\(\int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx\) [27]

Optimal result

Integrand size = 39, antiderivative size = 166 \[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\frac {\sqrt {2} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {g \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d) f}-\frac {2 \sqrt {c} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {c} \sqrt {g} \cos (e+f x)}{\sqrt {c+d} \sqrt {g \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d) \sqrt {c+d} f} \] Output:

2^(1/2)*g^(1/2)*arctan(1/2*a^(1/2)*g^(1/2)*cos(f*x+e)*2^(1/2)/(g*sin(f*x+e 
))^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(1/2)/(c-d)/f-2*c^(1/2)*g^(1/2)*arctan( 
a^(1/2)*c^(1/2)*g^(1/2)*cos(f*x+e)/(c+d)^(1/2)/(g*sin(f*x+e))^(1/2)/(a+a*s 
in(f*x+e))^(1/2))/a^(1/2)/(c-d)/(c+d)^(1/2)/f
 

Mathematica [A] (verified)

Time = 10.14 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.69 \[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=-\frac {\left (2 \sqrt {c} \sqrt {-c^2+d^2} \arctan \left (\sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}\right )-\sqrt {d-\sqrt {-c^2+d^2}} \left (-c+d+\sqrt {-c^2+d^2}\right ) \arctan \left (\frac {\sqrt {c} \sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}}{\sqrt {d-\sqrt {-c^2+d^2}}}\right )-\left (c-d+\sqrt {-c^2+d^2}\right ) \sqrt {d+\sqrt {-c^2+d^2}} \arctan \left (\frac {\sqrt {c} \sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}}{\sqrt {d+\sqrt {-c^2+d^2}}}\right )\right ) \sqrt {g \sin (e+f x)} \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {c} (c-d) \sqrt {-c^2+d^2} f \sqrt {a (1+\sin (e+f x))} \sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}} \] Input:

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

Output:

-(((2*Sqrt[c]*Sqrt[-c^2 + d^2]*ArcTan[Sqrt[Tan[(e + f*x)/2]]] - Sqrt[d - S 
qrt[-c^2 + d^2]]*(-c + d + Sqrt[-c^2 + d^2])*ArcTan[(Sqrt[c]*Sqrt[Tan[(e + 
 f*x)/2]])/Sqrt[d - Sqrt[-c^2 + d^2]]] - (c - d + Sqrt[-c^2 + d^2])*Sqrt[d 
 + Sqrt[-c^2 + d^2]]*ArcTan[(Sqrt[c]*Sqrt[Tan[(e + f*x)/2]])/Sqrt[d + Sqrt 
[-c^2 + d^2]]])*Sqrt[g*Sin[e + f*x]]*(1 + Tan[(e + f*x)/2]))/(Sqrt[c]*(c - 
 d)*Sqrt[-c^2 + d^2]*f*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[Tan[(e + f*x)/2]]))
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3042, 3415, 3042, 3261, 218, 3409, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3415

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3261

\(\displaystyle \frac {2 a g \int \frac {1}{\frac {\cos (e+f x) \cot (e+f x) a^3}{\sin (e+f x) a+a}+2 a^2}d\frac {a \cos (e+f x)}{\sqrt {g \sin (e+f x)} \sqrt {\sin (e+f x) a+a}}}{f (c-d)}+\frac {c g \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {g \sin (e+f x)} (c+d \sin (e+f x))}dx}{a (c-d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {c g \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {g \sin (e+f x)} (c+d \sin (e+f x))}dx}{a (c-d)}+\frac {\sqrt {2} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}\right )}{\sqrt {a} f (c-d)}\)

\(\Big \downarrow \) 3409

\(\displaystyle \frac {\sqrt {2} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}\right )}{\sqrt {a} f (c-d)}-\frac {2 c g \int \frac {1}{\frac {c \cos (e+f x) \cot (e+f x) a^2}{\sin (e+f x) a+a}+(c+d) a}d\frac {a \cos (e+f x)}{\sqrt {g \sin (e+f x)} \sqrt {\sin (e+f x) a+a}}}{f (c-d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\sqrt {2} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}\right )}{\sqrt {a} f (c-d)}-\frac {2 \sqrt {c} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {c} \sqrt {g} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}\right )}{\sqrt {a} f (c-d) \sqrt {c+d}}\)

Input:

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

Output:

(Sqrt[2]*Sqrt[g]*ArcTan[(Sqrt[a]*Sqrt[g]*Cos[e + f*x])/(Sqrt[2]*Sqrt[g*Sin 
[e + f*x]]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*(c - d)*f) - (2*Sqrt[c]*Sq 
rt[g]*ArcTan[(Sqrt[a]*Sqrt[c]*Sqrt[g]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[g*Si 
n[e + f*x]]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*(c - d)*Sqrt[c + d]*f)
 

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 3261

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

rule 3409

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

rule 3415

Int[Sqrt[(g_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_. 
)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(-a)*(g 
/(b*c - a*d))   Int[1/(Sqrt[g*Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]), x], 
x] + Simp[c*(g/(b*c - a*d))   Int[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[g*Sin[e + 
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && N 
eQ[b*c - a*d, 0] && (EqQ[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. \(1034\) vs. \(2(131)=262\).

Time = 22.90 (sec) , antiderivative size = 1035, normalized size of antiderivative = 6.23

method	result	size
default	\(\text {Expression too large to display}\)	\(1035\)

Input:

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

Output:

-2^(1/2)/f*(g*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e))^(1/2)*(g^(5/2)*((-d*g 
+(-g^2*(c-d)*(c+d))^(1/2))*c)^(1/2)*arctan((cos(1/2*f*x+1/2*e)*sin(1/2*f*x 
+1/2*e)/(cos(1/2*f*x+1/2*e)+1)^2*g)^(1/2)*sin(1/2*f*x+1/2*e)/(cos(1/2*f*x+ 
1/2*e)-1)*c/((d*g+(-g^2*(c-d)*(c+d))^(1/2))*c)^(1/2))*c^2-g^(5/2)*((-d*g+( 
-g^2*(c-d)*(c+d))^(1/2))*c)^(1/2)*arctan((cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1 
/2*e)/(cos(1/2*f*x+1/2*e)+1)^2*g)^(1/2)*sin(1/2*f*x+1/2*e)/(cos(1/2*f*x+1/ 
2*e)-1)*c/((d*g+(-g^2*(c-d)*(c+d))^(1/2))*c)^(1/2))*c*d+g^(5/2)*((d*g+(-g^ 
2*(c-d)*(c+d))^(1/2))*c)^(1/2)*arctanh((cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2 
*e)/(cos(1/2*f*x+1/2*e)+1)^2*g)^(1/2)*sin(1/2*f*x+1/2*e)/(cos(1/2*f*x+1/2* 
e)-1)*c/((-d*g+(-g^2*(c-d)*(c+d))^(1/2))*c)^(1/2))*c^2-g^(5/2)*((d*g+(-g^2 
*(c-d)*(c+d))^(1/2))*c)^(1/2)*arctanh((cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2* 
e)/(cos(1/2*f*x+1/2*e)+1)^2*g)^(1/2)*sin(1/2*f*x+1/2*e)/(cos(1/2*f*x+1/2*e 
)-1)*c/((-d*g+(-g^2*(c-d)*(c+d))^(1/2))*c)^(1/2))*c*d-g^(3/2)*((-d*g+(-g^2 
*(c-d)*(c+d))^(1/2))*c)^(1/2)*(-g^2*(c-d)*(c+d))^(1/2)*arctan((cos(1/2*f*x 
+1/2*e)*sin(1/2*f*x+1/2*e)/(cos(1/2*f*x+1/2*e)+1)^2*g)^(1/2)*sin(1/2*f*x+1 
/2*e)/(cos(1/2*f*x+1/2*e)-1)*c/((d*g+(-g^2*(c-d)*(c+d))^(1/2))*c)^(1/2))*c 
+g^(3/2)*((d*g+(-g^2*(c-d)*(c+d))^(1/2))*c)^(1/2)*(-g^2*(c-d)*(c+d))^(1/2) 
*arctanh((cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)/(cos(1/2*f*x+1/2*e)+1)^2*g 
)^(1/2)*sin(1/2*f*x+1/2*e)/(cos(1/2*f*x+1/2*e)-1)*c/((-d*g+(-g^2*(c-d)*(c+ 
d))^(1/2))*c)^(1/2))*c+2*arctan((cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)/...
 

Fricas [A] (verification not implemented)

Time = 1.15 (sec) , antiderivative size = 3048, normalized size of antiderivative = 18.36 \[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\int \frac {\sqrt {g \sin {\left (e + f x \right )}}}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (c + d \sin {\left (e + f x \right )}\right )}\, dx \] Input:

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

Output:

Integral(sqrt(g*sin(e + f*x))/(sqrt(a*(sin(e + f*x) + 1))*(c + d*sin(e + f 
*x))), x)
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\frac {\sqrt {g}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{2} d +\sin \left (f x +e \right ) c +\sin \left (f x +e \right ) d +c}d x \right )}{a} \] Input:

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

Output:

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