\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx\) [1391]

Optimal result

Integrand size = 35, antiderivative size = 266 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 \left (5 A b^2+a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^3 d}-\frac {2 A b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {2 b \left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^3 (a+b) d}+\frac {2 \left (5 A b^2+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a^3 d}-\frac {2 A b \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(642\) vs. \(2(266)=532\).

Time = 9.34 (sec) , antiderivative size = 642, normalized size of antiderivative = 2.41 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {2 \left (19 a^2 A b+45 A b^3+45 a^2 b C\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (18 a^3 A+40 a A b^2+30 a^3 C\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (9 a^2 A b+15 A b^3+15 a^2 b C\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{30 a^3 d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {2 \left (3 a^2 A+5 A b^2+5 a^2 C\right ) \sin (c+d x)}{5 a^3}-\frac {2 A b \tan (c+d x)}{3 a^2}+\frac {2 A \sec (c+d x) \tan (c+d x)}{5 a}\right )}{d} \] Input:

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2))/(a + b*Cos[c + d*x]) 
,x]
 

Output:

-1/30*((2*(19*a^2*A*b + 45*A*b^3 + 45*a^2*b*C)*Cos[c + d*x]^2*(EllipticF[A 
rcSin[Sqrt[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*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(18*a^3*A + 40*a*A*b^2 + 3 
0*a^3*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)) + ((9*a^2*A*b + 15*A*b^3 + 15*a^2*b*C)*Co 
s[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[Sqrt[Sec[c + d*x]]], -1]*Sqr 
t[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 - C 
os[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))/(a^3*d) + (Sqrt[S 
ec[c + d*x]]*((2*(3*a^2*A + 5*A*b^2 + 5*a^2*C)*Sin[c + d*x])/(5*a^3) - (2* 
A*b*Tan[c + d*x])/(3*a^2) + (2*A*Sec[c + d*x]*Tan[c + d*x])/(5*a)))/d
 

Rubi [A] (verified)

Time = 2.25 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.92, 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, 27, 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.

\(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^{7/2} \left (A+C \cos (c+d x)^2\right )}{a+b \cos (c+d x)}dx\)

\(\Big \downarrow \) 4709

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+A}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {3 b \left ((3 A+5 C) a^2+5 A b^2\right ) \cos ^2(c+d x)+a \left (3 (3 A+5 C) a^2+20 A b^2\right ) \cos (c+d x)+5 b \left ((A+3 C) a^2+3 A b^2\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {3 b \left ((3 A+5 C) a^2+5 A b^2\right ) \cos ^2(c+d x)+a \left (3 (3 A+5 C) a^2+20 A b^2\right ) \cos (c+d x)+5 b \left ((A+3 C) a^2+3 A b^2\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {3 b \left ((3 A+5 C) a^2+5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left (3 (3 A+5 C) a^2+20 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 b \left ((A+3 C) a^2+3 A b^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \int \frac {a A \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+\left ((A+3 C) a^2+3 A b^2\right ) b^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3481

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx+a A b^2 \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )}{b}+\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+a A b^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{b}+\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 a A b^2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )}{b}+\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3284

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {5 \left (\frac {6 b^2 \left (a^2 C+A b^2\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}+\frac {2 a A b^2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )}{b}}{a}}{3 a}}{5 a}\right )\)

Input:

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2))/(a + b*Cos[c + d*x]),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sin[c + d*x])/(5*a*d*Cos[c + d 
*x]^(5/2)) - ((10*A*b*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)) - (-(((6*(5 
*A*b^2 + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/d + (5*((2*a*A*b^2*El 
lipticF[(c + d*x)/2, 2])/d + (6*b^2*(A*b^2 + a^2*C)*EllipticPi[(2*b)/(a + 
b), (c + d*x)/2, 2])/((a + b)*d)))/b)/a) + (6*(5*A*b^2 + a^2*(3*A + 5*C))* 
Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]]))/(3*a))/(5*a))
 

Defintions of rubi rules used

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. \(758\) vs. \(2(243)=486\).

Time = 220.16 (sec) , antiderivative size = 759, normalized size of antiderivative = 2.85

method	result	size
default	\(\text {Expression too large to display}\)	\(759\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x,method=_RETURNV 
ERBOSE)
                                                                                    
                                                                                    
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/5*A/a/(8*sin 
(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 
2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(sin(1/2*d*x 
+1/2*c)^2)^(1/2)*EllipticE(cos(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)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c 
)^4+12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(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+8*sin(1/2*d*x+1/2*c) 
^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x 
+1/2*c),2^(1/2))*(2*sin(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)+2*(A*b^2+C*a^2)/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) 
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))-2 
*A/a^2*b*(-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)/(cos(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+4*b^2*(A*b^2+C*a^2)/a 
^3/(-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)))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x, algorith 
m="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2)/(a+b*cos(d*x+c)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x, algorith 
m="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(7/2)/(b*cos(d*x + c) + a), 
x)
 

Giac [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x, algorith 
m="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(7/2)/(b*cos(d*x + c) + a), 
x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{a+b\,\cos \left (c+d\,x\right )} \,d x \] Input:

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2))/(a + b*cos(c + d*x)),x 
)
 

Output:

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2))/(a + b*cos(c + d*x)), 
x)
 

Reduce [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right ) b +a}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right ) b +a}d x \right ) a \] Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x)
 

Output:

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