\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [1496]

Optimal result

Integrand size = 43, antiderivative size = 562 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^2 b \left (a^2-b^2\right )^2 d}-\frac {\left (15 A b^6-15 a^5 b B+6 a^3 b^3 B-3 a b^5 B+3 a^6 C-a^2 b^4 (38 A+C)+5 a^4 b^2 (7 A+2 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^3 (a-b)^2 b (a+b)^3 d}+\frac {\left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 7.30 (sec) , antiderivative size = 944, normalized size of antiderivative = 1.68 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

Integrate[((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,x]
 

Output:

-1/16*((2*(56*a^4*A*b - 95*a^2*A*b^3 + 45*A*b^5 - 16*a^5*B + 19*a^3*b^2*B 
- 9*a*b^4*B + 9*a^4*b*C - 3*a^2*b^3*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sq 
rt[Sec[c + d*x]]], -1] - EllipticPi[-(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*C 
os[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(16*a^5*A - 80*a^3*A*b^2 + 40*a*A* 
b^4 + 32*a^4*b*B - 8*a^2*b^3*B - 16*a^5*C - 8*a^3*b^2*C)*Cos[c + d*x]^2*El 
lipticPi[-(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])/(b*(a + b*Cos[c + d*x])*(1 - Cos[c + d* 
x]^2)) + ((8*a^4*A*b - 29*a^2*A*b^3 + 15*A*b^5 + 9*a^3*b^2*B - 3*a*b^4*B - 
 5*a^4*b*C - a^2*b^3*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*EllipticF[ArcSin[Sq 
rt[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]]*Sqr 
t[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)))/(a^3*(a - b)^2*(a + b)^2*d) + (Sqrt[Sec[c + d*x]]*(((8*a^4*A - 29 
*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B - 5*a^4*C - a^2*b^2*C)*Sin[c 
 + d*x])/(4*a^3*(a^2 - b^2)^2) + (A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*...
 

Rubi [A] (verified)

Time = 3.97 (sec) , antiderivative size = 507, normalized size of antiderivative = 0.90, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.465, Rules used = {3042, 4709, 3042, 3534, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\sin (c+d x) \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \int -\frac {-8 B a^5+b (24 A+7 C) a^4+5 b^2 B a^3-b^3 (33 A+C) a^2-3 b^4 B a+4 \left (2 (A-C) a^4+4 b B a^3-b^2 (10 A+C) a^2-b^3 B a+5 A b^4\right ) \cos (c+d x) a+15 A b^5+b \left ((8 A-5 C) a^4+9 b B a^3-b^2 (29 A+C) a^2-3 b^3 B a+15 A b^4\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {2 \sin (c+d x) \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right )}{a d \sqrt {\cos (c+d x)}}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\sin (c+d x) \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-8 B a^5+b (24 A+7 C) a^4+5 b^2 B a^3-b^3 (33 A+C) a^2-3 b^4 B a+4 \left (2 (A-C) a^4+4 b B a^3-b^2 (10 A+C) a^2-b^3 B a+5 A b^4\right ) \cos (c+d x) a+15 A b^5+b \left ((8 A-5 C) a^4+9 b B a^3-b^2 (29 A+C) a^2-3 b^3 B a+15 A b^4\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\sin (c+d x) \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-8 B a^5+b (24 A+7 C) a^4+5 b^2 B a^3-b^3 (33 A+C) a^2-3 b^4 B a+4 \left (2 (A-C) a^4+4 b B a^3-b^2 (10 A+C) a^2-b^3 B a+5 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+15 A b^5+b \left ((8 A-5 C) a^4+9 b B a^3-b^2 (29 A+C) a^2-3 b^3 B a+15 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((A*b^2 - a*(b*B - a*C))*Sin[c + d* 
x])/(2*a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^2) - (((5*A 
*b^4 + 7*a^3*b*B - a*b^3*B - 3*a^4*C - a^2*b^2*(11*A + 3*C))*Sin[c + d*x]) 
/(a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])) - (-(((2*(15*A* 
b^4 + 9*a^3*b*B - 3*a*b^3*B + a^4*(8*A - 5*C) - a^2*b^2*(29*A + C))*Ellipt 
icE[(c + d*x)/2, 2])/d + ((2*a*(5*A*b^4 + 7*a^3*b*B - a*b^3*B - 3*a^4*C - 
a^2*b^2*(11*A + 3*C))*EllipticF[(c + d*x)/2, 2])/d + (2*(15*A*b^6 - 15*a^5 
*b*B + 6*a^3*b^3*B - 3*a*b^5*B + 3*a^6*C - a^2*b^4*(38*A + C) + 5*a^4*b^2* 
(7*A + 2*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d))/b)/a) 
 + (2*(15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B + a^4*(8*A - 5*C) - a^2*b^2*(29*A 
+ C))*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]]))/(2*a*(a^2 - b^2)))/(4*a*(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 3534

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 3538

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

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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1998\) vs. \(2(537)=1074\).

Time = 8.06 (sec) , antiderivative size = 1999, normalized size of antiderivative = 3.56

method	result	size
default	\(\text {Expression too large to display}\)	\(1999\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+b*cos(d*x+c))^3,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*A/a^3/sin(1/ 
2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(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*cos(1/2*d*x+1/2*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)*EllipticE(cos(1/2*d*x+ 
1/2*c),2^(1/2)))-2*(A*b^2-C*a^2)/a^2/b*(-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*cos(1/2*d*x+1/ 
2*c)^2*b+a-b)-1/2/(a+b)/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)*Ellipt 
icF(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)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d* 
x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/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)*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)*(-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))+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)*(-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*b^2-B*a*b+C*a^2)/b/a*(-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...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sec(c + d*x))*cos(c + d*x)*sec(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)*b + int((sqrt 
(sec(c + d*x))*cos(c + d*x)**2*sec(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)*c + int((sqrt(sec(c 
 + d*x))*sec(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