\(\int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\) [1387]

Optimal result

Integrand size = 33, antiderivative size = 380 \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=-\frac {a^2 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}} \] Output:

-a^2*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(3/2) 
/(-a^2+b^2)^(3/4)/f/g^(1/2)-a^2*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2 
+b^2)^(1/4)/g^(1/2))/b^(3/2)/(-a^2+b^2)^(3/4)/f/g^(1/2)-2*(g*cos(f*x+e))^( 
1/2)/b/f/g-2*a*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))/b^2 
/f/(g*cos(f*x+e))^(1/2)+a^3*cos(f*x+e)^(1/2)*EllipticPi(sin(1/2*f*x+1/2*e) 
,2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))/b^2/(a^2-b*(b-(-a^2+b^2)^(1/2)))/f/(g*c 
os(f*x+e))^(1/2)+a^3*cos(f*x+e)^(1/2)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b 
+(-a^2+b^2)^(1/2)),2^(1/2))/b^2/(a^2-b*(b+(-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 5 vs. order 4 in optimal.

Time = 18.78 (sec) , antiderivative size = 572, normalized size of antiderivative = 1.51 \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=-\frac {\left (a^2-b^2\right ) \left (a^2-b^2+b^2 \cos ^2(e+f x)\right ) \sec (e+f x) \sqrt [4]{\sec ^2(e+f x)} \tan ^3(e+f x) \left (\frac {a^2 \arctan \left (\frac {\sqrt [4]{-a^2+b^2} \sqrt [4]{1+\tan ^2(e+f x)}}{\sqrt {b}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt [4]{-a^2+b^2} \sqrt [4]{1+\tan ^2(e+f x)}}{\sqrt {b}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}+\frac {a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{a^2-b^2}-\frac {a^3 \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {\sqrt {-a^2+b^2}}{b},\arcsin \left (\sqrt [4]{1+\tan ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}}{b^2 \left (a^2-b^2\right )}-\frac {a^3 \cot (e+f x) \operatorname {EllipticPi}\left (\frac {\sqrt {-a^2+b^2}}{b},\arcsin \left (\sqrt [4]{1+\tan ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}}{b^2 \left (a^2-b^2\right )}-\frac {2}{b \sqrt [4]{1+\tan ^2(e+f x)}}\right )}{f \sqrt {g \cos (e+f x)} (b+a \csc (e+f x)) \left (b \left (a^2-b^2\right ) \tan ^5(e+f x)+a^3 \sqrt {-\tan ^2(e+f x)} \sqrt {-\sec ^2(e+f x) \tan ^2(e+f x)}+\tan ^2(e+f x) \left (-a b^2 \sqrt {-\tan ^2(e+f x)} \sqrt {-\sec ^2(e+f x) \tan ^2(e+f x)}+a^3 \left (\sqrt {\sec ^2(e+f x)}+\sqrt {-\tan ^2(e+f x)} \sqrt {-\sec ^2(e+f x) \tan ^2(e+f x)}\right )\right )\right )} \] Input:

Integrate[Sin[e + f*x]^2/(Sqrt[g*Cos[e + f*x]]*(a + b*Sin[e + f*x])),x]
 

Output:

-(((a^2 - b^2)*(a^2 - b^2 + b^2*Cos[e + f*x]^2)*Sec[e + f*x]*(Sec[e + f*x] 
^2)^(1/4)*Tan[e + f*x]^3*((a^2*ArcTan[((-a^2 + b^2)^(1/4)*(1 + Tan[e + f*x 
]^2)^(1/4))/Sqrt[b]])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - (a^2*ArcTanh[((-a^2 + 
 b^2)^(1/4)*(1 + Tan[e + f*x]^2)^(1/4))/Sqrt[b]])/(b^(3/2)*(-a^2 + b^2)^(3 
/4)) + (a*Hypergeometric2F1[1/2, 3/4, 3/2, -Tan[e + f*x]^2]*Tan[e + f*x])/ 
(a^2 - b^2) - (a^3*Cot[e + f*x]*EllipticPi[-(Sqrt[-a^2 + b^2]/b), ArcSin[( 
1 + Tan[e + f*x]^2)^(1/4)], -1]*Sqrt[-Tan[e + f*x]^2])/(b^2*(a^2 - b^2)) - 
 (a^3*Cot[e + f*x]*EllipticPi[Sqrt[-a^2 + b^2]/b, ArcSin[(1 + Tan[e + f*x] 
^2)^(1/4)], -1]*Sqrt[-Tan[e + f*x]^2])/(b^2*(a^2 - b^2)) - 2/(b*(1 + Tan[e 
 + f*x]^2)^(1/4))))/(f*Sqrt[g*Cos[e + f*x]]*(b + a*Csc[e + f*x])*(b*(a^2 - 
 b^2)*Tan[e + f*x]^5 + a^3*Sqrt[-Tan[e + f*x]^2]*Sqrt[-(Sec[e + f*x]^2*Tan 
[e + f*x]^2)] + Tan[e + f*x]^2*(-(a*b^2*Sqrt[-Tan[e + f*x]^2]*Sqrt[-(Sec[e 
 + f*x]^2*Tan[e + f*x]^2)]) + a^3*(Sqrt[Sec[e + f*x]^2] + Sqrt[-Tan[e + f* 
x]^2]*Sqrt[-(Sec[e + f*x]^2*Tan[e + f*x]^2)])))))
 

Rubi [A] (warning: unable to verify)

Time = 1.87 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.01, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 3388, 3042, 3045, 15, 3346, 3042, 3121, 3042, 3120, 3181, 266, 756, 218, 221, 3042, 3286, 3042, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3388

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3045

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 3346

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3181

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

\(\displaystyle -\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\)

\(\Big \downarrow \) 3286

\(\displaystyle -\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\)

\(\Big \downarrow \) 3284

\(\displaystyle -\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}+\frac {a \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\)

Input:

Int[Sin[e + f*x]^2/(Sqrt[g*Cos[e + f*x]]*(a + b*Sin[e + f*x])),x]
 

Output:

(-2*Sqrt[g*Cos[e + f*x]])/(b*f*g) - (a*((2*Sqrt[Cos[e + f*x]]*EllipticF[(e 
 + f*x)/2, 2])/(b*f*Sqrt[g*Cos[e + f*x]]) - (a*((2*b*g*(-1/2*ArcTan[(Sqrt[ 
b]*Sqrt[g]*Cos[e + f*x])/(-a^2 + b^2)^(1/4)]/(Sqrt[b]*(-a^2 + b^2)^(3/4)*g 
^(3/2)) - ArcTanh[(Sqrt[b]*Sqrt[g]*Cos[e + f*x])/(-a^2 + b^2)^(1/4)]/(2*Sq 
rt[b]*(-a^2 + b^2)^(3/4)*g^(3/2))))/f + (a*Sqrt[Cos[e + f*x]]*EllipticPi[( 
2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(Sqrt[-a^2 + b^2]*(b - Sqrt[ 
-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - (a*Sqrt[Cos[e + f*x]]*EllipticPi[(2 
*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(Sqrt[-a^2 + b^2]*(b + Sqrt[- 
a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]])))/b))/b
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 3042

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

rule 3045

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> Simp[-(a*f)^(-1)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], 
x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && 
 !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
 

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 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3181

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[-a/(2*q)   Int[1/( 
Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Simp[b*(g/f)   Subst[ 
Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - S 
imp[a/(2*q)   Int[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x])] / 
; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]
 

rule 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3286

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt 
[c + d*Sin[e + f*x]]   Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + 
 d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]
 

rule 3346

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

rule 3388

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(959\) vs. \(2(335)=670\).

Time = 2.91 (sec) , antiderivative size = 960, normalized size of antiderivative = 2.53

method	result	size
default	\(\text {Expression too large to display}\)	\(960\)

Input:

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

Output:

(16*b*(-1/8/b^2/g*(g*(-1+2*cos(1/2*f*x+1/2*e)^2))^(1/2)-1/4*a^2/b^2*(g^2*( 
a^2-b^2)/b^2)^(1/4)*2^(1/2)*(-ln((-(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2* 
f*x+1/2*e)^2-g)^(1/2)*2^(1/2)-2*g*cos(1/2*f*x+1/2*e)^2-(g^2*(a^2-b^2)/b^2) 
^(1/2)+g)/((g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^ 
(1/2)-2*g*cos(1/2*f*x+1/2*e)^2-(g^2*(a^2-b^2)/b^2)^(1/2)+g))-2*arctan(2^(1 
/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+1)+2*arct 
an(-2^(1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+1 
))/(16*a^2-16*b^2)/g)+1/8*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e 
)^2)^(1/2)*a*(16*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^( 
1/2)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*b^2-sum(1/_alpha/(2*_alpha^2-1) 
*(8*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)* 
(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),-4*b^2/a^2* 
(_alpha^2-1),2^(1/2))*_alpha^3*b^2-8*b^2*_alpha*(sin(1/2*f*x+1/2*e)^2)^(1/ 
2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),-4*b^2/a 
^2*(_alpha^2-1),2^(1/2))*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)+a^2*2^(1 
/2)*arctanh(1/2*g*(4*_alpha^2-3)/(4*a^2-3*b^2)*(b^2*_alpha^2+4*a^2*cos(1/2 
*f*x+1/2*e)^2-3*b^2*cos(1/2*f*x+1/2*e)^2-3*a^2+2*b^2)*2^(1/2)/(g*(2*_alpha 
^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e 
)^2))^(1/2))*(-g*sin(1/2*f*x+1/2*e)^2*(-1+2*sin(1/2*f*x+1/2*e)^2))^(1/2))/ 
(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-g*sin(1/2*f*x+1/2*e)^2*(-1+2...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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