\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx\) [1494]

Optimal result

Integrand size = 43, antiderivative size = 505 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {\left (25 a^3 b B-20 a b^3 B-3 a^2 b^2 (5 A-8 C)-35 a^4 C+2 b^4 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^4 \left (a^2-b^2\right ) d}+\frac {\left (15 a^4 b B-16 a^2 b^3 B-2 b^5 B-a^3 b^2 (9 A-20 C)-21 a^5 C+4 a b^4 (3 A+C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^5 \left (a^2-b^2\right ) d}-\frac {a^2 \left (5 A b^4+5 a^3 b B-7 a b^3 B-3 a^2 b^2 (A-3 C)-7 a^4 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^5 (a+b)^2 d}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}} \] Output:

-1/5*(25*B*a^3*b-20*B*a*b^3-3*a^2*b^2*(5*A-8*C)-35*a^4*C+2*b^4*(5*A+3*C))* 
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^ 
4/(a^2-b^2)/d+1/3*(15*B*a^4*b-16*B*a^2*b^3-2*B*b^5-a^3*b^2*(9*A-20*C)-21*C 
*a^5+4*a*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^5/(a^2-b^2)/d-a^2*(5*A*b^4+5*B*a^3*b-7*B*a*b^3-3*a^ 
2*b^2*(A-3*C)-7*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^5/(a+b)^2/d-(A*b^2-a*(B*b-C*a))*si 
n(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))/sec(d*x+c)^(5/2)+1/5*(5*A*b^2-5*B* 
a*b+7*C*a^2-2*C*b^2)*sin(d*x+c)/b^2/(a^2-b^2)/d/sec(d*x+c)^(3/2)+1/3*(5*B* 
a^2*b-2*B*b^3-a*b^2*(3*A-4*C)-7*a^3*C)*sin(d*x+c)/b^3/(a^2-b^2)/d/sec(d*x+ 
c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 7.90 (sec) , antiderivative size = 831, normalized size of antiderivative = 1.65 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\frac {2 \left (15 a^2 A b^2-30 A b^4-25 a^3 b B+40 a b^3 B+35 a^4 C-32 a^2 b^2 C-18 b^4 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 (60 a A b^3-40 a^2 b^2 B-20 b^4 B+56 a^3 b C+4 a 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 (45 a^2 A b^2-30 A b^4-75 a^3 b B+60 a b^3 B+105 a^4 C-72 a^2 b^2 C-18 b^4 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 )}}{60 (a-b) b^3 (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (-\frac {\left (10 a^2 A b^2-10 a^3 b B+10 a^4 C-a^2 b^2 C+b^4 C\right ) \sin (c+d x)}{10 b^4 \left (a^2-b^2\right )}-\frac {a^3 A b^2 \sin (c+d x)-a^4 b B \sin (c+d x)+a^5 C \sin (c+d x)}{b^4 \left (-a^2+b^2\right ) (a+b \cos (c+d x))}+\frac {(b B-2 a C) \sin (2 (c+d x))}{3 b^3}+\frac {C \sin (3 (c+d x))}{10 b^2}\right )}{d} \] Input:

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

Output:

((2*(15*a^2*A*b^2 - 30*A*b^4 - 25*a^3*b*B + 40*a*b^3*B + 35*a^4*C - 32*a^2 
*b^2*C - 18*b^4*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[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*Cos[c + d*x])*(1 - Co 
s[c + d*x]^2)) + (2*(60*a*A*b^3 - 40*a^2*b^2*B - 20*b^4*B + 56*a^3*b*C + 4 
*a*b^3*C)*Cos[c + d*x]^2*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])/(b*(a + b*Co 
s[c + d*x])*(1 - Cos[c + d*x]^2)) + ((45*a^2*A*b^2 - 30*A*b^4 - 75*a^3*b*B 
 + 60*a*b^3*B + 105*a^4*C - 72*a^2*b^2*C - 18*b^4*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[S 
qrt[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]]*Sqr 
t[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*EllipticPi[-(a/ 
b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[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)))/(60*(a - b)*b^3*(a + b)*d) + (Sqrt[S 
ec[c + d*x]]*(-1/10*((10*a^2*A*b^2 - 10*a^3*b*B + 10*a^4*C - a^2*b^2*C + b 
^4*C)*Sin[c + d*x])/(b^4*(a^2 - b^2)) - (a^3*A*b^2*Sin[c + d*x] - a^4*b*B* 
Sin[c + d*x] + a^5*C*Sin[c + d*x])/(b^4*(-a^2 + b^2)*(a + b*Cos[c + d*x...
 

Rubi [A] (verified)

Time = 3.31 (sec) , antiderivative size = 452, normalized size of antiderivative = 0.90, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.465, Rules used = {3042, 4709, 3042, 3526, 27, 3042, 3528, 27, 3042, 3528, 27, 3042, 3538, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (-\left (\left (7 C a^2-5 b B a+5 A b^2-2 b^2 C\right ) \cos ^2(c+d x)\right )+2 b (b B-a (A+C)) \cos (c+d x)+5 \left (A b^2-a (b B-a C)\right )\right )}{2 (a+b \cos (c+d x))}dx}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (-\left (\left (7 C a^2-5 b B a+5 A b^2-2 b^2 C\right ) \cos ^2(c+d x)\right )+2 b (b B-a (A+C)) \cos (c+d x)+5 \left (A b^2-a (b B-a C)\right )\right )}{a+b \cos (c+d x)}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\left (-7 C a^2+5 b B a-5 A b^2+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b (b B-a (A+C)) \sin \left (c+d x+\frac {\pi }{2}\right )+5 \left (A b^2-a (b B-a C)\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3528

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {2 \int -\frac {\sqrt {\cos (c+d x)} \left (5 \left (-7 C a^3+5 b B a^2-b^2 (3 A-4 C) a-2 b^3 B\right ) \cos ^2(c+d x)-2 b \left (2 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \cos (c+d x)+3 a \left (7 C a^2-5 b B a+5 A b^2-2 b^2 C\right )\right )}{2 (a+b \cos (c+d x))}dx}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (5 \left (-7 C a^3+5 b B a^2-b^2 (3 A-4 C) a-2 b^3 B\right ) \cos ^2(c+d x)-2 b \left (2 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \cos (c+d x)+3 a \left (7 C a^2-5 b B a+5 A b^2-2 b^2 C\right )\right )}{a+b \cos (c+d x)}dx}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (5 \left (-7 C a^3+5 b B a^2-b^2 (3 A-4 C) a-2 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 b \left (2 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a \left (7 C a^2-5 b B a+5 A b^2-2 b^2 C\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3528

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {2 \int \frac {-3 \left (-35 C a^4+25 b B a^3-3 b^2 (5 A-8 C) a^2-20 b^3 B a+2 b^4 (5 A+3 C)\right ) \cos ^2(c+d x)-2 b \left (-14 C a^3+10 b B a^2-b^2 (15 A+C) a+5 b^3 B\right ) \cos (c+d x)+5 a \left (-7 C a^3+5 b B a^2-b^2 (3 A-4 C) a-2 b^3 B\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}+\frac {10 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b d}}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\int \frac {-3 \left (-35 C a^4+25 b B a^3-3 b^2 (5 A-8 C) a^2-20 b^3 B a+2 b^4 (5 A+3 C)\right ) \cos ^2(c+d x)-2 b \left (-14 C a^3+10 b B a^2-b^2 (15 A+C) a+5 b^3 B\right ) \cos (c+d x)+5 a \left (-7 C a^3+5 b B a^2-b^2 (3 A-4 C) a-2 b^3 B\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}+\frac {10 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b d}}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\int \frac {-3 \left (-35 C a^4+25 b B a^3-3 b^2 (5 A-8 C) a^2-20 b^3 B a+2 b^4 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 b \left (-14 C a^3+10 b B a^2-b^2 (15 A+C) a+5 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a \left (-7 C a^3+5 b B a^2-b^2 (3 A-4 C) a-2 b^3 B\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}+\frac {10 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b d}}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3538

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {-\frac {3 \left (-35 a^4 C+25 a^3 b B-3 a^2 b^2 (5 A-8 C)-20 a b^3 B+2 b^4 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {5 \left (a b \left (-7 C a^3+5 b B a^2-b^2 (3 A-4 C) a-2 b^3 B\right )+\left (-21 C a^5+15 b B a^4-b^2 (9 A-20 C) a^3-16 b^3 B a^2+4 b^4 (3 A+C) a-2 b^5 B\right ) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}+\frac {10 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b d}}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {5 \int \frac {a b \left (-7 C a^3+5 b B a^2-b^2 (3 A-4 C) a-2 b^3 B\right )+\left (-21 C a^5+15 b B a^4-b^2 (9 A-20 C) a^3-16 b^3 B a^2+4 b^4 (3 A+C) a-2 b^5 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {3 \left (-35 a^4 C+25 a^3 b B-3 a^2 b^2 (5 A-8 C)-20 a b^3 B+2 b^4 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{3 b}+\frac {10 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b d}}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {5 \int \frac {a b \left (-7 C a^3+5 b B a^2-b^2 (3 A-4 C) a-2 b^3 B\right )+\left (-21 C a^5+15 b B a^4-b^2 (9 A-20 C) a^3-16 b^3 B a^2+4 b^4 (3 A+C) a-2 b^5 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {3 \left (-35 a^4 C+25 a^3 b B-3 a^2 b^2 (5 A-8 C)-20 a b^3 B+2 b^4 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}+\frac {10 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b d}}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {5 \int \frac {a b \left (-7 C a^3+5 b B a^2-b^2 (3 A-4 C) a-2 b^3 B\right )+\left (-21 C a^5+15 b B a^4-b^2 (9 A-20 C) a^3-16 b^3 B a^2+4 b^4 (3 A+C) a-2 b^5 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-35 a^4 C+25 a^3 b B-3 a^2 b^2 (5 A-8 C)-20 a b^3 B+2 b^4 (5 A+3 C)\right )}{b d}}{3 b}+\frac {10 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b d}}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3481

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {5 \left (\frac {\left (-21 a^5 C+15 a^4 b B-a^3 b^2 (9 A-20 C)-16 a^2 b^3 B+4 a b^4 (3 A+C)-2 b^5 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {3 a^2 \left (-7 a^4 C+5 a^3 b B-3 a^2 b^2 (A-3 C)-7 a b^3 B+5 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}\right )}{b}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-35 a^4 C+25 a^3 b B-3 a^2 b^2 (5 A-8 C)-20 a b^3 B+2 b^4 (5 A+3 C)\right )}{b d}}{3 b}+\frac {10 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b d}}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {5 \left (\frac {\left (-21 a^5 C+15 a^4 b B-a^3 b^2 (9 A-20 C)-16 a^2 b^3 B+4 a b^4 (3 A+C)-2 b^5 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {3 a^2 \left (-7 a^4 C+5 a^3 b B-3 a^2 b^2 (A-3 C)-7 a b^3 B+5 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-35 a^4 C+25 a^3 b B-3 a^2 b^2 (5 A-8 C)-20 a b^3 B+2 b^4 (5 A+3 C)\right )}{b d}}{3 b}+\frac {10 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b d}}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\frac {5 \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-21 a^5 C+15 a^4 b B-a^3 b^2 (9 A-20 C)-16 a^2 b^3 B+4 a b^4 (3 A+C)-2 b^5 B\right )}{b d}-\frac {3 a^2 \left (-7 a^4 C+5 a^3 b B-3 a^2 b^2 (A-3 C)-7 a b^3 B+5 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-35 a^4 C+25 a^3 b B-3 a^2 b^2 (5 A-8 C)-20 a b^3 B+2 b^4 (5 A+3 C)\right )}{b d}}{3 b}+\frac {10 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b d}}{5 b}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\)

\(\Big \downarrow \) 3284

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b d}-\frac {\frac {10 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b d}+\frac {\frac {5 \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-21 a^5 C+15 a^4 b B-a^3 b^2 (9 A-20 C)-16 a^2 b^3 B+4 a b^4 (3 A+C)-2 b^5 B\right )}{b d}-\frac {6 a^2 \left (-7 a^4 C+5 a^3 b B-3 a^2 b^2 (A-3 C)-7 a b^3 B+5 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}\right )}{b}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-35 a^4 C+25 a^3 b B-3 a^2 b^2 (5 A-8 C)-20 a b^3 B+2 b^4 (5 A+3 C)\right )}{b d}}{3 b}}{5 b}}{2 b \left (a^2-b^2\right )}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-(((A*b^2 - a*(b*B - a*C))*Cos[c + 
d*x]^(5/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))) - ((-2*(5 
*A*b^2 - 5*a*b*B + 7*a^2*C - 2*b^2*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5* 
b*d) - (((-6*(25*a^3*b*B - 20*a*b^3*B - 3*a^2*b^2*(5*A - 8*C) - 35*a^4*C + 
 2*b^4*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(b*d) + (5*((2*(15*a^4*b*B 
- 16*a^2*b^3*B - 2*b^5*B - a^3*b^2*(9*A - 20*C) - 21*a^5*C + 4*a*b^4*(3*A 
+ C))*EllipticF[(c + d*x)/2, 2])/(b*d) - (6*a^2*(5*A*b^4 + 5*a^3*b*B - 7*a 
*b^3*B - 3*a^2*b^2*(A - 3*C) - 7*a^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x 
)/2, 2])/(b*(a + b)*d)))/b)/(3*b) + (10*(5*a^2*b*B - 2*b^3*B - a*b^2*(3*A 
- 4*C) - 7*a^3*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b*d))/(5*b))/(2*b*(a 
^2 - b^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 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 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 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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1381\) vs. \(2(484)=968\).

Time = 5.08 (sec) , antiderivative size = 1382, normalized size of antiderivative = 2.74

method	result	size
default	\(\text {Expression too large to display}\)	\(1382\)

Input:

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

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*a^2/b^4*(3* 
A*b^2-4*B*a*b+5*C*a^2)/(-2*a*b+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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2*(2*A*a*b^2+A*b^3 
-3*B*a^2*b-2*B*a*b^2-B*b^3+4*C*a^3+3*C*a^2*b+2*C*a*b^2+C*b^3)/b^5*(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))-4/ 
5*C/b^2/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2*d* 
x+1/2*c)*sin(1/2*d*x+1/2*c)^6-14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+6 
*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-5*(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(1/2*d*x+1/2*c),2^(1/2))+9*(si 
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1 
/2*d*x+1/2*c),2^(1/2)))+2/b^4*(A*b^2-2*B*a*b-2*B*b^2+3*C*a^2+4*C*a*b+3*C*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))+EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+4/3/b^3*(B*b-2*C*a-3*C* 
b)*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-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(1/2*d*x+1/2*c),2^(1/2))-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2* 
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3} b^{2}+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3} a b +\sec \left (d x +c \right )^{3} a^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3} b^{2}+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3} a b +\sec \left (d x +c \right )^{3} a^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3} b^{2}+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3} a b +\sec \left (d x +c \right )^{3} a^{2}}d x \right ) c \] Input:

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

Output:

int(sqrt(sec(c + d*x))/(cos(c + d*x)**2*sec(c + d*x)**3*b**2 + 2*cos(c + d 
*x)*sec(c + d*x)**3*a*b + sec(c + d*x)**3*a**2),x)*a + int((sqrt(sec(c + d 
*x))*cos(c + d*x))/(cos(c + d*x)**2*sec(c + d*x)**3*b**2 + 2*cos(c + d*x)* 
sec(c + d*x)**3*a*b + sec(c + d*x)**3*a**2),x)*b + int((sqrt(sec(c + d*x)) 
*cos(c + d*x)**2)/(cos(c + d*x)**2*sec(c + d*x)**3*b**2 + 2*cos(c + d*x)*s 
ec(c + d*x)**3*a*b + sec(c + d*x)**3*a**2),x)*c