3.7.18 \(\int \frac {A+B \cos (c+d x)}{\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx\) [618]

3.7.18.1 Optimal result
3.7.18.2 Mathematica [A] (verified)
3.7.18.3 Rubi [A] (verified)
3.7.18.4 Maple [B] (verified)
3.7.18.5 Fricas [F]
3.7.18.6 Sympy [F]
3.7.18.7 Maxima [F]
3.7.18.8 Giac [F]
3.7.18.9 Mupad [F(-1)]

3.7.18.1 Optimal result

Integrand size = 35, antiderivative size = 487 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=-\frac {(a-b) \sqrt {a+b} B \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a b d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} B \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} (2 A b-a B) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^2 d \sqrt {\sec (c+d x)}}+\frac {B \sin (c+d x)}{d \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {a B \sqrt {\sec (c+d x)} \sin (c+d x)}{b d \sqrt {a+b \cos (c+d x)}} \]

output
B*sin(d*x+c)/d/(a+b*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+a*B*sin(d*x+c)*sec( 
d*x+c)^(1/2)/b/d/(a+b*cos(d*x+c))^(1/2)-(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/b/d/sec(d*x+c)^(1/2)+B*csc(d*x+c)*EllipticF((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/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+sec(d*x+c))/(a-b)) 
^(1/2)/b^2/d/sec(d*x+c)^(1/2)
 
3.7.18.2 Mathematica [A] (verified)

Time = 9.36 (sec) , antiderivative size = 610, normalized size of antiderivative = 1.25 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (8 (a+b) B \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}}-16 A b \cos ^3\left (\frac {1}{2} (c+d x)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+24 A b \cos \left (\frac {1}{2} (c+d x)\right ) \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-12 a B \cos \left (\frac {1}{2} (c+d x)\right ) \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+8 A b \cos \left (\frac {3}{2} (c+d x)\right ) \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-4 a B \cos \left (\frac {3}{2} (c+d x)\right ) \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-2 a B \sin \left (\frac {1}{2} (c+d x)\right )+2 b B \sin \left (\frac {1}{2} (c+d x)\right )+2 a B \sin \left (\frac {3}{2} (c+d x)\right )-b B \sin \left (\frac {3}{2} (c+d x)\right )+b B \sin \left (\frac {5}{2} (c+d x)\right )\right )}{4 b d \sqrt {a+b \cos (c+d x)}} \]

input
Integrate[(A + B*Cos[c + d*x])/(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x] 
]),x]
 
output
(Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(8*(a + b)*B*Cos[(c + d*x)/2]^3*Sqrt[ 
(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c 
 + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)] - 16*A*b*Cos[ 
(c + d*x)/2]^3*EllipticF[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]))] + 24*A*b*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*B*Cos[(c + d*x)/2]*EllipticPi[-1, Arc 
Sin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqr 
t[(b + a*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + 8*A*b*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 
]))] - 4*a*B*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]))] - 2*a*B*Sin[(c + d*x)/2] + 2*b*B*Sin[(c + d 
*x)/2] + 2*a*B*Sin[(3*(c + d*x))/2] - b*B*Sin[(3*(c + d*x))/2] + b*B*Sin[( 
5*(c + d*x))/2]))/(4*b*d*Sqrt[a + b*Cos[c + d*x]])
 
3.7.18.3 Rubi [A] (verified)

Time = 1.96 (sec) , antiderivative size = 457, normalized size of antiderivative = 0.94, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 3440, 3042, 3482, 3042, 3530, 3042, 3288, 3472, 25, 27, 3042, 3280, 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.

\(\displaystyle \int \frac {A+B \cos (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3440

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3482

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \int \frac {(2 A b-a B) \cos ^2(c+d x)+2 a A \cos (c+d x)+a B}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}dx+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \int \frac {(2 A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 a A \sin \left (c+d x+\frac {\pi }{2}\right )+a B}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3530

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {\int \frac {B \cos (c+d x) a^2+b B a}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}dx}{b}+\frac {(2 A b-a B) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{b}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {\int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+b B a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{b}+\frac {(2 A b-a B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {\int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+b B a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3472

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {\frac {\int -\frac {a \left (a^2-b^2\right ) B}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}+\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {a \left (a^2-b^2\right ) B}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3280

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \left (\int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx\right )}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \left (\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \left (\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}\right )}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3473

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-a B \left (\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}\right )}{b}-\frac {2 \sqrt {a+b} (2 A b-a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a+b \cos (c+d x)}}\right )\)

input
Int[(A + B*Cos[c + d*x])/(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]),x]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((B*Sqrt[Cos[c + d*x]]*Sin[c + d*x]) 
/(d*Sqrt[a + b*Cos[c + d*x]]) + ((-2*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)])/(b^2*d) + (-(a*B*((2*(a - b)*Sqrt[a 
+ b]*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*S 
qrt[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^2*d) - (2*Sqrt[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)])/(a*d))) + (2*a*B*Sin[c + d*x])/(d*Sqrt[Cos[c 
 + d*x]]*Sqrt[a + b*Cos[c + d*x]]))/b)/2)
 

3.7.18.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3280
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin 
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[1/(a - b)   Int[1/(Sqrt[a + b*Sin 
[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x], x] - Simp[b/(a - b)   Int[(1 + Si 
n[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}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
 NeQ[c^2 - d^2, 0]
 

rule 3288
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]
 

rule 3295
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]
 

rule 3440
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Sin[e + f*x])^m*((c + 
d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

rule 3472
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(d_.)*sin[(e_.) + (f_.)*( 
x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)), x_Symbol] :> Simp[2*(A 
*b - a*B)*(Cos[e + f*x]/(f*(a^2 - b^2)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[d*Sin[ 
e + f*x]])), x] + Simp[d/(a^2 - b^2)   Int[(A*b - a*B + (a*A - b*B)*Sin[e + 
 f*x])/(Sqrt[a + b*Sin[e + f*x]]*(d*Sin[e + f*x])^(3/2)), x], x] /; FreeQ[{ 
a, b, d, e, f, A, B}, x] && NeQ[a^2 - b^2, 0]
 

rule 3473
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]
 

rule 3482
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]
 

rule 3530
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_ 
)])^(3/2)), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[d*Sin[e + f*x]]/Sqrt[a + b 
*Sin[e + f*x]], x], x] + Simp[1/b   Int[(A*b + (b*B - a*C)*Sin[e + f*x])/(( 
a + b*Sin[e + f*x])^(3/2)*Sqrt[d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, d, 
e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]
 
3.7.18.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(957\) vs. \(2(443)=886\).

Time = 16.82 (sec) , antiderivative size = 958, normalized size of antiderivative = 1.97

method result size
parts \(\text {Expression too large to display}\) \(958\)
default \(\text {Expression too large to display}\) \(1344\)

input
int((A+B*cos(d*x+c))/sec(d*x+c)^(1/2)/(a+cos(d*x+c)*b)^(1/2),x,method=_RET 
URNVERBOSE)
 
output
2*A/d/(a+cos(d*x+c)*b)^(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)))*((a 
+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)/(cos(d*x+c)/(1 
+cos(d*x+c)))^(1/2)-B/d/(1+cos(d*x+c))/(a+cos(d*x+c)*b)^(1/2)/sec(d*x+c)^( 
1/2)*(EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1 
+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a*cos(d* 
x+c)+EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+ 
cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*b*cos(d*x 
+c)-2*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*(cos(d*x+c 
)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a*co 
s(d*x+c)+2*(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)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a+2 
*(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)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*b-4*(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)/(1+cos(d*x+c))/(a+b))^(1/2)*a+sec(d*x+c)*(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)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a+sec(d*x+c)*(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)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*b-2*sec(d*x+c)*(co...
 
3.7.18.5 Fricas [F]

\[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{\sqrt {b \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}} \,d x } \]

input
integrate((A+B*cos(d*x+c))/sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2),x, algo 
rithm="fricas")
 
output
integral((B*cos(d*x + c) + A)/(sqrt(b*cos(d*x + c) + a)*sqrt(sec(d*x + c)) 
), x)
 
3.7.18.6 Sympy [F]

\[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\sqrt {a + b \cos {\left (c + d x \right )}} \sqrt {\sec {\left (c + d x \right )}}}\, dx \]

input
integrate((A+B*cos(d*x+c))/sec(d*x+c)**(1/2)/(a+b*cos(d*x+c))**(1/2),x)
 
output
Integral((A + B*cos(c + d*x))/(sqrt(a + b*cos(c + d*x))*sqrt(sec(c + d*x)) 
), x)
 
3.7.18.7 Maxima [F]

\[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{\sqrt {b \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}} \,d x } \]

input
integrate((A+B*cos(d*x+c))/sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2),x, algo 
rithm="maxima")
 
output
integrate((B*cos(d*x + c) + A)/(sqrt(b*cos(d*x + c) + a)*sqrt(sec(d*x + c) 
)), x)
 
3.7.18.8 Giac [F]

\[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{\sqrt {b \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}} \,d x } \]

input
integrate((A+B*cos(d*x+c))/sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2),x, algo 
rithm="giac")
 
output
integrate((B*cos(d*x + c) + A)/(sqrt(b*cos(d*x + c) + a)*sqrt(sec(d*x + c) 
)), x)
 
3.7.18.9 Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]

input
int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^(1/2 
)),x)
 
output
int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^(1/2 
)), x)