\(\int (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \, dx\) [748]

Optimal result

Integrand size = 25, antiderivative size = 503 \[ \int (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \, dx=-\frac {9 (a-b) b \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (8 a^2+9 a b+2 b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (15 a^2+4 b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 d \sqrt {\sec (c+d x)}}+\frac {b^2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {9 a b \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d} \] Output:

-9/4*(a-b)*b*(a+b)^(1/2)*cos(d*x+c)^(1/2)*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*(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)+1/4* 
(a+b)^(1/2)*(8*a^2+9*a*b+2*b^2)*cos(d*x+c)^(1/2)*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*( 
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)-1/4*(a+b)^(1/2)*(15*a^2+4*b^2)*cos(d*x+c)^(1/2)*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*(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)+1/2*b^2*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2) 
+9/4*a*b*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)/d
 

Mathematica [A] (warning: unable to verify)

Time = 8.35 (sec) , antiderivative size = 421, normalized size of antiderivative = 0.84 \[ \int (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \, dx=\frac {b^2 (a+b \cos (c+d x)) \sqrt {\sec (c+d x)} \sin (2 (c+d x))+\frac {-18 a b (a+b) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )-4 \left (4 a^3-12 a^2 b+a b^2-2 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-4 b \left (15 a^2+4 b^2\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-9 a b \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}}{4 d \sqrt {a+b \cos (c+d x)}} \] Input:

Integrate[(a + b*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]],x]
 

Output:

(b^2*(a + b*Cos[c + d*x])*Sqrt[Sec[c + d*x]]*Sin[2*(c + d*x)] + (-18*a*b*( 
a + b)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*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)] - 4*(4*a^3 - 12*a^2*b + a*b^2 - 2*b^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]))]*Elliptic 
F[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] - 4*b*(15*a^2 + 4*b^2)*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]))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] 
- 9*a*b*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x) 
/2])/(Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(-1 + 
 Tan[(c + d*x)/2]^2)))/(4*d*Sqrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 2.24 (sec) , antiderivative size = 473, normalized size of antiderivative = 0.94, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3042, 4710, 3042, 3272, 27, 3042, 3540, 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.

\(\displaystyle \int \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx\)

\(\Big \downarrow \) 4710

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(a+b \cos (c+d x))^{5/2}}{\sqrt {\cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3272

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \int \frac {9 a b^2 \cos ^2(c+d x)+2 b \left (6 a^2+b^2\right ) \cos (c+d x)+a \left (4 a^2+b^2\right )}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \int \frac {9 a b^2 \cos ^2(c+d x)+2 b \left (6 a^2+b^2\right ) \cos (c+d x)+a \left (4 a^2+b^2\right )}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \int \frac {9 a b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (6 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (4 a^2+b^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3540

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {\int -\frac {9 a^2 b^2-\left (15 a^2+4 b^2\right ) \cos ^2(c+d x) b^2-2 a \left (4 a^2+b^2\right ) \cos (c+d x) b}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {9 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {9 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {9 a^2 b^2-\left (15 a^2+4 b^2\right ) \cos ^2(c+d x) b^2-2 a \left (4 a^2+b^2\right ) \cos (c+d x) b}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}\right )+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {9 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {9 a^2 b^2-\left (15 a^2+4 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^2-2 a \left (4 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3532

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {9 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {9 a^2 b^2-2 a b \left (4 a^2+b^2\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-b^2 \left (15 a^2+4 b^2\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{2 b}\right )+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {9 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {9 a^2 b^2-2 a b \left (4 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-b^2 \left (15 a^2+4 b^2\right ) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {9 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {9 a^2 b^2-2 a b \left (4 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sqrt {a+b} \left (15 a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}\right )+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3477

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {9 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {9 a^2 b^2 \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a b \left (8 a^2+9 a b+2 b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {2 b \sqrt {a+b} \left (15 a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}\right )+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {9 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {9 a^2 b^2 \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a b \left (8 a^2+9 a b+2 b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sqrt {a+b} \left (15 a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}\right )+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {9 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {9 a^2 b^2 \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b \sqrt {a+b} \left (8 a^2+9 a b+2 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {2 b \sqrt {a+b} \left (15 a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}\right )+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3473

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {9 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {-\frac {2 b \sqrt {a+b} \left (8 a^2+9 a b+2 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {2 b \sqrt {a+b} \left (15 a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {18 b^2 (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{d}}{2 b}\right )+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

Input:

Int[(a + b*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]],x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((b^2*Sqrt[Cos[c + d*x]]*Sqrt[a + b* 
Cos[c + d*x]]*Sin[c + d*x])/(2*d) + (-1/2*((18*(a - b)*b^2*Sqrt[a + 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)])/d - (2*b*Sqrt[a + b]*(8*a^2 + 9*a*b + 2* 
b^2)*Cot[c + d*x]*EllipticF[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)])/d + (2*b*Sqrt[a + b]*(15*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)])/d)/b + (9*a*b*Sqrt[ 
a + b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))/4)
 

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 3272

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])))
 

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 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 3477

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]
 

rule 3532

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]
 

rule 3540

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]
 

rule 4710

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(944\) vs. \(2(440)=880\).

Time = 18.92 (sec) , antiderivative size = 945, normalized size of antiderivative = 1.88

method	result	size
default	\(\text {Expression too large to display}\)	\(945\)

Input:

int((a+cos(d*x+c)*b)^(5/2)*sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/4/d*((-30*cos(d*x+c)^2-60*cos(d*x+c)-30)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1 
)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*EllipticPi(cot(d*x+ 
c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))+(-8*cos(d*x+c)^2-16*cos(d*x+c)-8)*( 
(a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^( 
1/2)*b^3*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))+(-9*cos 
(d*x+c)^2-18*cos(d*x+c)-9)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*( 
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*EllipticE(cot(d*x+c)-csc(d*x+c),(-( 
a-b)/(a+b))^(1/2))+(-9*cos(d*x+c)^2-18*cos(d*x+c)-9)*((a+cos(d*x+c)*b)/(co 
s(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^2*EllipticE 
(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(-8*cos(d*x+c)^2-16*cos(d*x+c 
)-8)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c) 
+1))^(1/2)*a^3*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(24*c 
os(d*x+c)^2+48*cos(d*x+c)+24)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2 
)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*EllipticF(cot(d*x+c)-csc(d*x+c), 
(-(a-b)/(a+b))^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*((a+cos(d*x+c)*b)/( 
cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^2*Ellipti 
cF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(4*cos(d*x+c)^2+8*cos(d*x+c 
)+4)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c) 
+1))^(1/2)*b^3*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+9*a^2 
*b*cos(d*x+c)*sin(d*x+c)+sin(d*x+c)*cos(d*x+c)*(11*cos(d*x+c)+2)*b^2*a+...
 

Fricas [F]

\[ \int (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\sec \left (d x + c\right )} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(b*cos(d*x + 
c) + a)*sqrt(sec(d*x + c)), x)
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**(5/2)*sec(d*x+c)**(1/2),x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [F]

\[ \int (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\sec \left (d x + c\right )} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate((b*cos(d*x + c) + a)^(5/2)*sqrt(sec(d*x + c)), x)
 

Giac [F]

\[ \int (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\sec \left (d x + c\right )} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

integrate((b*cos(d*x + c) + a)^(5/2)*sqrt(sec(d*x + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \, dx=\int \sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:

int((1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^(5/2),x)
 

Output:

int((1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \, dx=2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )d x \right ) a b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2}d x \right ) b^{2}+\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}d x \right ) a^{2} \] Input:

int((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^(1/2),x)
 

Output:

2*int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a)*cos(c + d*x),x)*a*b + in 
t(sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**2,x)*b**2 + in 
t(sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a),x)*a**2