\(\int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [576]

Optimal result

Integrand size = 33, antiderivative size = 284 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}+\frac {\left (a^2 A b-2 A b^3-3 a^3 B+4 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}-\frac {a \left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}+\frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \sec (c+d x))} \] Output:

-(A*a*b-3*B*a^2+2*B*b^2)*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^2/(a^2-b^2)/d+(A*a^2*b-2*A*b^3-3*B*a^3+4*B*a*b^2) 
*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/ 
b^3/(a^2-b^2)/d-a*(A*a^2*b-3*A*b^3-3*B*a^3+5*B*a*b^2)*cos(d*x+c)^(1/2)*Ell 
ipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))*sec(d*x+c)^(1/2)/(a-b)/b^3/( 
a+b)^2/d+a*(A*b-B*a)*sec(d*x+c)^(1/2)*sin(d*x+c)/b/(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. \(655\) vs. \(2(284)=568\).

Time = 7.40 (sec) , antiderivative size = 655, normalized size of antiderivative = 2.31 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\frac {2 \left (-a A b-a^2 B+2 b^2 B\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (4 A b^2-4 a b B\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{4 b (-a+b) (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (-\frac {a (-A b+a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right )}+\frac {a^2 A b \sin (c+d x)-a^3 B \sin (c+d x)}{b^2 \left (-a^2+b^2\right ) (a+b \cos (c+d x))}\right )}{d} \] Input:

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

Output:

((2*(-(a*A*b) - a^2*B + 2*b^2*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 - Cos[c + d*x]^2)) + (2*(4*A*b^2 - 4*a*b*B)*Cos[c + d*x]^2*Ellip 
ticPi[-(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)) + ((a*A*b - 3*a^2*B + 2*b^2*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*Elliptic 
F[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), 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[Se 
c[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*b*(-a + b)*(a + b)*d) + (Sqrt[Sec[c + d*x]]*(-((a*( 
-(A*b) + a*B)*Sin[c + d*x])/(b^2*(a^2 - b^2))) + (a^2*A*b*Sin[c + d*x] - a 
^3*B*Sin[c + d*x])/(b^2*(-a^2 + b^2)*(a + b*Cos[c + d*x]))))/d
 

Rubi [A] (verified)

Time = 1.96 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.97, 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, 4518, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3439

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4518

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4594

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {\frac {2 b \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {\frac {2 b \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4336

\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b^2}+\frac {\frac {2 b \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}+\frac {\frac {2 b \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3284

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

Input:

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

Output:

-1/2*(((2*b*(a*A*b - 3*a^2*B + 2*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + 
d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (2*(a^2*A*b - 2*A*b^3 - 3*a^3*B + 4*a*b 
^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/ 
b^2 + (2*a*(a^2*A*b - 3*A*b^3 - 3*a^3*B + 5*a*b^2*B)*Sqrt[Cos[c + d*x]]*El 
lipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^2*(a + b)*d 
))/(b*(a^2 - b^2)) + (a*(A*b - a*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b*(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 4518

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

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. \(848\) vs. \(2(275)=550\).

Time = 9.05 (sec) , antiderivative size = 849, normalized size of antiderivative = 2.99

method	result	size
default	\(\text {Expression too large to display}\)	\(849\)

Input:

int((A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^2/sec(d*x+c)^(3/2),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^3/(-2*sin( 
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)* 
(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(A*b*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) 
)-2*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-B*b*EllipticE(cos(1/2*d*x+1/ 
2*c),2^(1/2)))+2*a^2*(A*b-B*a)/b^3*(-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(co 
s(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)*EllipticE(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)*E 
llipticPi(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*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)))+4*a/b^2*(2*A*b-3*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/...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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