\(\int \sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\) [1505]

Optimal result

Integrand size = 45, antiderivative size = 467 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {2 (a-b) \sqrt {a+b} (A b+3 a B) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {2 \sqrt {a+b} (b (A-3 B)-a (A-3 B+3 C)) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a d \sqrt {\sec (c+d x)}}-\frac {2 \sqrt {a+b} C \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{d \sqrt {\sec (c+d x)}}+\frac {2 A \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \] Output:

2/3*(a-b)*(a+b)^(1/2)*(A*b+3*B*a)*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)/a^2/d/sec(d*x+ 
c)^(1/2)-2/3*(a+b)^(1/2)*(b*(A-3*B)-a*(A-3*B+3*C))*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)/a/d/sec(d*x+c)^(1/2)-2*(a+b)^(1/2)*C*cos(d*x+c)^(1/2)*csc(d*x+c)*Ellipt 
icPi((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)+2/3*A*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(3/2)*sin(d*x+ 
c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 16.87 (sec) , antiderivative size = 672, normalized size of antiderivative = 1.44 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (12 a b C \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 )-2 (a+b) (A b+3 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 ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+2 a (b (A+3 B-3 C)+a (A+3 (B+C))) \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 ) \sec ^2\left (\frac {1}{2} (c+d x)\right )-a A b \tan \left (\frac {1}{2} (c+d x)\right )-A b^2 \tan \left (\frac {1}{2} (c+d x)\right )-3 a^2 B \tan \left (\frac {1}{2} (c+d x)\right )-3 a b B \tan \left (\frac {1}{2} (c+d x)\right )+12 a b C \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 ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 A b^2 \tan ^3\left (\frac {1}{2} (c+d x)\right )+6 a b B \tan ^3\left (\frac {1}{2} (c+d x)\right )+a A b \tan ^5\left (\frac {1}{2} (c+d x)\right )-A b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )+3 a^2 B \tan ^5\left (\frac {1}{2} (c+d x)\right )-3 a b B \tan ^5\left (\frac {1}{2} (c+d x)\right )\right )}{3 a d \sqrt {a+b \cos (c+d x)} \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2}}+\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 (A b+3 a B) \sin (c+d x)}{3 a}+\frac {2}{3} A \tan (c+d x)\right )}{d} \] Input:

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

Output:

(2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(12*a*b*C*Sqrt[Cos[c + d*x]/(1 + 
Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Ell 
ipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] - 2*(a + b)*(A*b + 
 3*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)]*Sec[(c + d*x)/2]^2 + 2*a*(b*(A + 3*B - 3*C) + a*(A + 3*(B + C)))*S 
qrt[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)]*S 
ec[(c + d*x)/2]^2 - a*A*b*Tan[(c + d*x)/2] - A*b^2*Tan[(c + d*x)/2] - 3*a^ 
2*B*Tan[(c + d*x)/2] - 3*a*b*B*Tan[(c + d*x)/2] + 12*a*b*C*Sqrt[Cos[c + d* 
x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x 
]))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d 
*x)/2]^2 + 2*A*b^2*Tan[(c + d*x)/2]^3 + 6*a*b*B*Tan[(c + d*x)/2]^3 + a*A*b 
*Tan[(c + d*x)/2]^5 - A*b^2*Tan[(c + d*x)/2]^5 + 3*a^2*B*Tan[(c + d*x)/2]^ 
5 - 3*a*b*B*Tan[(c + d*x)/2]^5))/(3*a*d*Sqrt[a + b*Cos[c + d*x]]*(Sec[(c + 
 d*x)/2]^2)^(3/2)) + (Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(A*b 
 + 3*a*B)*Sin[c + d*x])/(3*a) + (2*A*Tan[c + d*x])/3))/d
 

Rubi [A] (verified)

Time = 1.88 (sec) , antiderivative size = 429, normalized size of antiderivative = 0.92, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 4709, 3042, 3526, 27, 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.

\(\displaystyle \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (c+d x)^{5/2} \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4709

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {a+b \cos (c+d x)} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3532

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (\int \frac {A b+3 a B+(3 b B+a (A+3 C)) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 C \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3477

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (-(b (A-3 B)-a (A-3 B+3 C)) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+(3 a B+A b) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\frac {6 C \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (-(b (A-3 B)-a (A-3 B+3 C)) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+(3 a B+A b) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 C \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left ((3 a B+A b) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) (b (A-3 B)-a (A-3 B+3 C)) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}-\frac {6 C \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3473

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} (3 a B+A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}-\frac {2 \sqrt {a+b} \cot (c+d x) (b (A-3 B)-a (A-3 B+3 C)) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}-\frac {6 C \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((2*(a - b)*Sqrt[a + b]*(A*b + 3*a* 
B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqr 
t[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]*(b*(A - 3 
*B) - a*(A - 3*B + 3*C))*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) - (6*S 
qrt[a + b]*C*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)/3 + (2*A* 
Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)))
 

Defintions of rubi rules used

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 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 3526

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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - 
d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^(m - 
 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* 
d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 
1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x 
] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*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] && GtQ[m, 0] && LtQ[n, -1]
 

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 4709

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1232\) vs. \(2(412)=824\).

Time = 26.74 (sec) , antiderivative size = 1233, normalized size of antiderivative = 2.64

method	result	size
parts	\(\text {Expression too large to display}\)	\(1233\)
default	\(\text {Expression too large to display}\)	\(1373\)

Input:

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

Output:

-2/3*A/d/a*((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)*a*b*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1 
/2))*(-cos(d*x+c)^3-2*cos(d*x+c)^2-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)*b^2*EllipticE(-csc( 
d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-cos(d*x+c)^3-2*cos(d*x+c)^2-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)*a^2*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2)) 
*(cos(d*x+c)^3+2*cos(d*x+c)^2+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)*a*b*EllipticF(-csc(d*x+c 
)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)^3+2*cos(d*x+c)^2+cos(d*x+c) 
)+(-cos(d*x+c)-1)*sin(d*x+c)*a^2+sin(d*x+c)*cos(d*x+c)*(-cos(d*x+c)-2)*a*b 
-b^2*cos(d*x+c)^2*sin(d*x+c))*(a+b*cos(d*x+c))^(1/2)*cos(d*x+c)*sec(d*x+c) 
^(5/2)/(b*cos(d*x+c)^2+a*cos(d*x+c)+b*cos(d*x+c)+a)-2*B/d*((-cos(d*x+c)^2- 
2*cos(d*x+c)-1)*(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)*a*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^ 
(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(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)*b*EllipticE(-csc(d*x+c)+cot( 
d*x+c),(-(a-b)/(a+b))^(1/2))+(cos(d*x+c)^2+2*cos(d*x+c)+1)*(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)*a*Ellip 
ticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(cos(d*x+c)^2+2*cos(d...
 

Fricas [F]

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

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

Output:

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)* 
sec(d*x + c)^(5/2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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