\(\int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx\) [1429]

Optimal result

Integrand size = 37, antiderivative size = 273 \[ \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {2 \sqrt {2} b \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {2} b \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \cos (e+f x)}}+\frac {\operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {\sin (2 e+2 f x)}}{a f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \] Output:

2*2^(1/2)*b*cos(f*x+e)^(1/2)*EllipticPi((d*sin(f*x+e))^(1/2)/d^(1/2)/(1+co 
s(f*x+e))^(1/2),-a/(b-(-a^2+b^2)^(1/2)),I)/a/(-a^2+b^2)^(1/2)/d^(1/2)/f/(g 
*cos(f*x+e))^(1/2)-2*2^(1/2)*b*cos(f*x+e)^(1/2)*EllipticPi((d*sin(f*x+e))^ 
(1/2)/d^(1/2)/(1+cos(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)),I)/a/(-a^2+b^2) 
^(1/2)/d^(1/2)/f/(g*cos(f*x+e))^(1/2)+InverseJacobiAM(e-1/4*Pi+f*x,2^(1/2) 
)*sin(2*f*x+2*e)^(1/2)/a/f/(g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 17.85 (sec) , antiderivative size = 664, normalized size of antiderivative = 2.43 \[ \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx=-\frac {2 \sqrt {\cos (e+f x)} \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \sqrt {\cos (e+f x)}}{\left (1-\cos ^2(e+f x)\right )^{3/4} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+\left (-4 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+3 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )\right ) \cos ^2(e+f x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(e+f x)\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) b \left (2 \arctan \left (1-\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )+\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}-\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )-\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}+\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )\right )}{\sqrt {a} \left (-a^2+b^2\right )^{3/4}}\right ) \sin (e+f x)}{f \sqrt {g \cos (e+f x)} \sqrt [4]{1-\cos ^2(e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \] Input:

Integrate[1/(Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x] 
)),x]
 

Output:

(-2*Sqrt[Cos[e + f*x]]*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*a*(a^2 - b^2)* 
AppellF1[1/4, 3/4, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^ 
2)]*Sqrt[Cos[e + f*x]])/((1 - Cos[e + f*x]^2)^(3/4)*(5*(a^2 - b^2)*AppellF 
1[1/4, 3/4, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + ( 
-4*b^2*AppellF1[5/4, 3/4, 2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a 
^2 + b^2)] + 3*(a^2 - b^2)*AppellF1[5/4, 7/4, 1, 9/4, Cos[e + f*x]^2, (b^2 
*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)*(a^2 + b^2*(-1 + Cos[e + f 
*x]^2))) - ((1/8 - I/8)*b*(2*ArcTan[1 - ((1 + I)*Sqrt[a]*Sqrt[Cos[e + f*x] 
])/((-a^2 + b^2)^(1/4)*(-1 + Cos[e + f*x]^2)^(1/4))] - 2*ArcTan[1 + ((1 + 
I)*Sqrt[a]*Sqrt[Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*(-1 + Cos[e + f*x]^2)^( 
1/4))] + Log[Sqrt[-a^2 + b^2] + (I*a*Cos[e + f*x])/Sqrt[-1 + Cos[e + f*x]^ 
2] - ((1 + I)*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]])/(-1 + Cos[e + 
 f*x]^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] + (I*a*Cos[e + f*x])/Sqrt[-1 + Cos[ 
e + f*x]^2] + ((1 + I)*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]])/(-1 
+ Cos[e + f*x]^2)^(1/4)]))/(Sqrt[a]*(-a^2 + b^2)^(3/4)))*Sin[e + f*x])/(f* 
Sqrt[g*Cos[e + f*x]]*(1 - Cos[e + f*x]^2)^(1/4)*Sqrt[d*Sin[e + f*x]]*(a + 
b*Sin[e + f*x]))
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {3042, 3389, 3042, 3053, 3042, 3120, 3387, 3042, 3386, 1542}

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.

\(\displaystyle \int \frac {1}{\sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3389

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3053

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {b \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a d}\)

\(\Big \downarrow \) 3387

\(\displaystyle \frac {\sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\cos (e+f x)} \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))}dx}{a d \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\cos (e+f x)} \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))}dx}{a d \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3386

\(\displaystyle \frac {\sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\cos (e+f x)} \left (\frac {2 \sqrt {2} d \left (1-\frac {b}{\sqrt {b^2-a^2}}\right ) \int \frac {1}{\left (\left (b-\sqrt {b^2-a^2}\right ) d+\frac {a \sin (e+f x) d}{\cos (e+f x)+1}\right ) \sqrt {1-\frac {\sin ^2(e+f x)}{(\cos (e+f x)+1)^2}}}d\frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)+1}}}{f}+\frac {2 \sqrt {2} d \left (\frac {b}{\sqrt {b^2-a^2}}+1\right ) \int \frac {1}{\left (\left (b+\sqrt {b^2-a^2}\right ) d+\frac {a \sin (e+f x) d}{\cos (e+f x)+1}\right ) \sqrt {1-\frac {\sin ^2(e+f x)}{(\cos (e+f x)+1)^2}}}d\frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)+1}}}{f}\right )}{a d \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 1542

\(\displaystyle \frac {\sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\cos (e+f x)} \left (\frac {2 \sqrt {2} \sqrt {d} \left (1-\frac {b}{\sqrt {b^2-a^2}}\right ) \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right ),-1\right )}{f \left (b-\sqrt {b^2-a^2}\right )}+\frac {2 \sqrt {2} \sqrt {d} \left (\frac {b}{\sqrt {b^2-a^2}}+1\right ) \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right ),-1\right )}{f \left (\sqrt {b^2-a^2}+b\right )}\right )}{a d \sqrt {g \cos (e+f x)}}\)

Input:

Int[1/(Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])),x]
 

Output:

-((b*Sqrt[Cos[e + f*x]]*((2*Sqrt[2]*(1 - b/Sqrt[-a^2 + b^2])*Sqrt[d]*Ellip 
ticPi[-(a/(b - Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]*Sq 
rt[1 + Cos[e + f*x]])], -1])/((b - Sqrt[-a^2 + b^2])*f) + (2*Sqrt[2]*(1 + 
b/Sqrt[-a^2 + b^2])*Sqrt[d]*EllipticPi[-(a/(b + Sqrt[-a^2 + b^2])), ArcSin 
[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x]])], -1])/((b + Sqrt[- 
a^2 + b^2])*f)))/(a*d*Sqrt[g*Cos[e + f*x]])) + (EllipticF[e - Pi/4 + f*x, 
2]*Sqrt[Sin[2*e + 2*f*x]])/(a*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]])
 

Defintions of rubi rules used

rule 1542

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ 
{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x 
], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]
 

rule 3042

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

rule 3053

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ 
)]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b 
*Cos[e + f*x]])   Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f 
}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3386

Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]]*((a_ 
) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 
2]}, Simp[2*Sqrt[2]*d*((b + q)/(f*q))   Subst[Int[1/((d*(b + q) + a*x^2)*Sq 
rt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x] - 
 Simp[2*Sqrt[2]*d*((b - q)/(f*q))   Subst[Int[1/((d*(b - q) + a*x^2)*Sqrt[1 
 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x]] /; F 
reeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 3387

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

rule 3389

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( 
n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a   Int[(g 
*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] - Simp[b/(a*d)   Int[(g*Cos[e + 
 f*x])^p*((d*Sin[e + f*x])^(n + 1)/(a + b*Sin[e + f*x])), x], x] /; FreeQ[{ 
a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[-1 
, p, 1] && LtQ[n, 0]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(521\) vs. \(2(231)=462\).

Time = 2.78 (sec) , antiderivative size = 522, normalized size of antiderivative = 1.91

method	result	size
default	\(-\frac {\left (2 \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) a \sqrt {-a^{2}+b^{2}}-2 \sqrt {-a^{2}+b^{2}}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) b +\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right ) a b -\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right ) b^{2}+\sqrt {-a^{2}+b^{2}}\, \operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right ) b -\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) a b +\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) b^{2}+\sqrt {-a^{2}+b^{2}}\, \operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) b \right ) \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \left (1+\cos \left (f x +e \right )\right )}{f \sqrt {g \cos \left (f x +e \right )}\, \sqrt {d \sin \left (f x +e \right )}\, \sqrt {-a^{2}+b^{2}}\, \left (-b +\sqrt {-a^{2}+b^{2}}+a \right ) \left (b +\sqrt {-a^{2}+b^{2}}-a \right )}\)	\(522\)

Input:

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

Output:

-1/f*(2*EllipticF((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^(1/2))*a*(-a^2+b^2 
)^(1/2)-2*(-a^2+b^2)^(1/2)*EllipticF((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2 
^(1/2))*b+EllipticPi((csc(f*x+e)-cot(f*x+e)+1)^(1/2),a/(-b+(-a^2+b^2)^(1/2 
)+a),1/2*2^(1/2))*a*b-EllipticPi((csc(f*x+e)-cot(f*x+e)+1)^(1/2),a/(-b+(-a 
^2+b^2)^(1/2)+a),1/2*2^(1/2))*b^2+(-a^2+b^2)^(1/2)*EllipticPi((csc(f*x+e)- 
cot(f*x+e)+1)^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*b-EllipticPi((c 
sc(f*x+e)-cot(f*x+e)+1)^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*a*b+E 
llipticPi((csc(f*x+e)-cot(f*x+e)+1)^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^ 
(1/2))*b^2+(-a^2+b^2)^(1/2)*EllipticPi((csc(f*x+e)-cot(f*x+e)+1)^(1/2),-a/ 
(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*b)*(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*(-2 
*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*(1+cos(f* 
x+e))/(g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)/(-a^2+b^2)^(1/2)/(-b+(-a^2 
+b^2)^(1/2)+a)/(b+(-a^2+b^2)^(1/2)-a)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(1/(sqrt(g*cos(f*x + e))*(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) 
)), x)
 

Giac [F]

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

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

Output:

integrate(1/(sqrt(g*cos(f*x + e))*(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) 
)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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