Integrand size = 25, antiderivative size = 505 \[ \int \frac {1}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {3 (a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b^2 d \sqrt {\sec (c+d x)}}-\frac {(3 a-2 b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b^2 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (3 a^2+4 b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b^3 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 b d \sqrt {\sec (c+d x)}}-\frac {3 a \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^2 d} \]
1/2*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d/sec(d*x+c)^(1/2)-3/4*a*sin(d*x+c )*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)/b^2/d+3/4*(a-b)*csc(d*x+c)*Ellip ticE((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)/b^2/d/sec(d*x+c)^(1/2)-1/4*(3*a-2*b)*csc(d*x+c)*Elli pticF((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+se c(d*x+c))/(a-b))^(1/2)/b^2/d/sec(d*x+c)^(1/2)-1/4*(3*a^2+4*b^2)*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+sec(d*x+c))/(a-b))^(1/2)/b^3/d/sec(d*x+c)^(1/2)
Time = 12.71 (sec) , antiderivative size = 678, normalized size of antiderivative = 1.34 \[ \int \frac {1}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (2 (c+d x))}{4 b d}-\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (24 a (a+b) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )-16 (a-2 b) b \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-36 a^2 \cos \left (\frac {1}{2} (c+d x)\right ) \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-48 b^2 \cos \left (\frac {1}{2} (c+d x)\right ) \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-12 a^2 \cos \left (\frac {3}{2} (c+d x)\right ) \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-16 b^2 \cos \left (\frac {3}{2} (c+d x)\right ) \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-6 a^2 \sin \left (\frac {1}{2} (c+d x)\right )+6 a b \sin \left (\frac {1}{2} (c+d x)\right )+6 a^2 \sin \left (\frac {3}{2} (c+d x)\right )-3 a b \sin \left (\frac {3}{2} (c+d x)\right )+3 a b \sin \left (\frac {5}{2} (c+d x)\right )\right )}{16 b^2 d \sqrt {a+b \cos (c+d x)}} \]
(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[2*(c + d*x)])/(4*b*d) - ( Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(24*a*(a + b)*Cos[(c + d*x)/2]^3*Sqrt[ Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + C os[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] - 16* (a - 2*b)*b*Cos[(c + d*x)/2]^3*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[ (a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] - 36*a^2*Cos[(c + d*x)/2]*EllipticPi[-1, Ar cSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sq rt[(b + a*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] - 48*b^2*Cos[(c + d* x)/2]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(b + a*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]) )] - 12*a^2*Cos[(3*(c + d*x))/2]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(b + a*Sec[c + d*x])/ ((a + b)*(1 + Sec[c + d*x]))] - 16*b^2*Cos[(3*(c + d*x))/2]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)] *Sqrt[(b + a*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] - 6*a^2*Sin[(c + d*x)/2] + 6*a*b*Sin[(c + d*x)/2] + 6*a^2*Sin[(3*(c + d*x))/2] - 3*a*b*Sin[ (3*(c + d*x))/2] + 3*a*b*Sin[(5*(c + d*x))/2]))/(16*b^2*d*Sqrt[a + b*Cos[c + d*x]])
Time = 2.00 (sec) , antiderivative size = 468, normalized size of antiderivative = 0.93, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4710, 3042, 3272, 27, 3042, 3540, 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.
| \(\displaystyle \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4710 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3272 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {-3 a \cos ^2(c+d x)+2 b \cos (c+d x)+a}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {-3 a \cos ^2(c+d x)+2 b \cos (c+d x)+a}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{4 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {-3 a \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b \sin \left (c+d x+\frac {\pi }{2}\right )+a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
| \(\Big \downarrow \) 3540 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {3 a^2+2 b \cos (c+d x) a+\left (3 a^2+4 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {3 a \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {3 a^2+2 b \sin \left (c+d x+\frac {\pi }{2}\right ) a+\left (3 a^2+4 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {3 a \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
| \(\Big \downarrow \) 3532 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\left (3 a^2+4 b^2\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx+\int \frac {3 a^2+2 b \cos (c+d x) a}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {3 a \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\left (3 a^2+4 b^2\right ) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {3 a^2+2 b \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {3 a \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
| \(\Big \downarrow \) 3288 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {3 a^2+2 b \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (3 a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {3 a \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 a^2 \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a (3 a-2 b) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx-\frac {2 \sqrt {a+b} \left (3 a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {3 a \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 a^2 \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a (3 a-2 b) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (3 a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {3 a \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
| \(\Big \downarrow \) 3295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 a^2 \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (3 a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}-\frac {2 (3 a-2 b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {3 a \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
| \(\Big \downarrow \) 3473 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {-\frac {2 \sqrt {a+b} \left (3 a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}-\frac {2 (3 a-2 b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {6 (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {3 a \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[ c + d*x]]*Sin[c + d*x])/(2*b*d) + (((6*(a - b)*Sqrt[a + b]*Cot[c + d*x]*El lipticE[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*(3*a - 2*b)*Sqrt[a + b]*Cot[c + d*x]*Elliptic F[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*Sqrt[a + b]*(3*a^2 + 4*b^2)*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)])/(b*d))/(2*b) - (3*a*Sqrt[a + b*Cos[c + d*x]]*Sin [c + d*x])/(b*d*Sqrt[Cos[c + d*x]]))/(4*b))
3.8.56.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d *(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m ] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
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[((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[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(Sqrt[c + d*Sin[e + f *x]]/(d*f*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[1/(2*d) Int[(1/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]))*Simp[2*a*A*d - C*(b*c - a*d) - 2*(a*c*C - d*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - C*(b*c + a*d))*Sin[e + f*x]^2, 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[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Sin[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. \(1688\) vs. \(2(451)=902\).
Time = 8.53 (sec) , antiderivative size = 1689, normalized size of antiderivative = 3.34
1/4/d*sec(d*x+c)^(1/2)*(-2*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^ (1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+cos(d* x+c)))^(1/2)*a*b*cos(d*x+c)^2+4*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a +b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+c os(d*x+c)))^(1/2)*b^2*cos(d*x+c)^2+3*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a- b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c) /(1+cos(d*x+c)))^(1/2)*a^2*cos(d*x+c)^2+3*EllipticE(cot(d*x+c)-csc(d*x+c), (-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d *x+c)/(1+cos(d*x+c)))^(1/2)*a*b*cos(d*x+c)^2-6*EllipticPi(cot(d*x+c)-csc(d *x+c),-1,(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/ 2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^2*cos(d*x+c)^2-8*EllipticPi(cot(d*x +c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/( a+b))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*b^2*cos(d*x+c)^2+2*b^2*cos(d *x+c)^3*sin(d*x+c)-4*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2)) *((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c))) ^(1/2)*a*b*cos(d*x+c)+8*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/ 2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c )))^(1/2)*b^2*cos(d*x+c)+6*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^ (1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+cos(d* x+c)))^(1/2)*a^2*cos(d*x+c)+6*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(...
\[ \int \frac {1}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {1}{\sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {1}{\sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {1}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {1}{\sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {1}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]