\(\int \frac {(A+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{5/2}} \, dx\) [1448]

Optimal result

Integrand size = 37, antiderivative size = 456 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {4 b \left (A b^2-a^2 (3 A+2 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {2 \left (2 A b^2+3 a b (A+C)-a^2 (3 A+C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 \sqrt {a+b} \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {4 b \left (A b^2-a^2 (3 A+2 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \] Output:

-4/3*b*(A*b^2-a^2*(3*A+2*C))*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticE((a+b*co 
s(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-s 
ec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/(a-b)/(a+b)^(3/ 
2)/d/sec(d*x+c)^(1/2)-2/3*(2*A*b^2+3*a*b*(A+C)-a^2*(3*A+C))*cos(d*x+c)^(1/ 
2)*csc(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2 
),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/( 
a-b))^(1/2)/a^2/(a+b)^(1/2)/(a^2-b^2)/d/sec(d*x+c)^(1/2)+2/3*(A*b^2+C*a^2) 
*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2)+4/3*b*(A 
*b^2-a^2*(3*A+2*C))*sec(d*x+c)^(1/2)*sin(d*x+c)/a/(a^2-b^2)^2/d/(a+b*cos(d 
*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 15.43 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.19 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {4 b \left (3 a^2 A-A b^2+2 a^2 C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2}+\frac {2 \left (A b^2 \sin (c+d x)+a^2 C \sin (c+d x)\right )}{3 b \left (-a^2+b^2\right ) (a+b \cos (c+d x))^2}+\frac {2 \left (-5 a^2 A b^2 \sin (c+d x)+A b^4 \sin (c+d x)+a^4 C \sin (c+d x)-5 a^2 b^2 C \sin (c+d x)\right )}{3 a b \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}\right )}{d}+\frac {4 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 b (a+b) \left (A b^2-a^2 (3 A+2 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+a (a+b) \left (-2 A b^2+3 a b (A+C)+a^2 (3 A+C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )+b \left (A b^2-a^2 (3 A+2 C)\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 \left (a^3-a b^2\right )^2 d \sqrt {a+b \cos (c+d x)} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}} \] Input:

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

Output:

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((4*b*(3*a^2*A - A*b^2 + 2*a^ 
2*C)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)^2) + (2*(A*b^2*Sin[c + d*x] + a^2*C* 
Sin[c + d*x]))/(3*b*(-a^2 + b^2)*(a + b*Cos[c + d*x])^2) + (2*(-5*a^2*A*b^ 
2*Sin[c + d*x] + A*b^4*Sin[c + d*x] + a^4*C*Sin[c + d*x] - 5*a^2*b^2*C*Sin 
[c + d*x]))/(3*a*b*(a^2 - b^2)^2*(a + b*Cos[c + d*x]))))/d + (4*Sqrt[Cos[( 
c + d*x)/2]^2*Sec[c + d*x]]*(2*b*(a + b)*(A*b^2 - a^2*(3*A + 2*C))*Sqrt[Co 
s[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos 
[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + a*(a 
+ b)*(-2*A*b^2 + 3*a*b*(A + C) + a^2*(3*A + C))*Sqrt[Cos[c + d*x]/(1 + Cos 
[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Ellipt 
icF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + b*(A*b^2 - a^2*(3*A + 2* 
C))*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]) 
)/(3*(a^3 - a*b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2])
 

Rubi [A] (verified)

Time = 1.92 (sec) , antiderivative size = 452, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {3042, 4709, 3042, 3535, 27, 3042, 3472, 3042, 3477, 3042, 3295, 3473}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3472

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {2 b \left (A b^2-a^2 (3 A+2 C)\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (-\left (a^2 (3 A+C)\right )+3 a b (A+C)+2 A b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a^2-b^2}-\frac {4 b \left (A b^2-a^2 (3 A+2 C)\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3473

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(A*b^2 + a^2*C)*Sqrt[Cos[c + d*x 
]]*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) - (((4*(a 
- b)*b*Sqrt[a + b]*(A*b^2 - a^2*(3*A + 2*C))*Cot[c + d*x]*EllipticE[ArcSin 
[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a 
- b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a 
 - b)])/(a^2*d) + (2*(a - b)*Sqrt[a + b]*(2*A*b^2 + 3*a*b*(A + C) - a^2*(3 
*A + C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + 
b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/( 
a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d))/(a^2 - b^2) - (4*b*(A 
*b^2 - a^2*(3*A + 2*C))*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sq 
rt[a + b*Cos[c + d*x]]))/(3*a*(a^2 - b^2)))
 

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 3295

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 3472

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(d_.)*sin[(e_.) + (f_.)*( 
x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)), x_Symbol] :> Simp[2*(A 
*b - a*B)*(Cos[e + f*x]/(f*(a^2 - b^2)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[d*Sin[ 
e + f*x]])), x] + Simp[d/(a^2 - b^2)   Int[(A*b - a*B + (a*A - b*B)*Sin[e + 
 f*x])/(Sqrt[a + b*Sin[e + f*x]]*(d*Sin[e + f*x])^(3/2)), x], x] /; FreeQ[{ 
a, b, d, e, f, A, B}, x] && NeQ[a^2 - b^2, 0]
 

rule 3473

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) 
^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* 
(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] 
)/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + 
d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], 
x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && 
PosQ[(c + d)/b]
 

rule 3477

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S 
imp[(A - B)/(a - b)   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* 
x]]), x], x] - Simp[(A*b - a*B)/(a - b)   Int[(1 + Sin[e + f*x])/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e 
, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && NeQ[A, B]
 

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 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. \(2195\) vs. \(2(410)=820\).

Time = 26.32 (sec) , antiderivative size = 2196, normalized size of antiderivative = 4.82

method	result	size
default	\(\text {Expression too large to display}\)	\(2196\)
parts	\(\text {Expression too large to display}\)	\(2250\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(5/2),x,method=_R 
ETURNVERBOSE)
 

Output:

-2/3/d/(a-b)^2/(a+b)^2/a^2*((4*cos(d*x+c)^3+11*cos(d*x+c)^2+10*cos(d*x+c)+ 
3)*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d* 
x+c)))^(1/2)*a^3*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2) 
)+(-6*cos(d*x+c)^2-12*cos(d*x+c)-6)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1 
/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^4*b*EllipticE(-csc(d*x+c)+ 
cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(-6*cos(d*x+c)^3-18*cos(d*x+c)^2-18*cos(d 
*x+c)-6)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+ 
cos(d*x+c)))^(1/2)*a^3*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b)) 
^(1/2))+(-6*cos(d*x+c)^3-10*cos(d*x+c)^2-2*cos(d*x+c)+2)*A*(cos(d*x+c)/(1+ 
cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b^3 
*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(2*cos(d*x+c)^3+6* 
cos(d*x+c)^2+6*cos(d*x+c)+2)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)* 
(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b^4*EllipticE(-csc(d*x+c)+cot(d*x 
+c),(-(a-b)/(a+b))^(1/2))+C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+ 
b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b^3*EllipticE(-csc(d*x+c)+cot(d*x+ 
c),(-(a-b)/(a+b))^(1/2))*(-4*cos(d*x+c)^3-8*cos(d*x+c)^2-4*cos(d*x+c))+A*( 
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c))) 
^(1/2)*a*b^4*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-2*co 
s(d*x+c)^3-4*cos(d*x+c)^2-2*cos(d*x+c))+C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2 
)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b^3*EllipticF(-cs...
 

Fricas [F]

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

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(5/2),x, al 
gorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(5/2),x, al 
gorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(5/2),x, al 
gorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{5/2}} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) a \] Input:

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

Output:

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