\(\int \frac {1}{\sqrt {d \cos (e+f x)} (a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx\) [4]

Optimal result

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

-2*2^(1/2)*b*EllipticPi((d*cos(f*x+e))^(1/2)/d^(1/2)/(1+sin(f*x+e))^(1/2), 
-a/(b-(-a^2+b^2)^(1/2)),I)*sin(f*x+e)^(1/2)/a/(-a^2+b^2)^(1/2)/d^(1/2)/f/( 
g*sin(f*x+e))^(1/2)+2*2^(1/2)*b*EllipticPi((d*cos(f*x+e))^(1/2)/d^(1/2)/(1 
+sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)),I)*sin(f*x+e)^(1/2)/a/(-a^2+b^2 
)^(1/2)/d^(1/2)/f/(g*sin(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/(d*cos(f*x+e))^(1/2)/(g*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 = 22.01 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.82 \[ \int \frac {1}{\sqrt {d \cos (e+f x)} (a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx=\frac {18 (a+b) \sqrt {g \sin (e+f x)} \left (5 \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )+\operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{f g \sqrt {d \cos (e+f x)} (a+b \cos (e+f x)) \left (45 (a+b) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )+\tan ^2\left (\frac {1}{2} (e+f x)\right ) \left (45 (a+b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )-36 (a-b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )+18 (a+b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )+10 \left (-2 (a-b) \operatorname {AppellF1}\left (\frac {9}{4},\frac {1}{2},2,\frac {13}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )+(a+b) \operatorname {AppellF1}\left (\frac {9}{4},\frac {3}{2},1,\frac {13}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )} \] Input:

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

Output:

(18*(a + b)*Sqrt[g*Sin[e + f*x]]*(5*AppellF1[1/4, 1/2, 1, 5/4, Tan[(e + f* 
x)/2]^2, ((-a + b)*Tan[(e + f*x)/2]^2)/(a + b)] + AppellF1[5/4, 1/2, 1, 9/ 
4, Tan[(e + f*x)/2]^2, ((-a + b)*Tan[(e + f*x)/2]^2)/(a + b)]*Tan[(e + f*x 
)/2]^2))/(f*g*Sqrt[d*Cos[e + f*x]]*(a + b*Cos[e + f*x])*(45*(a + b)*Appell 
F1[1/4, 1/2, 1, 5/4, Tan[(e + f*x)/2]^2, ((-a + b)*Tan[(e + f*x)/2]^2)/(a 
+ b)] + Tan[(e + f*x)/2]^2*(45*(a + b)*AppellF1[5/4, 1/2, 1, 9/4, Tan[(e + 
 f*x)/2]^2, ((-a + b)*Tan[(e + f*x)/2]^2)/(a + b)] - 36*(a - b)*AppellF1[5 
/4, 1/2, 2, 9/4, Tan[(e + f*x)/2]^2, ((-a + b)*Tan[(e + f*x)/2]^2)/(a + b) 
] + 18*(a + b)*AppellF1[5/4, 3/2, 1, 9/4, Tan[(e + f*x)/2]^2, ((-a + b)*Ta 
n[(e + f*x)/2]^2)/(a + b)] + 10*(-2*(a - b)*AppellF1[9/4, 1/2, 2, 13/4, Ta 
n[(e + f*x)/2]^2, ((-a + b)*Tan[(e + f*x)/2]^2)/(a + b)] + (a + b)*AppellF 
1[9/4, 3/2, 1, 13/4, Tan[(e + f*x)/2]^2, ((-a + b)*Tan[(e + f*x)/2]^2)/(a 
+ b)])*Tan[(e + f*x)/2]^2)))
 

Rubi [A] (verified)

Time = 1.29 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3389

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3053

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3387

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3386

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

\(\Big \downarrow \) 1542

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

Input:

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

Output:

-((b*((-2*Sqrt[2]*(1 - b/Sqrt[-a^2 + b^2])*Sqrt[d]*EllipticPi[-(a/(b - Sqr 
t[-a^2 + b^2])), ArcSin[Sqrt[d*Cos[e + f*x]]/(Sqrt[d]*Sqrt[1 + Sin[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*Cos[e + f* 
x]]/(Sqrt[d]*Sqrt[1 + Sin[e + f*x]])], -1])/((b + Sqrt[-a^2 + b^2])*f))*Sq 
rt[Sin[e + f*x]])/(a*d*Sqrt[g*Sin[e + f*x]])) + (EllipticF[e - Pi/4 + f*x, 
 2]*Sqrt[Sin[2*e + 2*f*x]])/(a*f*Sqrt[d*Cos[e + f*x]]*Sqrt[g*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. \(515\) vs. \(2(231)=462\).

Time = 9.30 (sec) , antiderivative size = 516, normalized size of antiderivative = 1.89

method	result	size
default	\(-\frac {\left (2 \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) a \sqrt {-a^{2}+b^{2}}+\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) a b -\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b^{2}-\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b \sqrt {-a^{2}+b^{2}}-\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) a b +\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b^{2}-\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b \sqrt {-a^{2}+b^{2}}\right ) \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \left (\cos \left (f x +e \right )+1\right )}{f \sqrt {\cos \left (f x +e \right ) d}\, \sqrt {g \sin \left (f x +e \right )}\, \left (\sqrt {-a^{2}+b^{2}}-a +b \right ) \left (\sqrt {-a^{2}+b^{2}}+a -b \right ) \sqrt {-a^{2}+b^{2}}}\)	\(516\)

Input:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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