\(\int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\) [719]

Optimal result

Integrand size = 35, antiderivative size = 336 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=-\frac {b \left (5 A b^2-a^2 (4 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (5 A b^2-a^2 (2 A-3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 \left (a^2-b^2\right ) d}-\frac {\left (5 A b^4-a^2 b^2 (7 A-C)-3 a^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 (a-b) (a+b)^2 d}-\frac {\left (5 A b^2-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {b \left (5 A b^2-a^2 (4 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 6.46 (sec) , antiderivative size = 332, normalized size of antiderivative = 0.99 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {\frac {\frac {2 \left (-45 A b^4+a^2 b^2 (44 A-9 C)+4 a^4 (A+3 C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 \left (-10 a A b^2+a^3 (7 A-3 C)\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {6 \left (-5 A b^2+a^2 (4 A-C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}+4 \sqrt {\cos (c+d x)} \left (\frac {3 b^2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+2 A (-6 b+a \sec (c+d x)) \tan (c+d x)\right )}{12 a^3 d} \] Input:

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

Output:

(((2*(-45*A*b^4 + a^2*b^2*(44*A - 9*C) + 4*a^4*(A + 3*C))*EllipticPi[(2*b) 
/(a + b), (c + d*x)/2, 2])/(a + b) + (8*(-10*a*A*b^2 + a^3*(7*A - 3*C))*(( 
a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2 
, 2]))/(a + b) + (6*(-5*A*b^2 + a^2*(4*A - C))*(-2*a*b*EllipticE[ArcSin[Sq 
rt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], 
 -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])* 
Sin[c + d*x])/(a*Sqrt[Sin[c + d*x]^2]))/((a - b)*(a + b)) + 4*Sqrt[Cos[c + 
 d*x]]*((3*b^2*(A*b^2 + a^2*C)*Sin[c + d*x])/((a^2 - b^2)*(a + b*Cos[c + d 
*x])) + 2*A*(-6*b + a*Sec[c + d*x])*Tan[c + d*x]))/(12*a^3*d)
 

Rubi [A] (verified)

Time = 2.61 (sec) , antiderivative size = 316, normalized size of antiderivative = 0.94, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 3535, 27, 3042, 3534, 27, 3042, 3534, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\frac {2 \left (5 A b^2-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {-2 (A+3 C) a^4-b^2 (16 A-3 C) a^2+2 b \left (10 A b^2-a^2 (7 A-3 C)\right ) \cos (c+d x) a+15 A b^4+3 b^2 \left (5 A b^2-a^2 (4 A-C)\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {6 b \left (5 A b^2-a^2 (4 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\frac {2 \left (5 A b^2-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 b \left (5 A b^2-a^2 (4 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-2 (A+3 C) a^4-b^2 (16 A-3 C) a^2+2 b \left (10 A b^2-a^2 (7 A-3 C)\right ) \cos (c+d x) a+15 A b^4+3 b^2 \left (5 A b^2-a^2 (4 A-C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\frac {2 \left (5 A b^2-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 b \left (5 A b^2-a^2 (4 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-2 (A+3 C) a^4-b^2 (16 A-3 C) a^2+2 b \left (10 A b^2-a^2 (7 A-3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+15 A b^4+3 b^2 \left (5 A b^2-a^2 (4 A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

((A*b^2 + a^2*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a + b* 
Cos[c + d*x])) - ((2*(5*A*b^2 - a^2*(2*A - 3*C))*Sin[c + d*x])/(3*a*d*Cos[ 
c + d*x]^(3/2)) - (-(((6*b*(5*A*b^2 - a^2*(4*A - C))*EllipticE[(c + d*x)/2 
, 2])/d + ((2*a*b*(5*A*b^2 - a^2*(2*A - 3*C))*EllipticF[(c + d*x)/2, 2])/d 
 + (6*b*(5*A*b^4 - a^2*b^2*(7*A - C) - 3*a^4*C)*EllipticPi[(2*b)/(a + b), 
(c + d*x)/2, 2])/((a + b)*d))/b)/a) + (6*b*(5*A*b^2 - a^2*(4*A - C))*Sin[c 
 + d*x])/(a*d*Sqrt[Cos[c + d*x]]))/(3*a))/(2*a*(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 3534

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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 3535

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[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S 
in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m 
+ 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin 
[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n 
+ 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d 
*(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) || 
 !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 
 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(991\) vs. \(2(331)=662\).

Time = 7.91 (sec) , antiderivative size = 992, normalized size of antiderivative = 2.95

method	result	size
default	\(\text {Expression too large to display}\)	\(992\)

Input:

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

Output:

-(-(1-2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A/a^2*(-1/6*c 
os(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)/(co 
s(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d* 
x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell 
ipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(A*b^2+C*a^2)/a^2*(-b^2/a/(a^2-b^2)* 
cos(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 
*b*cos(1/2*d*x+1/2*c)^2+a-b)-1/2/(a+b)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2 
*cos(1/2*d*x+1/2*c)^2)^(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*b/a/(a^2-b^2)*(sin(1/2*d 
*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(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*b 
/a/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(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)*(1-2*cos(1/2*d*x+1/2*c)^2)^(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)*(1-2*cos(1/2*d*x+1/2 
*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic 
Pi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))-4*A/a^3*b/sin(1/2*d*x+1/2*c)^2/ 
(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algori 
thm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5} b^{2}+2 \cos \left (d x +c \right )^{4} a b +\cos \left (d x +c \right )^{3} a^{2}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3} b^{2}+2 \cos \left (d x +c \right )^{2} a b +\cos \left (d x +c \right ) a^{2}}d x \right ) c \] Input:

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

Output:

int(sqrt(cos(c + d*x))/(cos(c + d*x)**5*b**2 + 2*cos(c + d*x)**4*a*b + cos 
(c + d*x)**3*a**2),x)*a + int(sqrt(cos(c + d*x))/(cos(c + d*x)**3*b**2 + 2 
*cos(c + d*x)**2*a*b + cos(c + d*x)*a**2),x)*c