\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^3} \, dx\) [722]

Optimal result

Integrand size = 35, antiderivative size = 345 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {a \left (15 a^4 C+b^4 (7 A+24 C)-a^2 b^2 (A+33 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^4 \left (a^2-b^2\right )^2 d}+\frac {\left (3 A b^6+15 a^6 C+5 a^2 b^4 (2 A+7 C)-a^4 b^2 (A+38 C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 (a-b)^2 b^4 (a+b)^3 d}-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (3 A b^4-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

-1/4*(b^4*(5*A-8*C)-15*a^4*C+a^2*b^2*(A+29*C))*EllipticE(sin(1/2*d*x+1/2*c 
),2^(1/2))/b^3/(a^2-b^2)^2/d-1/4*a*(15*a^4*C+b^4*(7*A+24*C)-a^2*b^2*(A+33* 
C))*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^4/(a^2-b^2)^2/d+1/4*(3*A*b^6+ 
15*a^6*C+5*a^2*b^4*(2*A+7*C)-a^4*b^2*(A+38*C))*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^2+C*a^2)*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*(3*A*b^4-5*a^4*C+a^2* 
b^2*(3*A+11*C))*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 = 5.86 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {2 \sqrt {\cos (c+d x)} \left (3 a A b^4-5 a^5 C+a^3 b^2 (3 A+11 C)+\left (5 A b^5-7 a^4 b C+a^2 b^3 (A+13 C)\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\frac {\left (a^2 b^2 (5 A-7 C)+5 a^4 C+b^4 (A+8 C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 a \left (-3 A b^2+\left (a^2-4 b^2\right ) C\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {\left (15 a^4 C+b^4 (-5 A+8 C)-a^2 b^2 (A+29 C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 b^2 d} \] Input:

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

Output:

((2*Sqrt[Cos[c + d*x]]*(3*a*A*b^4 - 5*a^5*C + a^3*b^2*(3*A + 11*C) + (5*A* 
b^5 - 7*a^4*b*C + a^2*b^3*(A + 13*C))*Cos[c + d*x])*Sin[c + d*x])/((a^2 - 
b^2)^2*(a + b*Cos[c + d*x])^2) + (((a^2*b^2*(5*A - 7*C) + 5*a^4*C + b^4*(A 
 + 8*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*a*(-3*A*b 
^2 + (a^2 - 4*b^2)*C)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2 
*b)/(a + b), (c + d*x)/2, 2]))/(a + b) + ((15*a^4*C + b^4*(-5*A + 8*C) - a 
^2*b^2*(A + 29*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a 
*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*Ellipt 
icPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[Si 
n[c + d*x]^2]))/((a - b)^2*(a + b)^2))/(8*b^2*d)
 

Rubi [A] (verified)

Time = 2.31 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3527, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

-1/2*((A*b^2 + a^2*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a 
 + b*Cos[c + d*x])^2) - (-1/2*((-2*(b^4*(5*A - 8*C) - 15*a^4*C + a^2*b^2*( 
A + 29*C))*EllipticE[(c + d*x)/2, 2])/(b*d) + ((-2*a*(15*a^4*C + b^4*(7*A 
+ 24*C) - a^2*b^2*(A + 33*C))*EllipticF[(c + d*x)/2, 2])/(b*d) + (2*(3*A*b 
^6 + 15*a^6*C + 5*a^2*b^4*(2*A + 7*C) - a^4*b^2*(A + 38*C))*EllipticPi[(2* 
b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d))/b)/(b*(a^2 - b^2)) - ((3*A*b^4 
 - 5*a^4*C + a^2*b^2*(3*A + 11*C))*Sqrt[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 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 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 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 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. \(1965\) vs. \(2(336)=672\).

Time = 9.50 (sec) , antiderivative size = 1966, normalized size of antiderivative = 5.70

method	result	size
default	\(\text {Expression too large to display}\)	\(1966\)

Input:

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

Output:

-(-(1-2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*C/b^4/(-2*si 
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^ 
(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*a*EllipticF(cos(1/2*d*x+1/2*c),2^(1/ 
2))+b*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-4/b^3*(A*b^2+6*C*a^2)/(-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^2*(A*b^2+C*a^2)/b^4*(-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)*(1-2*c 
os(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))+1/4/(a+b)/(a^2-b^2)/a*(sin(1/2 
*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2* 
c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b+3 
/8/(a+b)/(a^2-b^2)/a^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))*b^2-9/8*b/(a^2-b^2)^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))+3/8*b^3/a^2/(a^2...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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