Integrand size = 33, antiderivative size = 380 \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=-\frac {a^2 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}} \]
-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*a*(cos(1/2*f*x+1 /2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*co s(f*x+e)^(1/2)/b^2/f/(g*cos(f*x+e))^(1/2)+a^3*(cos(1/2*f*x+1/2*e)^2)^(1/2) /cos(1/2*f*x+1/2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2)) ,2^(1/2))*cos(f*x+e)^(1/2)/b^2/f/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(g*cos(f*x+e ))^(1/2)+a^3*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticPi(si n(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos(f*x+e)^(1/2)/b^2/f/ (a^2-b*(b+(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)-2*(g*cos(f*x+e))^(1/2)/b /f/g
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 14.43 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.06 \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {\cos ^2(e+f x) \left (a^2-b^2+b^2 \cos ^2(e+f x)\right ) \sec ^2(e+f x)^{3/4} \left (\frac {a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{a^2-b^2}+\frac {\frac {a^2 \sqrt {b} \arctan \left (\frac {\sqrt [4]{-a^2+b^2} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt {b}}\right )}{\left (-a^2+b^2\right )^{3/4}}-\frac {a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt [4]{-a^2+b^2} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt {b}}\right )}{\left (-a^2+b^2\right )^{3/4}}-\frac {2 b}{\sqrt [4]{\sec ^2(e+f x)}}+\frac {a^3 \operatorname {EllipticPi}\left (-\frac {\sqrt {-a^2+b^2}}{b},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \tan (e+f x)}{\left (a^2-b^2\right ) \sqrt {-\tan ^2(e+f x)}}+\frac {a^3 \operatorname {EllipticPi}\left (\frac {\sqrt {-a^2+b^2}}{b},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \tan (e+f x)}{\left (a^2-b^2\right ) \sqrt {-\tan ^2(e+f x)}}}{b^2}\right )}{f \sqrt {g \cos (e+f x)} (a+b \sin (e+f x)) \left (a-b \cos (e+f x) \sqrt {\sec ^2(e+f x)} \sin (e+f x)\right )} \]
(Cos[e + f*x]^2*(a^2 - b^2 + b^2*Cos[e + f*x]^2)*(Sec[e + f*x]^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^2*Sqrt[b]*ArcTan[((-a^2 + b^2)^(1/4)*(Sec[e + f*x]^2)^(1/4))/Sqr t[b]])/(-a^2 + b^2)^(3/4) - (a^2*Sqrt[b]*ArcTanh[((-a^2 + b^2)^(1/4)*(Sec[ e + f*x]^2)^(1/4))/Sqrt[b]])/(-a^2 + b^2)^(3/4) - (2*b)/(Sec[e + f*x]^2)^( 1/4) + (a^3*EllipticPi[-(Sqrt[-a^2 + b^2]/b), ArcSin[(Sec[e + f*x]^2)^(1/4 )], -1]*Tan[e + f*x])/((a^2 - b^2)*Sqrt[-Tan[e + f*x]^2]) + (a^3*EllipticP i[Sqrt[-a^2 + b^2]/b, ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*Tan[e + f*x])/(( a^2 - b^2)*Sqrt[-Tan[e + f*x]^2]))/b^2))/(f*Sqrt[g*Cos[e + f*x]]*(a + b*Si n[e + f*x])*(a - b*Cos[e + f*x]*Sqrt[Sec[e + f*x]^2]*Sin[e + f*x]))
Time = 1.90 (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 |
| \(\displaystyle -\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}+\frac {a \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\) |
(-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
3.14.91.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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]
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]
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]
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]
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])
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[((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]
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]
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]
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]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.91 (sec) , antiderivative size = 995, normalized size of antiderivative = 2.62
(16*b*(-1/8/b^2/g*(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)+1/4*a^2/b^2*(g^2*(a ^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((2*g*cos(1/2*f*x+1/2*e)^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)+(g^2*(a^2-b^2)/b^2)^ (1/2))/(2*g*cos(1/2*f*x+1/2*e)^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)+(g^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2 )*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2- b^2)/b^2)^(1/4))+2*arctan((2^(1/2)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)-(g^2 *(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2)/b^2)^(1/4)))/(16*a^2-16*b^2)/g)+1/8* (g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*a*(16*EllipticF( cos(1/2*f*x+1/2*e),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)*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))*_alph a^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*(2*sin(1/2*f*x+1/2*e)^2-1))^(1/2))/(g*(2*_alpha^2*b^2+a^2-2*...
Timed out. \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
\[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
Timed out. \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^2}{\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]