Integrand size = 37, antiderivative size = 273 \[ \int \frac {1}{\sqrt {d \cos (e+f x)} (a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx=-\frac {2 \sqrt {2} b \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{a \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \sin (e+f x)}}+\frac {2 \sqrt {2} b \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{a \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \sin (e+f x)}}+\frac {\operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {\sin (2 e+2 f x)}}{a f \sqrt {d \cos (e+f x)} \sqrt {g \sin (e+f x)}} \]
-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)*2^(1/2)*sin(f*x+e)^(1/2)/a/f/(-a^2+b^2)^(1/2)/d^(1/2)/( g*sin(f*x+e))^(1/2)+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)*2^(1/2)*sin(f*x+e)^(1/2)/a/f/(-a^2+b ^2)^(1/2)/d^(1/2)/(g*sin(f*x+e))^(1/2)-(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1 /4*Pi+f*x)*EllipticF(cos(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)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 16.82 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.82 \[ \int \frac {1}{\sqrt {d \cos (e+f x)} (a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx=\frac {18 (a+b) \sqrt {g \sin (e+f x)} \left (5 \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )+\operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{f g \sqrt {d \cos (e+f x)} (a+b \cos (e+f x)) \left (45 (a+b) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )+\tan ^2\left (\frac {1}{2} (e+f x)\right ) \left (45 (a+b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )-36 (a-b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )+18 (a+b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )+10 \left (-2 (a-b) \operatorname {AppellF1}\left (\frac {9}{4},\frac {1}{2},2,\frac {13}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )+(a+b) \operatorname {AppellF1}\left (\frac {9}{4},\frac {3}{2},1,\frac {13}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )} \]
(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)))
Time = 1.43 (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 |
\(\displaystyle \frac {\sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a f \sqrt {d \cos (e+f x)} \sqrt {g \sin (e+f x)}}-\frac {b \sqrt {\sin (e+f x)} \left (-\frac {2 \sqrt {2} \sqrt {d} \left (1-\frac {b}{\sqrt {b^2-a^2}}\right ) \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {\sin (e+f x)+1}}\right ),-1\right )}{f \left (b-\sqrt {b^2-a^2}\right )}-\frac {2 \sqrt {2} \sqrt {d} \left (\frac {b}{\sqrt {b^2-a^2}}+1\right ) \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {\sin (e+f x)+1}}\right ),-1\right )}{f \left (\sqrt {b^2-a^2}+b\right )}\right )}{a d \sqrt {g \sin (e+f x)}}\) |
-((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]] )
3.1.4.3.1 Defintions of rubi rules used
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]
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]
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]
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]
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]
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]
Time = 4.84 (sec) , antiderivative size = 514, normalized size of antiderivative = 1.88
method | result | size |
default | \(\frac {\left (\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b \sqrt {-a^{2}+b^{2}}+\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b \sqrt {-a^{2}+b^{2}}-2 F\left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) a \sqrt {-a^{2}+b^{2}}+\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) a b -\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b^{2}-\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) a b +\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b^{2}\right ) \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )+1}\, \sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}\, \left (1+\cos \left (f x +e \right )\right ) \sqrt {2}}{f \sqrt {g \sin \left (f x +e \right )}\, \sqrt {\cos \left (f x +e \right ) d}\, \left (b +\sqrt {-a^{2}+b^{2}}-a \right ) \left (-b +\sqrt {-a^{2}+b^{2}}+a \right ) \sqrt {-a^{2}+b^{2}}}\) | \(514\) |
1/f*(EllipticPi((-cot(f*x+e)+csc(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((-cot(f*x+e)+csc(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)-2*EllipticF((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2*2^(1/2))*a*(-a^2+b^2)^ (1/2)+EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),-(a-b)/(-a+b+(-(a-b)*(a+ b))^(1/2)),1/2*2^(1/2))*a*b-EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),-( a-b)/(-a+b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b^2-EllipticPi((-cot(f*x+e)+ csc(f*x+e)+1)^(1/2),(a-b)/(a-b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a*b+Elli pticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),(a-b)/(a-b+(-(a-b)*(a+b))^(1/2)),1 /2*2^(1/2))*b^2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*(-csc(f*x+e)+cot(f*x+e)+1) ^(1/2)*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(1+cos(f*x+e))/(g*sin(f*x+e))^(1/2 )/(cos(f*x+e)*d)^(1/2)*2^(1/2)/(b+(-a^2+b^2)^(1/2)-a)/(-b+(-a^2+b^2)^(1/2) +a)/(-a^2+b^2)^(1/2)
Timed out. \[ \int \frac {1}{\sqrt {d \cos (e+f x)} (a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\sqrt {d \cos (e+f x)} (a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx=\int \frac {1}{\sqrt {d \cos {\left (e + f x \right )}} \sqrt {g \sin {\left (e + f x \right )}} \left (a + b \cos {\left (e + f x \right )}\right )}\, dx \]
\[ \int \frac {1}{\sqrt {d \cos (e+f x)} (a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx=\int { \frac {1}{{\left (b \cos \left (f x + e\right ) + a\right )} \sqrt {d \cos \left (f x + e\right )} \sqrt {g \sin \left (f x + e\right )}} \,d x } \]
\[ \int \frac {1}{\sqrt {d \cos (e+f x)} (a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx=\int { \frac {1}{{\left (b \cos \left (f x + e\right ) + a\right )} \sqrt {d \cos \left (f x + e\right )} \sqrt {g \sin \left (f x + e\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {d \cos (e+f x)} (a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx=\int \frac {1}{\sqrt {d\,\cos \left (e+f\,x\right )}\,\sqrt {g\,\sin \left (e+f\,x\right )}\,\left (a+b\,\cos \left (e+f\,x\right )\right )} \,d x \]