Integrand size = 54, antiderivative size = 445 \[ \int \frac {\left (a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {(a-b) \sqrt {a+b} B \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} (2 A+B) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} (2 A b+a B) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b d \sqrt {\sec (c+d x)}}+\frac {B \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]
B*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)/d-(a-b)*B*csc(d*x+c)* EllipticE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b ))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*( 1+sec(d*x+c))/(a-b))^(1/2)/a/d/sec(d*x+c)^(1/2)+(2*A+B)*csc(d*x+c)*Ellipti cF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2 ))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d *x+c))/(a-b))^(1/2)/d/sec(d*x+c)^(1/2)-(2*A*b+B*a)*csc(d*x+c)*EllipticPi(( a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,((-a-b)/(a-b))^ (1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+s ec(d*x+c))/(a-b))^(1/2)/b/d/sec(d*x+c)^(1/2)
Time = 6.06 (sec) , antiderivative size = 787, normalized size of antiderivative = 1.77 \[ \int \frac {\left (a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {-a B \tan \left (\frac {1}{2} (c+d x)\right )-b B \tan \left (\frac {1}{2} (c+d x)\right )+2 b B \tan ^3\left (\frac {1}{2} (c+d x)\right )+a B \tan ^5\left (\frac {1}{2} (c+d x)\right )-b B \tan ^5\left (\frac {1}{2} (c+d x)\right )-4 A b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-2 a B \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-4 A b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-2 a B \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-(a+b) B E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+2 (A b+a (-A+B)) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}}{d \sqrt {\frac {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (-1+\tan ^4\left (\frac {1}{2} (c+d x)\right )\right )} \]
Integrate[((a*A + (A*b + a*B)*Cos[c + d*x] + b*B*Cos[c + d*x]^2)*Sqrt[Sec[ c + d*x]])/Sqrt[a + b*Cos[c + d*x]],x]
(-(a*B*Tan[(c + d*x)/2]) - b*B*Tan[(c + d*x)/2] + 2*b*B*Tan[(c + d*x)/2]^3 + a*B*Tan[(c + d*x)/2]^5 - b*B*Tan[(c + d*x)/2]^5 - 4*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*S qrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*B *EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[( c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/ (a + b)] - 4*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b) ]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*B*EllipticPi[-1, ArcSin[ Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - (a + b)*B*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqr t[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*(A*b + a*(-A + B))*Ellipt icF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^ 2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)])/(d*Sqrt[(1 + Tan[(c + d*x)/2]^2)/(1 - Tan[(c + d*x) /2]^2)]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(1 + Ta n[(c + d*x)/2]^2)]*(-1 + Tan[(c + d*x)/2]^4))
Time = 2.09 (sec) , antiderivative size = 422, normalized size of antiderivative = 0.95, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3042, 4709, 3042, 3508, 3042, 3482, 25, 3042, 3532, 3042, 3288, 3477, 3042, 3295, 3473}
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.
| \(\displaystyle \int \frac {\sqrt {\sec (c+d x)} \left ((a B+A b) \cos (c+d x)+a A+b B \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\sec (c+d x)} \left ((a B+A b) \cos (c+d x)+a A+b B \cos (c+d x)^2\right )}{\sqrt {a+b \cos (c+d x)}}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {b B \cos ^2(c+d x)+(A b+a B) \cos (c+d x)+a A}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {b B \sin \left (c+d x+\frac {\pi }{2}\right )^2+(A b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )+a A}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {a+b \cos (c+d x)} \left (A b^2+B \cos (c+d x) b^2\right )}{\sqrt {\cos (c+d x)}}dx}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (A b^2+B \sin \left (c+d x+\frac {\pi }{2}\right ) b^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}\) |
| \(\Big \downarrow \) 3482 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \int -\frac {-\left ((2 A b+a B) \cos ^2(c+d x) b^2\right )+a B b^2-2 a A \cos (c+d x) b^2}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+\frac {b^2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )}{b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {b^2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {1}{2} \int \frac {-\left ((2 A b+a B) \cos ^2(c+d x) b^2\right )+a B b^2-2 a A \cos (c+d x) b^2}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx\right )}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {b^2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {1}{2} \int \frac {-\left ((2 A b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^2\right )+a B b^2-2 a A \sin \left (c+d x+\frac {\pi }{2}\right ) b^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{b^2}\) |
| \(\Big \downarrow \) 3532 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (b^2 (a B+2 A b) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx-\int \frac {a b^2 B-2 a A b^2 \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx\right )+\frac {b^2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (b^2 (a B+2 A b) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\int \frac {a b^2 B-2 a A b^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {b^2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )}{b^2}\) |
| \(\Big \downarrow \) 3288 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (-\int \frac {a b^2 B-2 a A b^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b \sqrt {a+b} (a B+2 A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {b^2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )}{b^2}\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (a b^2 (2 A+B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx-a b^2 B \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\frac {2 b \sqrt {a+b} (a B+2 A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {b^2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (a b^2 (2 A+B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a b^2 B \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b \sqrt {a+b} (a B+2 A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {b^2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )}{b^2}\) |
| \(\Big \downarrow \) 3295 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (-a b^2 B \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b^2 \sqrt {a+b} (2 A+B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}-\frac {2 b \sqrt {a+b} (a B+2 A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {b^2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )}{b^2}\) |
| \(\Big \downarrow \) 3473 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {2 b^2 \sqrt {a+b} (2 A+B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}-\frac {2 b \sqrt {a+b} (a B+2 A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}-\frac {2 b^2 B (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}\right )+\frac {b^2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )}{b^2}\) |
Int[((a*A + (A*b + a*B)*Cos[c + d*x] + b*B*Cos[c + d*x]^2)*Sqrt[Sec[c + d* x]])/Sqrt[a + b*Cos[c + d*x]],x]
(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((-2*(a - b)*b^2*Sqrt[a + b]*B*Cot [c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[ c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt [(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d) + (2*b^2*Sqrt[a + b]*(2*A + B)*Cot [c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[ c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt [(a*(1 + Sec[c + d*x]))/(a - b)])/d - (2*b*Sqrt[a + b]*(2*A*b + a*B)*Cot[c + d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b] *Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/d)/2 + (b^2*B*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/b^2
3.16.32.3.1 Defintions of rubi rules used
Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c *((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti cPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e , f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A, B]
Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[-2*B*Cos[e + f*x]*Sqrt[a + b*Sin[e + f*x]]*((c + d*Sin[e + f*x])^n/(f*(2* n + 3))), x] + Simp[1/(2*n + 3) Int[((c + d*Sin[e + f*x])^(n - 1)/Sqrt[a + b*Sin[e + f*x]])*Simp[a*A*c*(2*n + 3) + B*(b*c + 2*a*d*n) + (B*(a*c + b*d )*(2*n + 1) + A*(b*c + a*d)*(2*n + 3))*Sin[e + f*x] + (A*b*d*(2*n + 3) + B* (a*d + 2*b*c*n))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Eq Q[n^2, 1/4]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[C/b^2 Int[Sqrt[a + b*Sin[e + f*x]] /Sqrt[c + d*Sin[e + f*x]], x], x] + Simp[1/b^2 Int[(A*b^2 - a^2*C + b*(b* B - 2*a*C)*Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x ]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] & & NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1095\) vs. \(2(405)=810\).
Time = 9.57 (sec) , antiderivative size = 1096, normalized size of antiderivative = 2.46
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1096\) |
| default | \(\text {Expression too large to display}\) | \(2084\) |
int((a*A+(A*b+B*a)*cos(d*x+c)+b*B*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+b*cos( d*x+c))^(1/2),x,method=_RETURNVERBOSE)
2*(A*b+B*a)/d*sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2)*(EllipticF(cot(d*x+c )-csc(d*x+c),(-(a-b)/(a+b))^(1/2))-2*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,( -(a-b)/(a+b))^(1/2)))*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos (d*x+c)/(1+cos(d*x+c)))^(1/2)*(1+cos(d*x+c))-2*a*A/d*(1+cos(d*x+c))*(cos(d *x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2 )*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))/(a+b*cos(d*x+c))^( 1/2)*sec(d*x+c)^(1/2)+B/d*(-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+ b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b )/(a+b))^(1/2))*a*cos(d*x+c)^2-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)* (a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-( a-b)/(a+b))^(1/2))*b*cos(d*x+c)^2+2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/( a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+ c),-1,(-(a-b)/(a+b))^(1/2))*a*cos(d*x+c)^2-2*(cos(d*x+c)/(1+cos(d*x+c)))^( 1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(cot(d*x+c)- csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a*cos(d*x+c)-2*(cos(d*x+c)/(1+cos(d*x+c)) )^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(cot(d*x+ c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*b*cos(d*x+c)+4*(cos(d*x+c)/(1+cos(d*x+ c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticPi(cot( d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*a*cos(d*x+c)+b*cos(d*x+c)^2*sin (d*x+c)-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+...
\[ \int \frac {\left (a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (B b \cos \left (d x + c\right )^{2} + A a + {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sqrt {\sec \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]
integrate((a*A+(A*b+B*a)*cos(d*x+c)+b*B*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+ b*cos(d*x+c))^(1/2),x, algorithm="fricas")
\[ \int \frac {\left (a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \left (A + B \cos {\left (c + d x \right )}\right ) \sqrt {a + b \cos {\left (c + d x \right )}} \sqrt {\sec {\left (c + d x \right )}}\, dx \]
integrate((a*A+(A*b+B*a)*cos(d*x+c)+b*B*cos(d*x+c)**2)*sec(d*x+c)**(1/2)/( a+b*cos(d*x+c))**(1/2),x)
\[ \int \frac {\left (a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (B b \cos \left (d x + c\right )^{2} + A a + {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sqrt {\sec \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]
integrate((a*A+(A*b+B*a)*cos(d*x+c)+b*B*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+ b*cos(d*x+c))^(1/2),x, algorithm="maxima")
integrate((B*b*cos(d*x + c)^2 + A*a + (B*a + A*b)*cos(d*x + c))*sqrt(sec(d *x + c))/sqrt(b*cos(d*x + c) + a), x)
\[ \int \frac {\left (a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (B b \cos \left (d x + c\right )^{2} + A a + {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sqrt {\sec \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]
integrate((a*A+(A*b+B*a)*cos(d*x+c)+b*B*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+ b*cos(d*x+c))^(1/2),x, algorithm="giac")
integrate((B*b*cos(d*x + c)^2 + A*a + (B*a + A*b)*cos(d*x + c))*sqrt(sec(d *x + c))/sqrt(b*cos(d*x + c) + a), x)
Timed out. \[ \int \frac {\left (a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (B\,b\,{\cos \left (c+d\,x\right )}^2+\left (A\,b+B\,a\right )\,\cos \left (c+d\,x\right )+A\,a\right )}{\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]
int(((1/cos(c + d*x))^(1/2)*(A*a + cos(c + d*x)*(A*b + B*a) + B*b*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(1/2),x)