\(\int \sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)} \, dx\) [1507]

Optimal result

Integrand size = 45, antiderivative size = 543 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=-\frac {(a-b) \sqrt {a+b} (4 b B+a C) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 a b d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} (8 A b+a C+2 b (2 B+C)) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (8 A b^2+4 a b B-a^2 C+4 b^2 C\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b^2 d \sqrt {\sec (c+d x)}}+\frac {C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {(4 b B+a C) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b d} \] Output:

-1/4*(a-b)*(a+b)^(1/2)*(4*B*b+C*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/b/d/sec(d*x 
+c)^(1/2)+1/4*(a+b)^(1/2)*(8*A*b+a*C+2*b*(2*B+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 
)/b/d/sec(d*x+c)^(1/2)-1/4*(a+b)^(1/2)*(8*A*b^2+4*B*a*b-C*a^2+4*C*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)/b^2/d/sec(d*x+c)^(1/2)+1/2*C*(a+b*cos(d*x+ 
c))^(1/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+1/4*(4*B*b+C*a)*(a+b*cos(d*x+c))^( 
1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)/b/d
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1343\) vs. \(2(543)=1086\).

Time = 18.52 (sec) , antiderivative size = 1343, normalized size of antiderivative = 2.47 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(C*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[2*(c + d*x)])/(4*d) + ( 
-4*a*b*B*Tan[(c + d*x)/2] - 4*b^2*B*Tan[(c + d*x)/2] - a^2*C*Tan[(c + d*x) 
/2] - a*b*C*Tan[(c + d*x)/2] + 8*b^2*B*Tan[(c + d*x)/2]^3 + 2*a*b*C*Tan[(c 
 + d*x)/2]^3 + 4*a*b*B*Tan[(c + d*x)/2]^5 - 4*b^2*B*Tan[(c + d*x)/2]^5 + a 
^2*C*Tan[(c + d*x)/2]^5 - a*b*C*Tan[(c + d*x)/2]^5 - 16*A*b^2*EllipticPi[- 
1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2 
]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - 8* 
a*b*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - 
Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2 
]^2)/(a + b)] + 2*a^2*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/ 
(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - 
 b*Tan[(c + d*x)/2]^2)/(a + b)] - 8*b^2*C*EllipticPi[-1, ArcSin[Tan[(c + d 
*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Ta 
n[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - 16*A*b^2*EllipticPi[-1 
, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - 
Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2 
]^2)/(a + b)] - 8*a*b*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/ 
(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*T 
an[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a^2*C*EllipticPi[-1 
, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1...
 

Rubi [A] (verified)

Time = 2.51 (sec) , antiderivative size = 502, normalized size of antiderivative = 0.92, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.356, Rules used = {3042, 4709, 3042, 3528, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3540

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {\int -\frac {-\left (\left (-C a^2+4 b B a+8 A b^2+4 b^2 C\right ) \cos ^2(c+d x)\right )-2 a b (4 A+C) \cos (c+d x)+a (4 b B+a C)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-\left (\left (-C a^2+4 b B a+8 A b^2+4 b^2 C\right ) \cos ^2(c+d x)\right )-2 a b (4 A+C) \cos (c+d x)+a (4 b B+a C)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {\left (C a^2-4 b B a-8 A b^2-4 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a b (4 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )+a (4 b B+a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3532

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a (4 b B+a C)-2 a b (4 A+C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a (4 b B+a C)-2 a b (4 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )\)

\(\Big \downarrow \) 3288

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

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

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

\(\Big \downarrow \) 3473

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Co 
s[c + d*x]]*Sin[c + d*x])/(2*d) + (-1/2*((2*(a - b)*Sqrt[a + b]*(4*b*B + a 
*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sq 
rt[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*Sqrt[a + b]*(8*A*b + a 
*C + 2*b*(2*B + 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)])/d + (2*Sqrt[a + b 
]*(8*A*b^2 + 4*a*b*B - a^2*C + 4*b^2*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)])/(b*d))/b + ((4*b*B + a*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + 
d*x])/(b*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 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 3528

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)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

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 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. \(1397\) vs. \(2(480)=960\).

Time = 16.57 (sec) , antiderivative size = 1398, normalized size of antiderivative = 2.57

method	result	size
parts	\(\text {Expression too large to display}\)	\(1398\)
default	\(\text {Expression too large to display}\)	\(1490\)

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)^(1/2 
),x,method=_RETURNVERBOSE)
 

Output:

-2*A/d*sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2)*(EllipticF(-csc(d*x+c)+cot( 
d*x+c),(-(a-b)/(a+b))^(1/2))*a-EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a 
+b))^(1/2))*b+2*b*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(-(a-b)/(a+b))^(1/2 
)))*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x 
+c)))^(1/2)*(1+cos(d*x+c))+B/d*((-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*(cos(d*x+ 
c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a 
*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(-(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*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))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2)*(cos(d*x+c)/ 
(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*El 
lipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+cos(d*x+c)^2*sin(d*x+ 
c)*b+cos(d*x+c)*sin(d*x+c)*a)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)/(b*c 
os(d*x+c)^2+a*cos(d*x+c)+b*cos(d*x+c)+a)-1/4*C/d/b*((-2*cos(d*x+c)^2-4*cos 
(d*x+c)-2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+ 
cos(d*x+c)))^(1/2)*a^2*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(-(a-b)/(a+b)) 
^(1/2))+(8*cos(d*x+c)^2+16*cos(d*x+c)+8)*(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*EllipticPi(-csc(d*...
 

Fricas [F]

\[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )} \,d x } \] 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 
)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\int \sqrt {a + b \cos {\left (c + d x \right )}} \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}\, dx \] 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)**(1/2),x)
 

Output:

Integral(sqrt(a + b*cos(c + d*x))*(A + B*cos(c + d*x) + C*cos(c + d*x)**2) 
*sqrt(sec(c + d*x)), x)
 

Maxima [F]

\[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )} \,d x } \] 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 
)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )} \,d x } \] 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 
)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Reduce [F]

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

Output:

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