\(\int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [1402]

Optimal result

Integrand size = 35, antiderivative size = 352 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {a \left (A b^2+5 a^2 C-4 b^2 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2 b^2 (3 A-16 C)+15 a^4 C-2 b^4 (3 A+C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^4 \left (a^2-b^2\right ) d}+\frac {a \left (3 A b^4-a^2 b^2 (A-7 C)-5 a^4 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^4 (a+b)^2 d}-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 A b^2+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}} \] Output:

-a*(A*b^2+5*C*a^2-4*C*b^2)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2 
^(1/2))*sec(d*x+c)^(1/2)/b^3/(a^2-b^2)/d+1/3*(a^2*b^2*(3*A-16*C)+15*a^4*C- 
2*b^4*(3*A+C))*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec 
(d*x+c)^(1/2)/b^4/(a^2-b^2)/d+a*(3*A*b^4-a^2*b^2*(A-7*C)-5*a^4*C)*cos(d*x+ 
c)^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))*sec(d*x+c)^(1/2) 
/(a-b)/b^4/(a+b)^2/d-(A*b^2+C*a^2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c 
))/sec(d*x+c)^(3/2)+1/3*(3*A*b^2+5*C*a^2-2*C*b^2)*sin(d*x+c)/b^2/(a^2-b^2) 
/d/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 7.43 (sec) , antiderivative size = 692, normalized size of antiderivative = 1.97 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\frac {2 \left (-3 a A b^2+5 a^3 C-8 a b^2 C\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (12 A b^3+8 a^2 b C+4 b^3 C\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (3 a A b^2+15 a^3 C-12 a b^2 C\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{12 b^2 (-a+b) (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {a \left (A b^2+a^2 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right )}-\frac {-a^2 A b^2 \sin (c+d x)-a^4 C \sin (c+d x)}{b^3 \left (-a^2+b^2\right ) (a+b \cos (c+d x))}+\frac {C \sin (2 (c+d x))}{3 b^2}\right )}{d} \] Input:

Integrate[(A + C*Cos[c + d*x]^2)/((a + b*Cos[c + d*x])^2*Sec[c + d*x]^(3/2 
)),x]
 

Output:

((2*(-3*a*A*b^2 + 5*a^3*C - 8*a*b^2*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sq 
rt[Sec[c + d*x]]], -1] - EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1 
])*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(a + b*C 
os[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(12*A*b^3 + 8*a^2*b*C + 4*b^3*C)*C 
os[c + d*x]^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(b + a*Se 
c[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(a + b*Cos[c + d*x]) 
*(1 - Cos[c + d*x]^2)) + ((3*a*A*b^2 + 15*a^3*C - 12*a*b^2*C)*Cos[2*(c + d 
*x)]*(b + a*Sec[c + d*x])*(-4*a*b + 4*a*b*Sec[c + d*x]^2 - 4*a*b*EllipticE 
[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^ 
2] + 2*(2*a - b)*b*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + 
d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*a^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[ 
c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*b^2*Ellipt 
icPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - S 
ec[c + d*x]^2])*Sin[c + d*x])/(a*b^2*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x 
]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))/(12*b^2*(-a + b)*(a + b)*d) 
 + (Sqrt[Sec[c + d*x]]*((a*(A*b^2 + a^2*C)*Sin[c + d*x])/(b^3*(a^2 - b^2)) 
 - (-(a^2*A*b^2*Sin[c + d*x]) - a^4*C*Sin[c + d*x])/(b^3*(-a^2 + b^2)*(a + 
 b*Cos[c + d*x])) + (C*Sin[2*(c + d*x)])/(3*b^2)))/d
 

Rubi [A] (verified)

Time = 2.19 (sec) , antiderivative size = 304, normalized size of antiderivative = 0.86, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4709, 3042, 3527, 27, 3042, 3528, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

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.

\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \cos (c+d x)^2}{\sec (c+d x)^{3/2} (a+b \cos (c+d x))^2}dx\)

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\left (\left (5 C a^2+3 A b^2-2 b^2 C\right ) \cos ^2(c+d x)\right )-2 a b (A+C) \cos (c+d x)+3 \left (C a^2+A b^2\right )\right )}{2 (a+b \cos (c+d x))}dx}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\left (\left (5 C a^2+3 A b^2-2 b^2 C\right ) \cos ^2(c+d x)\right )-2 a b (A+C) \cos (c+d x)+3 \left (C a^2+A b^2\right )\right )}{a+b \cos (c+d x)}dx}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\left (-5 C a^2-3 A b^2+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (C a^2+A b^2\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3528

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {2 \int -\frac {-3 a \left (5 C a^2+A b^2-4 b^2 C\right ) \cos ^2(c+d x)-2 b \left (3 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x)+a \left (5 C a^2+3 A b^2-2 b^2 C\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}-\frac {2 \left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\int \frac {-3 a \left (5 C a^2+A b^2-4 b^2 C\right ) \cos ^2(c+d x)-2 b \left (3 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x)+a \left (5 C a^2+3 A b^2-2 b^2 C\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}-\frac {2 \left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\int \frac {-3 a \left (5 C a^2+A b^2-4 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 b \left (3 A b^2+\left (2 a^2+b^2\right ) C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (5 C a^2+3 A b^2-2 b^2 C\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}-\frac {2 \left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3538

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {3 a \left (5 a^2 C+A b^2-4 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {a b \left (5 C a^2+b^2 (3 A-2 C)\right )+\left (15 C a^4+b^2 (3 A-16 C) a^2-2 b^4 (3 A+C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}-\frac {2 \left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\int \frac {a b \left (5 C a^2+b^2 (3 A-2 C)\right )+\left (15 C a^4+b^2 (3 A-16 C) a^2-2 b^4 (3 A+C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {3 a \left (5 a^2 C+A b^2-4 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{3 b}-\frac {2 \left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\int \frac {a b \left (5 C a^2+b^2 (3 A-2 C)\right )+\left (15 C a^4+b^2 (3 A-16 C) a^2-2 b^4 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {3 a \left (5 a^2 C+A b^2-4 b^2 C\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}-\frac {2 \left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\int \frac {a b \left (5 C a^2+b^2 (3 A-2 C)\right )+\left (15 C a^4+b^2 (3 A-16 C) a^2-2 b^4 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {6 a \left (5 a^2 C+A b^2-4 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3481

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {\left (15 a^4 C+a^2 b^2 (3 A-16 C)-2 b^4 (3 A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}+\frac {3 a \left (-5 a^4 C-a^2 b^2 (A-7 C)+3 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{b}-\frac {6 a \left (5 a^2 C+A b^2-4 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {\left (15 a^4 C+a^2 b^2 (3 A-16 C)-2 b^4 (3 A+C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {3 a \left (-5 a^4 C-a^2 b^2 (A-7 C)+3 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {6 a \left (5 a^2 C+A b^2-4 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {3 a \left (-5 a^4 C-a^2 b^2 (A-7 C)+3 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 \left (15 a^4 C+a^2 b^2 (3 A-16 C)-2 b^4 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}-\frac {6 a \left (5 a^2 C+A b^2-4 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3284

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {-\frac {2 \left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}-\frac {\frac {\frac {2 \left (15 a^4 C+a^2 b^2 (3 A-16 C)-2 b^4 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}+\frac {6 a \left (-5 a^4 C-a^2 b^2 (A-7 C)+3 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}}{b}-\frac {6 a \left (5 a^2 C+A b^2-4 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}}{2 b \left (a^2-b^2\right )}\right )\)

Input:

Int[(A + C*Cos[c + d*x]^2)/((a + b*Cos[c + d*x])^2*Sec[c + d*x]^(3/2)),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-(((A*b^2 + a^2*C)*Cos[c + d*x]^(3/ 
2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))) - (-1/3*((-6*a*(A 
*b^2 + 5*a^2*C - 4*b^2*C)*EllipticE[(c + d*x)/2, 2])/(b*d) + ((2*(a^2*b^2* 
(3*A - 16*C) + 15*a^4*C - 2*b^4*(3*A + C))*EllipticF[(c + d*x)/2, 2])/(b*d 
) + (6*a*(3*A*b^4 - a^2*b^2*(A - 7*C) - 5*a^4*C)*EllipticPi[(2*b)/(a + b), 
 (c + d*x)/2, 2])/(b*(a + b)*d))/b)/b - (2*(3*A*b^2 + 5*a^2*C - 2*b^2*C)*S 
qrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b*d))/(2*b*(a^2 - b^2)))
 

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 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], 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] && GtQ[c + d, 0]
 

rule 3481

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

rule 3527

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b 
*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( 
A*d^2*(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, 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 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 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{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. \(1101\) vs. \(2(337)=674\).

Time = 6.38 (sec) , antiderivative size = 1102, normalized size of antiderivative = 3.13

method	result	size
default	\(\text {Expression too large to display}\)	\(1102\)

Input:

int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(3/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*(A*b^2+3*C*a 
^2+2*C*a*b+C*b^2)/b^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^ 
2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF( 
cos(1/2*d*x+1/2*c),2^(1/2))+2*a^2*(A*b^2+C*a^2)/b^4*(-1/a*b^2/(a^2-b^2)*co 
s(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*c 
os(1/2*d*x+1/2*c)^2*b+a-b)-1/2/(a+b)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co 
s(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2) 
^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/a*b/(a^2-b^2)*(sin(1/2*d* 
x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c) 
^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/a 
*b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2 
)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d 
*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^( 
1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d* 
x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a 
^2-b^2)/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/ 
2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip 
ticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))+4/3*C/b^2*(2*cos(1/2*d*x+1/2 
*c)*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+2*(sin(1/ 
2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(3/2),x, algori 
thm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**2/sec(d*x+c)**(3/2),x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(3/2),x, algori 
thm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)/((b*cos(d*x + c) + a)^2*sec(d*x + c)^(3/2 
)), x)
 

Giac [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(3/2),x, algori 
thm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)/((b*cos(d*x + c) + a)^2*sec(d*x + c)^(3/2 
)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:

int((A + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^2) 
,x)
 

Output:

int((A + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^2) 
, x)
 

Reduce [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2} a b +\sec \left (d x +c \right )^{2} a^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2} a b +\sec \left (d x +c \right )^{2} a^{2}}d x \right ) c \] Input:

int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(3/2),x)
 

Output:

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