\(\int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx\) [623]

Optimal result

Integrand size = 35, antiderivative size = 476 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx=-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a b \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a b \sqrt {a+b} d \sqrt {\sec (c+d x)}}-\frac {2 \sqrt {a+b} B \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^2 d \sqrt {\sec (c+d x)}}+\frac {2 a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \] Output:

-2*(A*b-B*a)*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticE((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/b/(a+b)^(1/2)/d/sec(d*x+c)^(1/2)+ 
2*(A*b-B*a)*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/b/(a+b)^(1/2)/d/sec(d*x+c)^(1/2)-2 
*(a+b)^(1/2)*B*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/ 
2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,(-(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)/b^2/d/sec(d*x+c)^(1/2)+2 
*a*(A*b-B*a)*sec(d*x+c)^(1/2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1 
/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1050\) vs. \(2(476)=952\).

Time = 14.53 (sec) , antiderivative size = 1050, normalized size of antiderivative = 2.21 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(A*b - a*B)*Sin[c + d*x]) 
/(b*(-a^2 + b^2)) - (2*(a*A*b*Sin[c + d*x] - a^2*B*Sin[c + d*x]))/(b*(-a^2 
 + b^2)*(a + b*Cos[c + d*x]))))/d + (2*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 
- b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(a*A*b*Tan[(c + d*x)/2] 
+ A*b^2*Tan[(c + d*x)/2] - a^2*B*Tan[(c + d*x)/2] - a*b*B*Tan[(c + d*x)/2] 
 - 2*A*b^2*Tan[(c + d*x)/2]^3 + 2*a*b*B*Tan[(c + d*x)/2]^3 - a*A*b*Tan[(c 
+ d*x)/2]^5 + A*b^2*Tan[(c + d*x)/2]^5 + a^2*B*Tan[(c + d*x)/2]^5 - a*b*B* 
Tan[(c + d*x)/2]^5 + 2*a^2*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a 
+ b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2 
]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*b^2*B*EllipticPi[-1, ArcSin[Tan[( 
c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + 
 a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a^2*B*EllipticP 
i[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[ 
1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d* 
x)/2]^2)/(a + b)] - 2*b^2*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + 
 b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + 
 a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - (a + b)*(-(A*b) + 
 a*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[( 
c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 
 - b*Tan[(c + d*x)/2]^2)/(a + b)] - b*(a + b)*(A - B)*EllipticF[ArcSin[...
 

Rubi [A] (verified)

Time = 1.72 (sec) , antiderivative size = 441, normalized size of antiderivative = 0.93, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 3440, 3042, 3471, 3042, 3273, 3042, 3274, 3042, 3288, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3440

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3471

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3273

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3288

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

\(\Big \downarrow \) 3295

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

\(\Big \downarrow \) 3473

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*Sqrt[a + b]*B*Cot[c + d*x]*Elli 
pticPi[(a + b)/b, 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)])/(b^2*d) + ((A*b - a*B)*(-(((2*(a - b)*Sqrt 
[a + b]*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*d) - (2*(a - b)*Sqrt[a + b 
]*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)) + (2*a*Sin[c + 
d*x])/((a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])))/b)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3273

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

rule 3274

Int[Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]/((a_.) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)])^(3/2), x_Symbol] :> Simp[(c - d)/(a - b)   Int[1/(Sqrt[a + b*Si 
n[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x], x] - Simp[(b*c - a*d)/(a - b) 
Int[(1 + Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]] 
), x], 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]
 

rule 3288

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c 
*((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti 
cPi[(c + d)/d, 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}, x] && NeQ[c^2 - 
 d^2, 0] && PosQ[(c + d)/b]
 

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 3440

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Sin[e + f*x])^m*((c + 
d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

rule 3471

Int[(((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)]])/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> S 
imp[B/b   Int[Sqrt[c + d*Sin[e + f*x]]/Sqrt[a + b*Sin[e + f*x]], x], x] + S 
imp[(A*b - a*B)/b   Int[Sqrt[c + d*Sin[e + f*x]]/(a + b*Sin[e + f*x])^(3/2) 
, 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]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1097\) vs. \(2(427)=854\).

Time = 8.88 (sec) , antiderivative size = 1098, normalized size of antiderivative = 2.31

method	result	size
default	\(\text {Expression too large to display}\)	\(1098\)
parts	\(\text {Expression too large to display}\)	\(1146\)

Input:

int((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

2/d/b/(a-b)/(a+b)*(a+b*cos(d*x+c))^(1/2)/(cos(d*x+c)^2*b+a*cos(d*x+c)+b*co 
s(d*x+c)+a)/sec(d*x+c)^(1/2)*(B*(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*EllipticPi(-csc(d*x+c)+cot(d*x 
+c),-1,(-(a-b)/(a+b))^(1/2))*(-2*cos(d*x+c)-4-2*sec(d*x+c))+B*(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)*b^2* 
EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*(2*cos(d*x+c)+4 
+2*sec(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*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b 
))^(1/2))*(-cos(d*x+c)-2-sec(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)*b^2*EllipticE(-csc(d*x+c)+c 
ot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-cos(d*x+c)-2-sec(d*x+c))+B*(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*E 
llipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)+2+sec(d* 
x+c))+B*(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*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2)) 
*(cos(d*x+c)+2+sec(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*EllipticF(-csc(d*x+c)+cot(d*x+c), 
(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)+2+sec(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)*b^2*EllipticF(-c 
sc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)+2+sec(d*x+c))+B*...
 

Fricas [F]

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

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algo 
rithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))**(3/2)/sec(d*x+c)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algo 
rithm="maxima")
 

Output:

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

Giac [F]

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

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algo 
rithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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