\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [885]

Optimal result

Integrand size = 42, antiderivative size = 303 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\left (3 a^2 b B-2 b^3 B-5 a^3 C+4 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (9 a^3 b B-12 a b^3 B-15 a^4 C+16 a^2 b^2 C+2 b^4 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 \left (a^2-b^2\right ) d}+\frac {a^2 \left (3 a^2 b B-5 b^3 B-5 a^3 C+7 a b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^4 (a+b)^2 d}-\frac {\left (3 a b B-5 a^2 C+2 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (b B-a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

(3*B*a^2*b-2*B*b^3-5*C*a^3+4*C*a*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2) 
)/b^3/(a^2-b^2)/d-1/3*(9*B*a^3*b-12*B*a*b^3-15*C*a^4+16*C*a^2*b^2+2*C*b^4) 
*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^4/(a^2-b^2)/d+a^2*(3*B*a^2*b-5*B 
*b^3-5*C*a^3+7*C*a*b^2)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/( 
a-b)/b^4/(a+b)^2/d-1/3*(3*B*a*b-5*C*a^2+2*C*b^2)*cos(d*x+c)^(1/2)*sin(d*x+ 
c)/b^2/(a^2-b^2)/d+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))
 

Mathematica [A] (warning: unable to verify)

Time = 3.68 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {4 \sqrt {\cos (c+d x)} \left (2 C+\frac {3 a^2 (-b B+a C)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}\right ) \sin (c+d x)-\frac {\frac {2 \left (-3 a^2 b B+6 b^3 B+5 a^3 C-8 a b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 \left (-3 a b B+2 a^2 C+b^2 C\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {6 \left (-3 a^2 b B+2 b^3 B+5 a^3 C-4 a b^2 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{12 b^2 d} \] 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])^2,x]
 

Output:

(4*Sqrt[Cos[c + d*x]]*(2*C + (3*a^2*(-(b*B) + a*C))/((a^2 - b^2)*(a + b*Co 
s[c + d*x])))*Sin[c + d*x] - ((2*(-3*a^2*b*B + 6*b^3*B + 5*a^3*C - 8*a*b^2 
*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*(-3*a*b*B + 2* 
a^2*C + b^2*C)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a 
+ b), (c + d*x)/2, 2]))/(a + b) + (6*(-3*a^2*b*B + 2*b^3*B + 5*a^3*C - 4*a 
*b^2*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*El 
lipticF[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)*(a + b)))/(12*b^2*d)
 

Rubi [A] (verified)

Time = 2.01 (sec) , antiderivative size = 295, normalized size of antiderivative = 0.97, 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, 3528, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3468

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

\(\displaystyle \frac {-\frac {-\frac {3 \left (-5 a^3 C+3 a^2 b B+4 a b^2 C-2 b^3 B\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {a b \left (-5 C a^2+3 b B a+2 b^2 C\right )+\left (-15 C a^4+9 b B a^3+16 b^2 C a^2-12 b^3 B a+2 b^4 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}-\frac {2 \left (-5 a^2 C+3 a b B+2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\frac {\int \frac {a b \left (-5 C a^2+3 b B a+2 b^2 C\right )+\left (-15 C a^4+9 b B a^3+16 b^2 C a^2-12 b^3 B a+2 b^4 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {3 \left (-5 a^3 C+3 a^2 b B+4 a b^2 C-2 b^3 B\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{3 b}-\frac {2 \left (-5 a^2 C+3 a b B+2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {\int \frac {a b \left (-5 C a^2+3 b B a+2 b^2 C\right )+\left (-15 C a^4+9 b B a^3+16 b^2 C a^2-12 b^3 B a+2 b^4 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {3 \left (-5 a^3 C+3 a^2 b B+4 a b^2 C-2 b^3 B\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}-\frac {2 \left (-5 a^2 C+3 a b B+2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {-\frac {\frac {\int \frac {a b \left (-5 C a^2+3 b B a+2 b^2 C\right )+\left (-15 C a^4+9 b B a^3+16 b^2 C a^2-12 b^3 B a+2 b^4 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {6 \left (-5 a^3 C+3 a^2 b B+4 a b^2 C-2 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (-5 a^2 C+3 a b B+2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {-\frac {\frac {\frac {\left (-15 a^4 C+9 a^3 b B+16 a^2 b^2 C-12 a b^3 B+2 b^4 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {3 a^2 \left (-5 a^3 C+3 a^2 b B+7 a b^2 C-5 b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{b}-\frac {6 \left (-5 a^3 C+3 a^2 b B+4 a b^2 C-2 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (-5 a^2 C+3 a b B+2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {\frac {\left (-15 a^4 C+9 a^3 b B+16 a^2 b^2 C-12 a b^3 B+2 b^4 C\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {3 a^2 \left (-5 a^3 C+3 a^2 b B+7 a b^2 C-5 b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {6 \left (-5 a^3 C+3 a^2 b B+4 a b^2 C-2 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (-5 a^2 C+3 a b B+2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {-\frac {\frac {\frac {2 \left (-15 a^4 C+9 a^3 b B+16 a^2 b^2 C-12 a b^3 B+2 b^4 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {3 a^2 \left (-5 a^3 C+3 a^2 b B+7 a b^2 C-5 b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {6 \left (-5 a^3 C+3 a^2 b B+4 a b^2 C-2 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (-5 a^2 C+3 a b B+2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {a (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {-\frac {2 \left (-5 a^2 C+3 a b B+2 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}-\frac {\frac {\frac {2 \left (-15 a^4 C+9 a^3 b B+16 a^2 b^2 C-12 a b^3 B+2 b^4 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {6 a^2 \left (-5 a^3 C+3 a^2 b B+7 a b^2 C-5 b^3 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}}{b}-\frac {6 \left (-5 a^3 C+3 a^2 b B+4 a b^2 C-2 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}}{2 b \left (a^2-b^2\right )}\)

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])^2,x]
 

Output:

(a*(b*B - a*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Co 
s[c + d*x])) + (-1/3*((-6*(3*a^2*b*B - 2*b^3*B - 5*a^3*C + 4*a*b^2*C)*Elli 
pticE[(c + d*x)/2, 2])/(b*d) + ((2*(9*a^3*b*B - 12*a*b^3*B - 15*a^4*C + 16 
*a^2*b^2*C + 2*b^4*C)*EllipticF[(c + d*x)/2, 2])/(b*d) - (6*a^2*(3*a^2*b*B 
 - 5*b^3*B - 5*a^3*C + 7*a*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2 
])/(b*(a + b)*d))/b)/b - (2*(3*a*b*B - 5*a^2*C + 2*b^2*C)*Sqrt[Cos[c + d*x 
]]*Sin[c + d*x])/(3*b*d))/(2*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 3528

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

rule 3538

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1065\) vs. \(2(300)=600\).

Time = 8.71 (sec) , antiderivative size = 1066, normalized size of antiderivative = 3.52

method	result	size
default	\(\text {Expression too large to display}\)	\(1066\)

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

Fricas [F(-1)]

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

Output:

Timed out
 

Sympy [F(-1)]

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

Output:

Timed out
 

Maxima [F]

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

Giac [F]

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

Mupad [F(-1)]

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

Reduce [F]

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

Output:

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