\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx\) [378]

Optimal result

Integrand size = 33, antiderivative size = 344 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}+\frac {\left (a^3 A b-7 a A b^3+3 a^4 B-5 a^2 b^2 B+8 b^4 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {\left (a^4 A b-10 a^2 A b^3-3 A b^5+3 a^5 B-6 a^3 b^2 B+15 a b^4 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 (a-b)^2 b^3 (a+b)^3 d}+\frac {a (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

-1/4*(A*a^2*b+5*A*b^3+3*B*a^3-9*B*a*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1 
/2))/b^2/(a^2-b^2)^2/d+1/4*(A*a^3*b-7*A*a*b^3+3*B*a^4-5*B*a^2*b^2+8*B*b^4) 
*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^3/(a^2-b^2)^2/d-1/4*(A*a^4*b-10* 
A*a^2*b^3-3*A*b^5+3*B*a^5-6*B*a^3*b^2+15*B*a*b^4)*EllipticPi(sin(1/2*d*x+1 
/2*c),2*b/(a+b),2^(1/2))/(a-b)^2/b^3/(a+b)^3/d+1/2*a*(A*b-B*a)*cos(d*x+c)^ 
(1/2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^2+1/4*(A*a^2*b+5*A*b^3+3*B 
*a^3-9*B*a*b^2)*cos(d*x+c)^(1/2)*sin(d*x+c)/b/(a^2-b^2)^2/d/(a+b*cos(d*x+c 
))
 

Mathematica [A] (warning: unable to verify)

Time = 4.34 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {2 \sqrt {\cos (c+d x)} \left (a \left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right )+b \left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-\frac {\frac {\left (-5 a^2 A b-A b^3+a^3 B+5 a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}-\frac {8 \left (-3 a A b+a^2 B+2 b^2 B\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 b d} \] Input:

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

Output:

((2*Sqrt[Cos[c + d*x]]*(a*(3*a^2*A*b + 3*A*b^3 + a^3*B - 7*a*b^2*B) + b*(a 
^2*A*b + 5*A*b^3 + 3*a^3*B - 9*a*b^2*B)*Cos[c + d*x])*Sin[c + d*x])/((a^2 
- b^2)^2*(a + b*Cos[c + d*x])^2) - (((-5*a^2*A*b - A*b^3 + a^3*B + 5*a*b^2 
*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) - (8*(-3*a*A*b + a^ 
2*B + 2*b^2*B)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a 
+ b), (c + d*x)/2, 2]))/(a + b) + ((a^2*A*b + 5*A*b^3 + 3*a^3*B - 9*a*b^2* 
B)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*Ellipti 
cF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), Arc 
Sin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[Sin[c + d*x]^2]))/ 
((a - b)^2*(a + b)^2))/(8*b*d)
 

Rubi [A] (verified)

Time = 2.08 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.01, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3468, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3468

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {\int \frac {-a \left (3 B a^3+A b a^2-9 b^2 B a+5 A b^3\right ) \cos ^2(c+d x)-4 a b \left (-B a^2+3 A b a-2 b^2 B\right ) \cos (c+d x)+a \left (B a^3+3 A b a^2-7 b^2 B a+3 A b^3\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {\left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {-a \left (3 B a^3+A b a^2-9 b^2 B a+5 A b^3\right ) \cos ^2(c+d x)-4 a b \left (-B a^2+3 A b a-2 b^2 B\right ) \cos (c+d x)+a \left (B a^3+3 A b a^2-7 b^2 B a+3 A b^3\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}+\frac {\left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {-a \left (3 B a^3+A b a^2-9 b^2 B a+5 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-4 a b \left (-B a^2+3 A b a-2 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (B a^3+3 A b a^2-7 b^2 B a+3 A b^3\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}+\frac {\left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {\frac {-\frac {a \left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {a b \left (B a^3+3 A b a^2-7 b^2 B a+3 A b^3\right )+a \left (3 B a^4+A b a^3-5 b^2 B a^2-7 A b^3 a+8 b^4 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{2 a \left (a^2-b^2\right )}+\frac {\left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {a b \left (B a^3+3 A b a^2-7 b^2 B a+3 A b^3\right )+a \left (3 B a^4+A b a^3-5 b^2 B a^2-7 A b^3 a+8 b^4 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {a \left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{2 a \left (a^2-b^2\right )}+\frac {\left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {a b \left (B a^3+3 A b a^2-7 b^2 B a+3 A b^3\right )+a \left (3 B a^4+A b a^3-5 b^2 B a^2-7 A b^3 a+8 b^4 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {a \left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{2 a \left (a^2-b^2\right )}+\frac {\left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {\int \frac {a b \left (B a^3+3 A b a^2-7 b^2 B a+3 A b^3\right )+a \left (3 B a^4+A b a^3-5 b^2 B a^2-7 A b^3 a+8 b^4 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 a \left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}+\frac {\left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

(a*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b* 
Cos[c + d*x])^2) + (((-2*a*(a^2*A*b + 5*A*b^3 + 3*a^3*B - 9*a*b^2*B)*Ellip 
ticE[(c + d*x)/2, 2])/(b*d) + ((2*a*(a^3*A*b - 7*a*A*b^3 + 3*a^4*B - 5*a^2 
*b^2*B + 8*b^4*B)*EllipticF[(c + d*x)/2, 2])/(b*d) - (2*a*(a^4*A*b - 10*a^ 
2*A*b^3 - 3*A*b^5 + 3*a^5*B - 6*a^3*b^2*B + 15*a*b^4*B)*EllipticPi[(2*b)/( 
a + b), (c + d*x)/2, 2])/(b*(a + b)*d))/b)/(2*a*(a^2 - b^2)) + ((a^2*A*b + 
 5*A*b^3 + 3*a^3*B - 9*a*b^2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b 
^2)*d*(a + b*Cos[c + d*x])))/(4*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 3468

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((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 - 2)*(c + d*Sin[e + f*x])^(n 
+ 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a 
*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) 
 - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - 
a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] 
/; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2 
, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
 

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 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. \(1936\) vs. \(2(335)=670\).

Time = 8.41 (sec) , antiderivative size = 1937, normalized size of antiderivative = 5.63

method	result	size
default	\(\text {Expression too large to display}\)	\(1937\)

Input:

int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^3,x,method=_RETURNV 
ERBOSE)
                                                                                    
                                                                                    
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B/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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))- 
4/b^2*(A*b-3*B*a)/(-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*a^2*(A*b-B*a)/b^3*(-1 
/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)^2-3/4*b^2*(3*a^2-b^2)/a^2/ 
(a^2-b^2)^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)-7/8/(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))+1/4/(a+ 
b)/(a^2-b^2)/a*(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))*b+3/8/(a+b)/(a^2-b^2)/a^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))*b^2-9/8*b/(a^2-b^2)^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))+3/8*b^3/a^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2...
 

Fricas [F(-1)]

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

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,x, algorith 
m="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,x, algorith 
m="maxima")
 

Output:

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

Giac [F]

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

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,x, algorith 
m="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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