\(\int \frac {\sqrt {d \cos (e+f x)}}{(a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx\) [2]

Optimal result

Integrand size = 37, antiderivative size = 209 \[ \int \frac {\sqrt {d \cos (e+f x)}}{(a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx=\frac {2 \sqrt {2} \sqrt {d} \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{\sqrt {-a^2+b^2} f \sqrt {g \sin (e+f x)}}-\frac {2 \sqrt {2} \sqrt {d} \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{\sqrt {-a^2+b^2} f \sqrt {g \sin (e+f x)}} \] Output:

2*2^(1/2)*d^(1/2)*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^2+b^2)^(1/2)/f/(g*sin 
(f*x+e))^(1/2)-2*2^(1/2)*d^(1/2)*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^2+b^2) 
^(1/2)/f/(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 = 21.87 (sec) , antiderivative size = 594, normalized size of antiderivative = 2.84 \[ \int \frac {\sqrt {d \cos (e+f x)}}{(a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx=\frac {2 \sqrt {d \cos (e+f x)} \sec ^2(e+f x) \sqrt {\tan (e+f x)} \left (b+a \sqrt {1+\tan ^2(e+f x)}\right ) \left (\frac {\sqrt {a} \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\tan (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\tan (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}+a \tan (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}+a \tan (e+f x)\right )\right )}{4 \sqrt {2} \left (a^2-b^2\right )^{3/4}}+\frac {5 b \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2(e+f x),-\frac {a^2 \tan ^2(e+f x)}{a^2-b^2}\right ) \sqrt {\tan (e+f x)}}{\sqrt {1+\tan ^2(e+f x)} \left (-5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2(e+f x),-\frac {a^2 \tan ^2(e+f x)}{a^2-b^2}\right )+2 \left (2 a^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},-\tan ^2(e+f x),-\frac {a^2 \tan ^2(e+f x)}{a^2-b^2}\right )+\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},-\tan ^2(e+f x),-\frac {a^2 \tan ^2(e+f x)}{a^2-b^2}\right )\right ) \tan ^2(e+f x)\right ) \left (-b^2+a^2 \left (1+\tan ^2(e+f x)\right )\right )}\right )}{f (a+b \cos (e+f x)) \sqrt {g \sin (e+f x)} \left (1+\tan ^2(e+f x)\right )^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3387

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3386

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

\(\Big \downarrow \) 1542

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

Input:

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

Output:

(((-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) - (2*Sqrt[2]*(1 + b/Sqrt[-a^2 + b^2])*Sq 
rt[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))*Sqrt[S 
in[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 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]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(506\) vs. \(2(173)=346\).

Time = 9.89 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.43

method	result	size
default	\(\frac {\left (2 \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}+a \operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b -\sqrt {-a^{2}+b^{2}}\, \operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )-a \operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b -\sqrt {-a^{2}+b^{2}}\, \operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \sqrt {\cos \left (f x +e \right ) d}\, \left (\sec \left (f x +e \right )+1\right ) a}{f \sqrt {g \sin \left (f x +e \right )}\, \sqrt {-a^{2}+b^{2}}\, \left (\sqrt {-a^{2}+b^{2}}+a -b \right ) \left (\sqrt {-a^{2}+b^{2}}-a +b \right )}\)	\(507\)

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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