\(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx\) [657]

Optimal result

Integrand size = 35, antiderivative size = 375 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{5 b^4 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (5 A b^2+2 \left (4 a^2+b^2\right ) C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{5 b^4 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \] Output:

2/5*(2*a^2*b^2*(5*A-4*C)+16*a^4*C-b^4*(5*A+3*C))*(a+b*cos(d*x+c))^(1/2)*El 
lipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))/b^4/(a^2-b^2)/d/((a+b* 
cos(d*x+c))/(a+b))^(1/2)-4/5*a*(5*A*b^2+2*(4*a^2+b^2)*C)*((a+b*cos(d*x+c)) 
/(a+b))^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/b^4/d 
/(a+b*cos(d*x+c))^(1/2)-2*(A*b^2+C*a^2)*cos(d*x+c)^2*sin(d*x+c)/b/(a^2-b^2 
)/d/(a+b*cos(d*x+c))^(1/2)-2/5*a*(5*A*b^2+8*C*a^2-3*C*b^2)*(a+b*cos(d*x+c) 
)^(1/2)*sin(d*x+c)/b^3/(a^2-b^2)/d+2/5*(5*A*b^2+6*C*a^2-C*b^2)*cos(d*x+c)* 
(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/b^2/(a^2-b^2)/d
 

Mathematica [A] (verified)

Time = 2.92 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {\frac {2 a b^2 \left (5 A b^2+\left (4 a^2+b^2\right ) C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{(a-b) (a+b)}+\frac {2 \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )}{(a-b) (a+b)}+\frac {10 a^2 b \left (A b^2+a^2 C\right ) \sin (c+d x)}{-a^2+b^2}-6 a b C (a+b \cos (c+d x)) \sin (c+d x)+b^2 C (a+b \cos (c+d x)) \sin (2 (c+d x))}{5 b^4 d \sqrt {a+b \cos (c+d x)}} \] Input:

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

Output:

((2*a*b^2*(5*A*b^2 + (4*a^2 + b^2)*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*E 
llipticF[(c + d*x)/2, (2*b)/(a + b)])/((a - b)*(a + b)) + (2*(2*a^2*b^2*(5 
*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)] 
*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, 
 (2*b)/(a + b)]))/((a - b)*(a + b)) + (10*a^2*b*(A*b^2 + a^2*C)*Sin[c + d* 
x])/(-a^2 + b^2) - 6*a*b*C*(a + b*Cos[c + d*x])*Sin[c + d*x] + b^2*C*(a + 
b*Cos[c + d*x])*Sin[2*(c + d*x)])/(5*b^4*d*Sqrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 2.10 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 3527, 27, 3042, 3528, 27, 3042, 3502, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

\(\displaystyle -\frac {\frac {2 \int -\frac {-3 a \left (8 C a^2+5 A b^2-3 b^2 C\right ) \cos ^2(c+d x)-b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \cos (c+d x)+2 a \left (6 C a^2+b^2 (5 A-C)\right )}{2 \sqrt {a+b \cos (c+d x)}}dx}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\int \frac {-3 a \left (8 C a^2+5 A b^2-3 b^2 C\right ) \cos ^2(c+d x)-b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \cos (c+d x)+2 a \left (6 C a^2+b^2 (5 A-C)\right )}{\sqrt {a+b \cos (c+d x)}}dx}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\int \frac {-3 a \left (8 C a^2+5 A b^2-3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a \left (6 C a^2+b^2 (5 A-C)\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3502

\(\displaystyle -\frac {-\frac {\frac {2 \int \frac {3 \left (a b \left (5 A b^2+\left (4 a^2+b^2\right ) C\right )+\left (16 C a^4+2 b^2 (5 A-4 C) a^2-b^4 (5 A+3 C)\right ) \cos (c+d x)\right )}{2 \sqrt {a+b \cos (c+d x)}}dx}{3 b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\frac {\int \frac {a b \left (5 A b^2+\left (4 a^2+b^2\right ) C\right )+\left (16 C a^4+2 b^2 (5 A-4 C) a^2-b^4 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\int \frac {a b \left (5 A b^2+\left (4 a^2+b^2\right ) C\right )+\left (16 C a^4+2 b^2 (5 A-4 C) a^2-b^4 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3231

\(\displaystyle -\frac {-\frac {\frac {\frac {\left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}-\frac {2 a \left (a^2-b^2\right ) \left (8 a^2 C+5 A b^2+2 b^2 C\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\frac {\left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 a \left (a^2-b^2\right ) \left (8 a^2 C+5 A b^2+2 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3134

\(\displaystyle -\frac {-\frac {\frac {\frac {\left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (a^2-b^2\right ) \left (8 a^2 C+5 A b^2+2 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\frac {\left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (a^2-b^2\right ) \left (8 a^2 C+5 A b^2+2 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3132

\(\displaystyle -\frac {-\frac {\frac {\frac {2 \left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (a^2-b^2\right ) \left (8 a^2 C+5 A b^2+2 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3142

\(\displaystyle -\frac {-\frac {\frac {\frac {2 \left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (a^2-b^2\right ) \left (8 a^2 C+5 A b^2+2 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\frac {2 \left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (a^2-b^2\right ) \left (8 a^2 C+5 A b^2+2 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3140

\(\displaystyle -\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}-\frac {\frac {\frac {2 \left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (a^2-b^2\right ) \left (8 a^2 C+5 A b^2+2 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}}{b \left (a^2-b^2\right )}\)

Input:

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

Output:

(-2*(A*b^2 + a^2*C)*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + 
 b*Cos[c + d*x]]) - ((-2*(5*A*b^2 + 6*a^2*C - b^2*C)*Cos[c + d*x]*Sqrt[a + 
 b*Cos[c + d*x]]*Sin[c + d*x])/(5*b*d) - (((2*(2*a^2*b^2*(5*A - 4*C) + 16* 
a^4*C - b^4*(5*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, ( 
2*b)/(a + b)])/(b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (4*a*(a^2 - b^2) 
*(5*A*b^2 + 8*a^2*C + 2*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Elliptic 
F[(c + d*x)/2, (2*b)/(a + b)])/(b*d*Sqrt[a + b*Cos[c + d*x]]))/b - (2*a*(5 
*A*b^2 + 8*a^2*C - 3*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*d))/ 
(5*b))/(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 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3231

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b 
, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

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])))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1292\) vs. \(2(362)=724\).

Time = 13.64 (sec) , antiderivative size = 1293, normalized size of antiderivative = 3.45

method	result	size
default	\(\text {Expression too large to display}\)	\(1293\)
parts	\(\text {Expression too large to display}\)	\(1817\)

Input:

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

Output:

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*a^2*(A* 
b^2+C*a^2)/b^4/sin(1/2*d*x+1/2*c)^2/(2*b*sin(1/2*d*x+1/2*c)^2-a-b)/(a^2-b^ 
2)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b*sin(1 
/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/ 
(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(sin(1/2*d*x 
+1/2*c)^2)^(1/2)*a-(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*Ell 
ipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2) 
*b)-2*(A*a*b^2+A*b^3+C*a^3+C*a^2*b+C*a*b^2+C*b^3)/b^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)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/ 
2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2* 
b/(a-b))^(1/2))+16*C/b*(-1/10/b*cos(1/2*d*x+1/2*c)^3*(-2*b*sin(1/2*d*x+1/2 
*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)-1/60/b^2*(-4*a+12*b)*cos(1/2*d*x+1 
/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/60/b^ 
2*(-4*a+12*b)*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^ 
2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2) 
^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/60*(4*a^2-15*a*b 
+27*b^2)/b^3*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2 
+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^ 
(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2* 
d*x+1/2*c),(-2*b/(a-b))^(1/2))))-2/b^4*(A*b^2+C*a^2+2*C*a*b+3*C*b^2)*(a...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 798, normalized size of antiderivative = 2.13 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

2/15*(sqrt(1/2)*(32*I*C*a^6 + 4*I*(5*A - 7*C)*a^4*b^2 - I*(25*A + 9*C)*a^2 
*b^4 + (32*I*C*a^5*b + 4*I*(5*A - 7*C)*a^3*b^3 - I*(25*A + 9*C)*a*b^5)*cos 
(d*x + c))*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a 
^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) + 
sqrt(1/2)*(-32*I*C*a^6 - 4*I*(5*A - 7*C)*a^4*b^2 + I*(25*A + 9*C)*a^2*b^4 
+ (-32*I*C*a^5*b - 4*I*(5*A - 7*C)*a^3*b^3 + I*(25*A + 9*C)*a*b^5)*cos(d*x 
 + c))*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 
 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b) + 3*sq 
rt(1/2)*(16*I*C*a^5*b + 2*I*(5*A - 4*C)*a^3*b^3 - I*(5*A + 3*C)*a*b^5 + (1 
6*I*C*a^4*b^2 + 2*I*(5*A - 4*C)*a^2*b^4 - I*(5*A + 3*C)*b^6)*cos(d*x + c)) 
*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/ 
b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/ 
b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b)) + 3*sqrt(1/2)*( 
-16*I*C*a^5*b - 2*I*(5*A - 4*C)*a^3*b^3 + I*(5*A + 3*C)*a*b^5 + (-16*I*C*a 
^4*b^2 - 2*I*(5*A - 4*C)*a^2*b^4 + I*(5*A + 3*C)*b^6)*cos(d*x + c))*sqrt(b 
)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, we 
ierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/ 
3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b)) - 3*(8*C*a^4*b^2 + (5* 
A - 3*C)*a^2*b^4 - (C*a^2*b^4 - C*b^6)*cos(d*x + c)^2 + 2*(C*a^3*b^3 - C*a 
*b^5)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/((a^2*b^6 - ...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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