\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^3} \, dx\) [892]

Optimal result

Integrand size = 42, antiderivative size = 367 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (3 a^3 b B-9 a b^3 B-15 a^4 C+29 a^2 b^2 C-8 b^4 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4 b B-5 a^2 b^3 B+8 b^5 B-15 a^5 C+33 a^3 b^2 C-24 a b^4 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^4 \left (a^2-b^2\right )^2 d}-\frac {a \left (3 a^4 b B-6 a^2 b^3 B+15 b^5 B-15 a^5 C+38 a^3 b^2 C-35 a b^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 (a-b)^2 b^4 (a+b)^3 d}+\frac {a (b B-a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {a \left (a^2 b B-7 b^3 B-5 a^3 C+11 a b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

-1/4*(3*B*a^3*b-9*B*a*b^3-15*C*a^4+29*C*a^2*b^2-8*C*b^4)*EllipticE(sin(1/2 
*d*x+1/2*c),2^(1/2))/b^3/(a^2-b^2)^2/d+1/4*(3*B*a^4*b-5*B*a^2*b^3+8*B*b^5- 
15*C*a^5+33*C*a^3*b^2-24*C*a*b^4)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b 
^4/(a^2-b^2)^2/d-1/4*a*(3*B*a^4*b-6*B*a^2*b^3+15*B*b^5-15*C*a^5+38*C*a^3*b 
^2-35*C*a*b^4)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/(a-b)^2/b^ 
4/(a+b)^3/d+1/2*a*(B*b-C*a)*cos(d*x+c)^(3/2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b 
*cos(d*x+c))^2+1/4*a*(B*a^2*b-7*B*b^3-5*C*a^3+11*C*a*b^2)*cos(d*x+c)^(1/2) 
*sin(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 5.77 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.06 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\frac {-\frac {2 a \sqrt {\cos (c+d x)} \left (a \left (-a^2 b B+7 b^3 B+5 a^3 C-11 a b^2 C\right )+b \left (-3 a^2 b B+9 b^3 B+7 a^3 C-13 a b^2 C\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\frac {\left (-a^3 b B-5 a b^3 B+5 a^4 C-7 a^2 b^2 C+8 b^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 \left (a^2 b B+2 b^3 B+a^3 C-4 a b^2 C\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {\left (-3 a^3 b B+9 a b^3 B+15 a^4 C-29 a^2 b^2 C+8 b^4 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 b^2 d} \] Input:

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

Output:

((-2*a*Sqrt[Cos[c + d*x]]*(a*(-(a^2*b*B) + 7*b^3*B + 5*a^3*C - 11*a*b^2*C) 
 + b*(-3*a^2*b*B + 9*b^3*B + 7*a^3*C - 13*a*b^2*C)*Cos[c + d*x])*Sin[c + d 
*x])/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2) + (((-(a^3*b*B) - 5*a*b^3*B + 
5*a^4*C - 7*a^2*b^2*C + 8*b^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2] 
)/(a + b) + (8*(a^2*b*B + 2*b^3*B + a^3*C - 4*a*b^2*C)*((a + b)*EllipticF[ 
(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(a + b) + 
((-3*a^3*b*B + 9*a*b^3*B + 15*a^4*C - 29*a^2*b^2*C + 8*b^4*C)*(-2*a*b*Elli 
pticE[ArcSin[Sqrt[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*b^2*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + 
 b)^2))/(8*b^2*d)
 

Rubi [A] (verified)

Time = 2.30 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {3042, 3508, 3042, 3468, 27, 3042, 3526, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) (B+C \cos (c+d x))}{(a+b \cos (c+d x))^3}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3468

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

\(\displaystyle \frac {\frac {a \left (-5 a^3 C+a^2 b B+11 a b^2 C-7 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int -\frac {-\left (\left (-15 C a^4+3 b B a^3+29 b^2 C a^2-9 b^3 B a-8 b^4 C\right ) \cos ^2(c+d x)\right )+4 b \left (C a^3+b B a^2-4 b^2 C a+2 b^3 B\right ) \cos (c+d x)+a \left (-5 C a^3+b B a^2+11 b^2 C a-7 b^3 B\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b \left (a^2-b^2\right )}}{4 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {-\left (\left (-15 C a^4+3 b B a^3+29 b^2 C a^2-9 b^3 B a-8 b^4 C\right ) \cos ^2(c+d x)\right )+4 b \left (C a^3+b B a^2-4 b^2 C a+2 b^3 B\right ) \cos (c+d x)+a \left (-5 C a^3+b B a^2+11 b^2 C a-7 b^3 B\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{2 b \left (a^2-b^2\right )}+\frac {a \left (-5 a^3 C+a^2 b B+11 a b^2 C-7 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\left (15 C a^4-3 b B a^3-29 b^2 C a^2+9 b^3 B a+8 b^4 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+4 b \left (C a^3+b B a^2-4 b^2 C a+2 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (-5 C a^3+b B a^2+11 b^2 C a-7 b^3 B\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \left (a^2-b^2\right )}+\frac {a \left (-5 a^3 C+a^2 b B+11 a b^2 C-7 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {\frac {-\frac {\left (-15 a^4 C+3 a^3 b B+29 a^2 b^2 C-9 a b^3 B-8 b^4 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {a b \left (-5 C a^3+b B a^2+11 b^2 C a-7 b^3 B\right )+\left (-15 C a^5+3 b B a^4+33 b^2 C a^3-5 b^3 B a^2-24 b^4 C a+8 b^5 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-5 a^3 C+a^2 b B+11 a b^2 C-7 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {a b \left (-5 C a^3+b B a^2+11 b^2 C a-7 b^3 B\right )+\left (-15 C a^5+3 b B a^4+33 b^2 C a^3-5 b^3 B a^2-24 b^4 C a+8 b^5 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {\left (-15 a^4 C+3 a^3 b B+29 a^2 b^2 C-9 a b^3 B-8 b^4 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-5 a^3 C+a^2 b B+11 a b^2 C-7 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {a b \left (-5 C a^3+b B a^2+11 b^2 C a-7 b^3 B\right )+\left (-15 C a^5+3 b B a^4+33 b^2 C a^3-5 b^3 B a^2-24 b^4 C a+8 b^5 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {\left (-15 a^4 C+3 a^3 b B+29 a^2 b^2 C-9 a b^3 B-8 b^4 C\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-5 a^3 C+a^2 b B+11 a b^2 C-7 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {\int \frac {a b \left (-5 C a^3+b B a^2+11 b^2 C a-7 b^3 B\right )+\left (-15 C a^5+3 b B a^4+33 b^2 C a^3-5 b^3 B a^2-24 b^4 C a+8 b^5 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 \left (-15 a^4 C+3 a^3 b B+29 a^2 b^2 C-9 a b^3 B-8 b^4 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-5 a^3 C+a^2 b B+11 a b^2 C-7 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {\frac {\frac {\left (-15 a^5 C+3 a^4 b B+33 a^3 b^2 C-5 a^2 b^3 B-24 a b^4 C+8 b^5 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {a \left (-15 a^5 C+3 a^4 b B+38 a^3 b^2 C-6 a^2 b^3 B-35 a b^4 C+15 b^5 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{b}-\frac {2 \left (-15 a^4 C+3 a^3 b B+29 a^2 b^2 C-9 a b^3 B-8 b^4 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-5 a^3 C+a^2 b B+11 a b^2 C-7 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {\left (-15 a^5 C+3 a^4 b B+33 a^3 b^2 C-5 a^2 b^3 B-24 a b^4 C+8 b^5 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {a \left (-15 a^5 C+3 a^4 b B+38 a^3 b^2 C-6 a^2 b^3 B-35 a b^4 C+15 b^5 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {2 \left (-15 a^4 C+3 a^3 b B+29 a^2 b^2 C-9 a b^3 B-8 b^4 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-5 a^3 C+a^2 b B+11 a b^2 C-7 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {\frac {2 \left (-15 a^5 C+3 a^4 b B+33 a^3 b^2 C-5 a^2 b^3 B-24 a b^4 C+8 b^5 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {a \left (-15 a^5 C+3 a^4 b B+38 a^3 b^2 C-6 a^2 b^3 B-35 a b^4 C+15 b^5 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {2 \left (-15 a^4 C+3 a^3 b B+29 a^2 b^2 C-9 a b^3 B-8 b^4 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-5 a^3 C+a^2 b B+11 a b^2 C-7 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\frac {a \left (-5 a^3 C+a^2 b B+11 a b^2 C-7 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\frac {2 \left (-15 a^5 C+3 a^4 b B+33 a^3 b^2 C-5 a^2 b^3 B-24 a b^4 C+8 b^5 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {2 a \left (-15 a^5 C+3 a^4 b B+38 a^3 b^2 C-6 a^2 b^3 B-35 a b^4 C+15 b^5 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}}{b}-\frac {2 \left (-15 a^4 C+3 a^3 b B+29 a^2 b^2 C-9 a b^3 B-8 b^4 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}}{4 b \left (a^2-b^2\right )}\)

Input:

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

Output:

(a*(b*B - a*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b* 
Cos[c + d*x])^2) + (((-2*(3*a^3*b*B - 9*a*b^3*B - 15*a^4*C + 29*a^2*b^2*C 
- 8*b^4*C)*EllipticE[(c + d*x)/2, 2])/(b*d) + ((2*(3*a^4*b*B - 5*a^2*b^3*B 
 + 8*b^5*B - 15*a^5*C + 33*a^3*b^2*C - 24*a*b^4*C)*EllipticF[(c + d*x)/2, 
2])/(b*d) - (2*a*(3*a^4*b*B - 6*a^2*b^3*B + 15*b^5*B - 15*a^5*C + 38*a^3*b 
^2*C - 35*a*b^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d 
))/b)/(2*b*(a^2 - b^2)) + (a*(a^2*b*B - 7*b^3*B - 5*a^3*C + 11*a*b^2*C)*Sq 
rt[Cos[c + d*x]]*Sin[c + d*x])/(b*(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 3508

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[1/b^2   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - 
a*b*B + a^2*C, 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 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. \(1976\) vs. \(2(358)=716\).

Time = 8.66 (sec) , antiderivative size = 1977, normalized size of antiderivative = 5.39

method	result	size
default	\(\text {Expression too large to display}\)	\(1977\)

Input:

int(cos(d*x+c)^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x,me 
thod=_RETURNVERBOSE)
 

Output:

-(-(1-2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/b^4/(-2*sin(1 
/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*( 
2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(B*b*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)) 
-3*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-C*b*EllipticE(cos(1/2*d*x+1/2 
*c),2^(1/2)))+2*a^2/b^4*(3*B*b-4*C*a)*(-1/a*b^2/(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)*Elliptic 
F(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/a*b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*(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/a*b/(a^2-b^2)*(si 
n(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)*EllipticPi(cos(1/2*d*x+1 
/2*c),-2*b/(a-b),2^(1/2)))+12*a*(B*b-2*C*a)/b^3/(-2*a*b+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*...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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