\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx\) [1492]

Optimal result

Integrand size = 43, antiderivative size = 311 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {\left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}+\frac {\left (a^2 b B-2 b^3 B-3 a^3 C+a b^2 (A+4 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (A b^4+a^3 b B-3 a b^3 B-3 a^4 C+a^2 b^2 (A+5 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \] Output:

(A*b^2-B*a*b+3*C*a^2-2*C*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+(B*a^2*b-2*B*b^3-3*a^3*C+a*b^2 
*(A+4*C))*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*b^4+B*a^3*b-3*B*a*b^3-3*a^4*C+a^2*b^2*(A+5*C)) 
*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^3/(a+b)^2/d-(A*b^2-a*(B*b-C*a))*sin(d*x+c)/b/(a^2-b^2)/d 
/(a+b*cos(d*x+c))/sec(d*x+c)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 6.83 (sec) , antiderivative size = 482, normalized size of antiderivative = 1.55 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {\frac {4 b^2 \left (A b^2+a (-b B+a C)\right ) \sin (c+d x)}{-a^2+b^2}-\frac {\cos (c+d x) \cot (c+d x) (b+a \sec (c+d x)) \left (2 b^2 \left (A b^2-a b B-a^2 C+2 b^2 C\right ) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-8 a b^2 (-b B+a (A+C)) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+\left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \left (4 a b-4 a b \sec ^2(c+d x)+4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )\right )}{a (a-b) (a+b)}}{4 b^3 d (a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \] Input:

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

Output:

((4*b^2*(A*b^2 + a*(-(b*B) + a*C))*Sin[c + d*x])/(-a^2 + b^2) - (Cos[c + d 
*x]*Cot[c + d*x]*(b + a*Sec[c + d*x])*(2*b^2*(A*b^2 - a*b*B - a^2*C + 2*b^ 
2*C)*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-(a/b), ArcSi 
n[Sqrt[Sec[c + d*x]]], -1])*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 8*a 
*b^2*(-(b*B) + a*(A + C))*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], - 
1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + (A*b^2 - a*b*B + 3*a^2*C - 2 
*b^2*C)*(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[-Tan[c + d*x]^2] - 2*(2*a - b)*b*Ell 
ipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d* 
x]^2] + 4*a^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[ 
c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 2*b^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec 
[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2])))/(a*(a - b)*(a 
 + b)))/(4*b^3*d*(a + b*Cos[c + d*x])*Sqrt[Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 1.80 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.86, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4709, 3042, 3526, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3481

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

rule 3526

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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - 
d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^(m - 
 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* 
d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 
1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x 
] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f 
*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d 
, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 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. \(861\) vs. \(2(302)=604\).

Time = 4.08 (sec) , antiderivative size = 862, normalized size of antiderivative = 2.77

method	result	size
default	\(\text {Expression too large to display}\)	\(862\)

Input:

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

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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 
/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(B*b*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) 
)-2*C*a*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-C*b*EllipticE(cos(1/2*d*x+1/ 
2*c),2^(1/2)))-4/b^2*(A*b^2-2*B*a*b+3*C*a^2)/(-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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/ 
2))-2*a*(A*b^2-B*a*b+C*a^2)/b^3*(-1/a*b^2/(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*cos(1/2*d*x+1/2*c)^2* 
b+a-b)-1/2/(a+b)/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/a*b/(a^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/a*b/(a^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)*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)*El 
lipticPi(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)/(...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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