\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [1441]

Optimal result

Integrand size = 37, antiderivative size = 432 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 b \left (8 A b^2-a^2 (5 A-3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {2 \left (6 a A b+8 A b^2+a^2 (A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^2+a^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (4 A b^2-a^2 (A-3 C)\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d} \] Output:

2/3*b*(8*A*b^2-a^2*(5*A-3*C))*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticE((a+b*c 
os(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^4/(a+b)^(1/2)/d/ 
sec(d*x+c)^(1/2)+2/3*(6*a*A*b+8*A*b^2+a^2*(A+3*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^3/(a+b)^(1/2)/d/sec(d*x+c)^(1/2)+2*(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))^(1/2)-2/3*(4*A*b^2-a^2*(A-3*C))*(a+ 
b*cos(d*x+c))^(1/2)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/(a^2-b^2)/d
 

Mathematica [A] (warning: unable to verify)

Time = 15.80 (sec) , antiderivative size = 472, normalized size of antiderivative = 1.09 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 b (a+b) \left (8 A b^2+a^2 (-5 A+3 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )-2 a (a+b) \left (-6 a A b+8 A b^2+a^2 (A+3 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )+b \left (8 A b^2+a^2 (-5 A+3 C)\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {2 b \left (5 a^2 A-8 A b^2-3 a^2 C\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )}-\frac {2 \left (A b^3 \sin (c+d x)+a^2 b C \sin (c+d x)\right )}{a^2 \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {2 A \tan (c+d x)}{3 a^2}\right )}{d} \] Input:

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/(a + b*Cos[c + d*x]) 
^(3/2),x]
 

Output:

(-2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*b*(a + b)*(8*A*b^2 + a^2*(-5* 
A + 3*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]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b) 
/(a + b)] - 2*a*(a + b)*(-6*a*A*b + 8*A*b^2 + a^2*(A + 3*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]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + b*(8*A*b^2 
+ a^2*(-5*A + 3*C))*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*T 
an[(c + d*x)/2]))/(3*a^3*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[( 
c + d*x)/2]^2]) + (Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*b*(5*a 
^2*A - 8*A*b^2 - 3*a^2*C)*Sin[c + d*x])/(3*a^3*(a^2 - b^2)) - (2*(A*b^3*Si 
n[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 = 1.90 (sec) , antiderivative size = 433, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 4709, 3042, 3535, 27, 3042, 3534, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (4 A b^2-a^2 (A-3 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {b \left (8 A b^2-a^2 (5 A-3 C)\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+(a-b) \left (a^2 (A+3 C)+6 a A b+8 A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{3 a}}{a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (4 A b^2-a^2 (A-3 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(a-b) \left (a^2 (A+3 C)+6 a A b+8 A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (8 A b^2-a^2 (5 A-3 C)\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}}{a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (4 A b^2-a^2 (A-3 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {b \left (8 A b^2-a^2 (5 A-3 C)\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (a^2 (A+3 C)+6 a A b+8 A b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{3 a}}{a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3473

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (4 A b^2-a^2 (A-3 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (a-b) \sqrt {a+b} \left (a^2 (A+3 C)+6 a A b+8 A b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}+\frac {2 b (a-b) \sqrt {a+b} \left (8 A b^2-a^2 (5 A-3 C)\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}}{3 a}}{a \left (a^2-b^2\right )}\right )\)

Input:

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/(a + b*Cos[c + d*x])^(3/2) 
,x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(A*b^2 + a^2*C)*Sin[c + d*x])/(a 
*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*Sqrt[a + b*Cos[c + d*x]]) - (-1/3*((2*(a 
 - b)*b*Sqrt[a + b]*(8*A*b^2 - a^2*(5*A - 3*C))*Cot[c + d*x]*EllipticE[Arc 
Sin[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]*(6*a*A*b + 8*A*b^2 + a^2*(A + 
3*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*(4*A*b^2 - a^2*(A 
 - 3*C))*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)) 
)/(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 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 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 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. \(1593\) vs. \(2(388)=776\).

Time = 37.34 (sec) , antiderivative size = 1594, normalized size of antiderivative = 3.69

method	result	size
default	\(\text {Expression too large to display}\)	\(1594\)
parts	\(\text {Expression too large to display}\)	\(1600\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(3/2),x,method=_R 
ETURNVERBOSE)
 

Output:

-2/3/d/(a+b)/(a-b)/a^3*(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*EllipticE(-csc(d*x+c)+cot(d*x+c),( 
-(a-b)/(a+b))^(1/2))*(5*cos(d*x+c)^3+10*cos(d*x+c)^2+5*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^2*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(5*cos(d 
*x+c)^3+10*cos(d*x+c)^2+5*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^3*EllipticE(-csc(d*x+c 
)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-8*cos(d*x+c)^3-16*cos(d*x+c)^2-8*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+c 
os(d*x+c)))^(1/2)*b^4*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2 
))*(-8*cos(d*x+c)^3-16*cos(d*x+c)^2-8*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^3*b*Elliptic 
E(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-3*cos(d*x+c)^3-6*cos(d*x+ 
c)^2-3*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^2*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a 
-b)/(a+b))^(1/2))*(-3*cos(d*x+c)^3-6*cos(d*x+c)^2-3*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^4*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)^3+2 
*cos(d*x+c)^2+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^3*b*EllipticF(-csc(d*x+c)+cot(d*x...
 

Fricas [F]

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

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(3/2),x, al 
gorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**(5/2)/(a+b*cos(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(3/2),x, al 
gorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(5/2)/(b*cos(d*x + c) + a)^( 
3/2), x)
 

Giac [F]

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

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(3/2),x, al 
gorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(5/2)/(b*cos(d*x + c) + a)^( 
3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2))/(a + b*cos(c + d*x))^( 
3/2),x)
 

Output:

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2))/(a + b*cos(c + d*x))^( 
3/2), x)
 

Reduce [F]

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

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(3/2),x)
 

Output:

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