\(\int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\) [1428]

Optimal result

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

-2*2^(1/2)*d^(1/2)*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^2+b^2)^(1/2)/f/(g*co 
s(f*x+e))^(1/2)+2*2^(1/2)*d^(1/2)*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^2+b^2 
)^(1/2)/f/(g*cos(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 = 16.85 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.78 \[ \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {2 \sqrt {d \sin (e+f x)} \left (a \sqrt {\sec ^2(e+f x)}+b \tan (e+f x)\right ) \left (\frac {\sqrt {a} \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}}{\sqrt {a}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}}{\sqrt {a}}\right )+\log \left (-a+\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}-\sqrt {a^2-b^2} \tan (e+f x)\right )-\log \left (a+\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}+\sqrt {a^2-b^2} \tan (e+f x)\right )\right )}{4 \sqrt {2} \left (a^2-b^2\right )^{3/4}}-\frac {b \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right ) \tan ^{\frac {5}{2}}(e+f x)}{5 a^2}\right )}{f \sqrt {g \cos (e+f x)} \sqrt {\sec ^2(e+f x)} (a+b \sin (e+f x)) \sqrt {\tan (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.71 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3387

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3386

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

\(\Big \downarrow \) 1542

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

Input:

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

Output:

(Sqrt[Cos[e + f*x]]*((2*Sqrt[2]*(1 - b/Sqrt[-a^2 + b^2])*Sqrt[d]*EllipticP 
i[-(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) + (2*Sqrt[2]*(1 + b/Sq 
rt[-a^2 + b^2])*Sqrt[d]*EllipticPi[-(a/(b + Sqrt[-a^2 + b^2])), ArcSin[Sqr 
t[d*Sin[e + f*x]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x]])], -1])/((b + Sqrt[-a^2 
+ b^2])*f)))/Sqrt[g*Cos[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. \(446\) vs. \(2(173)=346\).

Time = 2.85 (sec) , antiderivative size = 447, normalized size of antiderivative = 2.14

method	result	size
default	\(-\frac {\sqrt {d \sin \left (f x +e \right )}\, \left (\sqrt {-a^{2}+b^{2}}\, \operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right )+\sqrt {-a^{2}+b^{2}}\, \operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right ) a -\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right ) b -\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) a +\operatorname {EllipticPi}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) b \right ) \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \sin \left (f x +e \right ) a}{f \left (\cos \left (f x +e \right )-1\right ) \sqrt {g \cos \left (f x +e \right )}\, \sqrt {-a^{2}+b^{2}}\, \left (-b +\sqrt {-a^{2}+b^{2}}+a \right ) \left (b +\sqrt {-a^{2}+b^{2}}-a \right )}\)	\(447\)

Input:

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

Output:

-1/f*(d*sin(f*x+e))^(1/2)*((-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))+(-a^2+b^2)^(1/2)*Ellip 
ticPi((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 
))+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-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((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+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))^(1/2) 
*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*sin( 
f*x+e)/(cos(f*x+e)-1)/(g*cos(f*x+e))^(1/2)*a/(-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. \[ \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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