\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [1499]

Optimal result

Integrand size = 43, antiderivative size = 483 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 b^3 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4 b B-5 a^2 b^3 B+8 b^5 B-15 a^5 C-a b^4 (7 A+24 C)+a^3 b^2 (A+33 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 b^4 \left (a^2-b^2\right )^2 d}+\frac {\left (3 A b^6-3 a^5 b B+6 a^3 b^3 B-15 a b^5 B+15 a^6 C+5 a^2 b^4 (2 A+7 C)-a^4 b^2 (A+38 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 (a-b)^2 b^4 (a+b)^3 d}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 7.42 (sec) , antiderivative size = 915, normalized size of antiderivative = 1.89 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

((2*(5*a^2*A*b^2 + A*b^4 - a^3*b*B - 5*a*b^3*B + 5*a^4*C - 7*a^2*b^2*C + 8 
*b^4*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - Ellipt 
icPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(b + a*Sec[c + d*x])*Sqrt[1 
- Sec[c + d*x]^2]*Sin[c + d*x])/(a*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^ 
2)) + (2*(-24*a*A*b^3 + 8*a^2*b^2*B + 16*b^4*B + 8*a^3*b*C - 32*a*b^3*C)*C 
os[c + d*x]^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(b + a*Se 
c[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(a + b*Cos[c + d*x]) 
*(1 - Cos[c + d*x]^2)) + ((-(a^2*A*b^2) - 5*A*b^4 - 3*a^3*b*B + 9*a*b^3*B 
+ 15*a^4*C - 29*a^2*b^2*C + 8*b^4*C)*Cos[2*(c + d*x)]*(b + a*Sec[c + d*x]) 
*(-4*a*b + 4*a*b*Sec[c + d*x]^2 - 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x] 
]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*(2*a - b)*b*Ellipt 
icF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d* 
x]^2] - 4*a^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[ 
c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*b^2*EllipticPi[-(a/b), ArcSin[Sqrt[ 
Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d 
*x])/(a*b^2*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*( 
2 - Sec[c + d*x]^2)))/(16*(a - b)^2*b^2*(a + b)^2*d) + (Sqrt[Sec[c + d*x]] 
*(-1/4*((-(a^2*A*b^2) - 5*A*b^4 - 3*a^3*b*B + 9*a*b^3*B + 7*a^4*C - 13*a^2 
*b^2*C)*Sin[c + d*x])/(b^3*(a^2 - b^2)^2) - (-(a^2*A*b^2*Sin[c + d*x]) + a 
^3*b*B*Sin[c + d*x] - a^4*C*Sin[c + d*x])/(2*b^3*(-a^2 + b^2)*(a + b*Co...
 

Rubi [A] (verified)

Time = 3.00 (sec) , antiderivative size = 450, normalized size of antiderivative = 0.93, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4709, 3042, 3526, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A*b^2 - a*(b*B - a*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*(3*a^3*b*B - 9*a*b^3*B + 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*(3*a^4*b*B - 5*a^2*b^3*B + 
8*b^5*B - 15*a^5*C - a*b^4*(7*A + 24*C) + a^3*b^2*(A + 33*C))*EllipticF[(c 
 + d*x)/2, 2])/(b*d) + (2*(3*A*b^6 - 3*a^5*b*B + 6*a^3*b^3*B - 15*a*b^5*B 
+ 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 + 
a^3*b*B - 7*a*b^3*B - 5*a^4*C + a^2*b^2*(3*A + 11*C))*Sqrt[Cos[c + d*x]]*S 
in[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 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]
 

rule 4709

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1999\) vs. \(2(462)=924\).

Time = 8.07 (sec) , antiderivative size = 2000, normalized size of antiderivative = 4.14

method	result	size
default	\(\text {Expression too large to display}\)	\(2000\)

Input:

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

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

Fricas [F]

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

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

Output:

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((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)*sec(d*x + c)^(3/2)), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(3/2)*(a + b 
*cos(c + d*x))^3),x)
 

Output:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(3/2)*(a + b 
*cos(c + d*x))^3), x)
 

Reduce [F]

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

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

Output:

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