\(\int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx\) [575]

Optimal result

Integrand size = 33, antiderivative size = 258 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {(A b-a B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b \left (a^2-b^2\right ) d}+\frac {\left (a A b+a^2 B-2 b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}-\frac {\left (a^2 A b+A b^3+a^3 B-3 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^2 (a+b)^2 d}-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \sec (c+d x))} \] Output:

(A*b-B*a)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c 
)^(1/2)/b/(a^2-b^2)/d+(A*a*b+B*a^2-2*B*b^2)*cos(d*x+c)^(1/2)*InverseJacobi 
AM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/b^2/(a^2-b^2)/d-(A*a^2*b+A*b^3+ 
B*a^3-3*B*a*b^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-b)/b^2/(a+b)^2/d-(A*b-B*a)*sec(d*x+c)^(1/2)*s 
in(d*x+c)/(a^2-b^2)/d/(b+a*sec(d*x+c))
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(626\) vs. \(2(258)=516\).

Time = 7.19 (sec) , antiderivative size = 626, normalized size of antiderivative = 2.43 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {\frac {2 (-A b+a B) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 (4 a A-4 b B) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {(A b-a B) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{4 (a-b) (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {(A b-a B) \sin (c+d x)}{b \left (-a^2+b^2\right )}+\frac {-a A b \sin (c+d x)+a^2 B \sin (c+d x)}{b \left (-a^2+b^2\right ) (a+b \cos (c+d x))}\right )}{d} \] Input:

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

Output:

((2*(-(A*b) + a*B)*Cos[c + d*x]^2*(EllipticF[ArcSin[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 - Co 
s[c + d*x]^2)) + (2*(4*a*A - 4*b*B)*Cos[c + d*x]^2*EllipticPi[-(a/b), ArcS 
in[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)) + ((A*b - a*B) 
*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]*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]* 
Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*a^2*EllipticPi[-(a/b), Arc 
Sin[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)))/(4*(a - b)*(a 
+ b)*d) + (Sqrt[Sec[c + d*x]]*(((A*b - a*B)*Sin[c + d*x])/(b*(-a^2 + b^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
 

Rubi [A] (verified)

Time = 1.76 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.96, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3439, 3042, 4515, 27, 3042, 4594, 3042, 4274, 3042, 4258, 3042, 3119, 3120, 4336, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3439

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4515

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4594

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 4336

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

(((2*b*(A*b - a*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c 
 + d*x]])/d + (2*(a*A*b + a^2*B - 2*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c 
 + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/b^2 - (2*(a^2*A*b + A*b^3 + a^3*B - 3 
*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqr 
t[Sec[c + d*x]])/(b^2*(a + b)*d))/(2*(a^2 - b^2)) - ((A*b - a*B)*Sqrt[Sec[ 
c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*d*(b + a*Sec[c + d*x]))
 

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 3439

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^(m + n)   Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(d + 
c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4336

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

rule 4515

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-d)*(A 
*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 
1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2))   Int[(a + b 
*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[d*(n - 1)*(A*b - a*B) 
+ d*(a*A - b*B)*(m + 1)*Csc[e + f*x] - d*(A*b - a*B)*(m + n + 1)*Csc[e + f* 
x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && 
 NeQ[a^2 - b^2, 0] && LtQ[m, -1] && LtQ[0, n, 1]
 

rule 4594

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2)   Int[(d*Csc[e + 
f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Simp[1/a^2   Int[(a*A - (A*b - a 
*B)*Csc[e + f*x])/Sqrt[d*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, 
B, C}, x] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(807\) vs. \(2(249)=498\).

Time = 6.85 (sec) , antiderivative size = 808, normalized size of antiderivative = 3.13

method	result	size
default	\(\text {Expression too large to display}\)	\(808\)

Input:

int((A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^2/sec(d*x+c)^(1/2),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*B/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))- 
4/b*(A*b-2*B*a)/(-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)*E 
llipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2*a*(A*b-B*a)/b^2*(-b^2/a 
/(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*b*cos(1/2*d*x+1/2*c)^2+a-b)-1/2/a/(a+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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*b/(a^2-b^2)/ 
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*b/(a^2-b^2)/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)*Ellipt 
icE(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*c 
os(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...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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