\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [1398]

Optimal result

Integrand size = 35, antiderivative size = 396 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {b \left (5 A b^2-a^2 (4 A-C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (5 A b^2-a^2 (2 A-3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 \left (a^2-b^2\right ) d}-\frac {\left (5 A b^4-a^2 b^2 (7 A-C)-3 a^4 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^3 (a-b) (a+b)^2 d}+\frac {b \left (5 A b^2-a^2 (4 A-C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (5 A b^2-a^2 (2 A-3 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

-b*(5*A*b^2-a^2*(4*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)/d-1/3*(5*A*b^2-a^2*(2*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/(a^2 
-b^2)/d-(5*A*b^4-a^2*b^2*(7*A-C)-3*a^4*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)/(a+b)^2/d+b*( 
5*A*b^2-a^2*(4*A-C))*sec(d*x+c)^(1/2)*sin(d*x+c)/a^3/(a^2-b^2)/d-1/3*(5*A* 
b^2-a^2*(2*A-3*C))*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/(a^2-b^2)/d+(A*b^2+C*a^ 
2)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 9.87 (sec) , antiderivative size = 718, normalized size of antiderivative = 1.81 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 \left (-4 a^4 A-44 a^2 A b^2+45 A b^4-12 a^4 C+9 a^2 b^2 C\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (-28 a^3 A b+40 a A b^3+12 a^3 b C\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (-12 a^2 A b^2+15 A b^4+3 a^2 b^2 C\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{12 a^3 (-a+b) (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (-\frac {b \left (4 a^2 A-5 A b^2-a^2 C\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right )}+\frac {-A b^3 \sin (c+d x)-a^2 b C \sin (c+d x)}{a^2 \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {2 A \tan (c+d x)}{3 a^2}\right )}{d} \] Input:

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/(a + b*Cos[c + d*x]) 
^2,x]
 

Output:

((2*(-4*a^4*A - 44*a^2*A*b^2 + 45*A*b^4 - 12*a^4*C + 9*a^2*b^2*C)*Cos[c + 
d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-(a/b), Arc 
Sin[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*(-28*a^ 
3*A*b + 40*a*A*b^3 + 12*a^3*b*C)*Cos[c + d*x]^2*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])/(b*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((-12*a^2*A*b^2 
 + 15*A*b^4 + 3*a^2*b^2*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]*Sq 
rt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*(2*a - b)*b*EllipticF[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)))/(12*a^3*(-a + b)*(a + b)*d) + (Sqrt[Sec[c + d*x]]*(-((b*(4*a^2 
*A - 5*A*b^2 - a^2*C)*Sin[c + d*x])/(a^3*(a^2 - b^2))) + (-(A*b^3*Sin[c + 
d*x]) - a^2*b*C*Sin[c + d*x])/(a^2*(a^2 - b^2)*(a + b*Cos[c + d*x])) + (2* 
A*Tan[c + d*x])/(3*a^2)))/d
 

Rubi [A] (verified)

Time = 2.90 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.85, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4709, 3042, 3535, 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.

\(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^{5/2} \left (A+C \cos (c+d x)^2\right )}{(a+b \cos (c+d x))^2}dx\)

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\frac {2 \left (5 A b^2-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {-2 (A+3 C) a^4-b^2 (16 A-3 C) a^2+2 b \left (10 A b^2-a^2 (7 A-3 C)\right ) \cos (c+d x) a+15 A b^4+3 b^2 \left (5 A b^2-a^2 (4 A-C)\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {6 b \left (5 A b^2-a^2 (4 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}}{3 a}}{2 a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\frac {2 \left (5 A b^2-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 b \left (5 A b^2-a^2 (4 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-2 (A+3 C) a^4-b^2 (16 A-3 C) a^2+2 b \left (10 A b^2-a^2 (7 A-3 C)\right ) \cos (c+d x) a+15 A b^4+3 b^2 \left (5 A b^2-a^2 (4 A-C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{3 a}}{2 a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\frac {2 \left (5 A b^2-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 b \left (5 A b^2-a^2 (4 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-2 (A+3 C) a^4-b^2 (16 A-3 C) a^2+2 b \left (10 A b^2-a^2 (7 A-3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+15 A b^4+3 b^2 \left (5 A b^2-a^2 (4 A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{3 a}}{2 a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3538

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\frac {2 \left (5 A b^2-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 b \left (5 A b^2-a^2 (4 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 b \left (5 A b^2-a^2 (4 A-C)\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {a \left (5 A b^2-a^2 (2 A-3 C)\right ) \cos (c+d x) b^2+\left (-2 (A+3 C) a^4-b^2 (16 A-3 C) a^2+15 A b^4\right ) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\frac {2 \left (5 A b^2-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 b \left (5 A b^2-a^2 (4 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 b \left (5 A b^2-a^2 (4 A-C)\right ) \int \sqrt {\cos (c+d x)}dx+\frac {\int \frac {a \left (5 A b^2-a^2 (2 A-3 C)\right ) \cos (c+d x) b^2+\left (-2 (A+3 C) a^4-b^2 (16 A-3 C) a^2+15 A b^4\right ) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\frac {2 \left (5 A b^2-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 b \left (5 A b^2-a^2 (4 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 b \left (5 A b^2-a^2 (4 A-C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\int \frac {a \left (5 A b^2-a^2 (2 A-3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (-2 (A+3 C) a^4-b^2 (16 A-3 C) a^2+15 A b^4\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/(a + b*Cos[c + d*x])^2,x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((A*b^2 + a^2*C)*Sin[c + d*x])/(a*( 
a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])) - ((2*(5*A*b^2 - a^2 
*(2*A - 3*C))*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)) - (-(((6*b*(5*A*b^2 
 - a^2*(4*A - C))*EllipticE[(c + d*x)/2, 2])/d + ((2*a*b*(5*A*b^2 - a^2*(2 
*A - 3*C))*EllipticF[(c + d*x)/2, 2])/d + (6*b*(5*A*b^4 - a^2*b^2*(7*A - C 
) - 3*a^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d))/b)/a) 
 + (6*b*(5*A*b^2 - a^2*(4*A - C))*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]]))/ 
(3*a))/(2*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 3535

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[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S 
in[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[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n 
+ 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d 
*(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, 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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(991\) vs. \(2(379)=758\).

Time = 98.61 (sec) , antiderivative size = 992, normalized size of antiderivative = 2.51

method	result	size
default	\(\text {Expression too large to display}\)	\(992\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x,method=_RETUR 
NVERBOSE)
 

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^2*(-1/6* 
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)/(c 
os(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(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)*E 
llipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(A*b^2+C*a^2)/a^2*(-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)*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)*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)))-4*A/a^3*b/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*...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algori 
thm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**(5/2)/(a+b*cos(d*x+c))**2,x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algori 
thm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(5/2)/(b*cos(d*x + c) + a)^2 
, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2))/(a + b*cos(c + d*x))^2 
,x)
 

Output:

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2))/(a + b*cos(c + d*x))^2 
, x)
 

Reduce [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) a \] Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x)
 

Output:

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