)]*(Ellipt 
icE[ArcSin[Sqrt[(-b + Sqrt[-a^2 + b^2] - a*Tan[(e + f*x)/2])/Sqrt[-a^2 + b 
^2]]/Sqrt[2]], (2*Sqrt[-a^2 + b^2])/(-b + Sqrt[-a^2 + b^2])]*Tan[(e + f*x) 
/2] + EllipticF[ArcSin[Sqrt[(b + Sqrt[-a^2 + b^2] + a*Tan[(e + f*x)/2])/Sq 
rt[-a^2 + b^2]]/Sqrt[2]], (2*Sqrt[-a^2 + b^2])/(b + Sqrt[-a^2 + b^2])]*Sqr 
t[(a*Tan[(e + f*x)/2])/(-b + Sqrt[-a^2 + b^2])]*Sqrt[-((a*Tan[(e + f*x)/2] 
)/(b + Sqrt[-a^2 + b^2]))]))/(Sqrt[(a*Tan[(e + f*x)/2])/(-b + Sqrt[-a^2 + 
b^2])]*(a + 2*b*Tan[(e + f*x)/2] + a*Tan[(e + f*x)/2]^2))))/(f*Sqrt[g*Sin[ 
e + f*x]]*(c + c*Sin[e + f*x])*Sqrt[Tan[(e + f*x)/2]/(2 + 2*Tan[(e + f*...
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {3042, 3411}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3411

\(\displaystyle -\frac {\sqrt {\frac {\sin (e+f x)}{\sin (e+f x)+1}} \sqrt {a+b \sin (e+f x)} E\left (\arcsin \left (\frac {\cos (e+f x)}{\sin (e+f x)+1}\right )|-\frac {a-b}{a+b}\right )}{c f \sqrt {g \sin (e+f x)} \sqrt {\frac {a+b \sin (e+f x)}{(a+b) (\sin (e+f x)+1)}}}\)

Input:

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

Output:

-((EllipticE[ArcSin[Cos[e + f*x]/(1 + Sin[e + f*x])], -((a - b)/(a + b))]* 
Sqrt[Sin[e + f*x]/(1 + Sin[e + f*x])]*Sqrt[a + b*Sin[e + f*x]])/(c*f*Sqrt[ 
g*Sin[e + f*x]]*Sqrt[(a + b*Sin[e + f*x])/((a + b)*(1 + 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 3411

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1578\) vs. \(2(109)=218\).

Time = 3.48 (sec) , antiderivative size = 1579, normalized size of antiderivative = 13.61

method	result	size
default	\(\text {Expression too large to display}\)	\(1579\)

Input:

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

Output:

1/2/c/f*(a+b*sin(f*x+e))^(1/2)*(((2*cos(f*x+e)+2)*sin(f*x+e)+2*(1+cos(f*x+ 
e))^2)*(-a^2+b^2)^(1/2)*(1/(-a^2+b^2)^(1/2)*(a*cot(f*x+e)+(-a^2+b^2)^(1/2) 
-b-a*csc(f*x+e)))^(1/2)*(1/(b+(-a^2+b^2)^(1/2))*a*(-csc(f*x+e)+cot(f*x+e)) 
)^(1/2)*EllipticE(((-a*cot(f*x+e)+(-a^2+b^2)^(1/2)+b+a*csc(f*x+e))/(b+(-a^ 
2+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^( 
1/2))*((-a*cot(f*x+e)+(-a^2+b^2)^(1/2)+b+a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2) 
))^(1/2)*b+((-2*cos(f*x+e)-2)*sin(f*x+e)-2*(1+cos(f*x+e))^2)*(1/(-a^2+b^2) 
^(1/2)*(a*cot(f*x+e)+(-a^2+b^2)^(1/2)-b-a*csc(f*x+e)))^(1/2)*(1/(b+(-a^2+b 
^2)^(1/2))*a*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*EllipticE(((-a*cot(f*x+e)+(-a 
^2+b^2)^(1/2)+b+a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+ 
(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*((-a*cot(f*x+e)+(-a^2+b^2)^(1/2 
)+b+a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2)))^(1/2)*a^2+((2*cos(f*x+e)+2)*sin(f* 
x+e)+2*(1+cos(f*x+e))^2)*(1/(-a^2+b^2)^(1/2)*(a*cot(f*x+e)+(-a^2+b^2)^(1/2 
)-b-a*csc(f*x+e)))^(1/2)*(1/(b+(-a^2+b^2)^(1/2))*a*(-csc(f*x+e)+cot(f*x+e) 
))^(1/2)*EllipticE(((-a*cot(f*x+e)+(-a^2+b^2)^(1/2)+b+a*csc(f*x+e))/(b+(-a 
^2+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^ 
(1/2))*((-a*cot(f*x+e)+(-a^2+b^2)^(1/2)+b+a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2 
)))^(1/2)*b^2+((1+cos(f*x+e))*sin(f*x+e)+(1+cos(f*x+e))^2)*(-a^2+b^2)^(1/2 
)*(1/(-a^2+b^2)^(1/2)*(a*cot(f*x+e)+(-a^2+b^2)^(1/2)-b-a*csc(f*x+e)))^(1/2 
)*(1/(b+(-a^2+b^2)^(1/2))*a*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*EllipticF((...
 

Fricas [F]

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

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

Output:

integral(-sqrt(b*sin(f*x + e) + a)*sqrt(g*sin(f*x + e))/(c*g*cos(f*x + e)^ 
2 - c*g*sin(f*x + e) - c*g), x)
 

Sympy [F]

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

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

Output:

Integral(sqrt(a + b*sin(e + f*x))/(sqrt(g*sin(e + f*x))*sin(e + f*x) + sqr 
t(g*sin(e + f*x))), x)/c
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {a+b \sin (e+f x)}}{\sqrt {g \sin (e+f x)} (c+c \sin (e+f x))} \, dx=\frac {\sqrt {g}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right ) b +a}}{\sin \left (f x +e \right )^{2}+\sin \left (f x +e \right )}d x \right )}{c g} \] Input:

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

Output:

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