\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [1487]

Optimal result

Integrand size = 43, antiderivative size = 270 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^3 d}+\frac {2 \left (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^4 d}+\frac {2 a^2 \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^4 (a+b) d}+\frac {2 C \sin (c+d x)}{5 b d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (b B-a C) \sin (c+d x)}{3 b^2 d \sqrt {\sec (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 7.26 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.19 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4 b^2 (5 b B-5 a C+3 b C \cos (c+d x)) \sin (c+d x)+\frac {4 \cos (c+d x) \cot (c+d x) (b+a \sec (c+d x)) \left (3 b \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \tan ^2(c+d x)-3 b \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+b \left (15 A b^2+15 a^2 C-5 a b (3 B+C)+b^2 (5 B+9 C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-15 a \left (A b^2+a (-b B+a C)\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )}{a+b \cos (c+d x)}}{30 b^4 d \sqrt {\sec (c+d x)}} \] Input:

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

Output:

(4*b^2*(5*b*B - 5*a*C + 3*b*C*Cos[c + d*x])*Sin[c + d*x] + (4*Cos[c + d*x] 
*Cot[c + d*x]*(b + a*Sec[c + d*x])*(3*b*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b 
^2*C)*Tan[c + d*x]^2 - 3*b*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b^2*C)*Ellipti 
cE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2 
] + b*(15*A*b^2 + 15*a^2*C - 5*a*b*(3*B + C) + b^2*(5*B + 9*C))*EllipticF[ 
ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 
 15*a*(A*b^2 + a*(-(b*B) + 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*Cos[c + d*x])) 
/(30*b^4*d*Sqrt[Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 1.92 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.92, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4709, 3042, 3528, 27, 3042, 3528, 27, 3042, 3538, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int \frac {3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \cos ^2(c+d x)+b (5 b B+4 a C) \cos (c+d x)+5 a (b B-a C)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \cos ^2(c+d x)+b (5 b B+4 a C) \cos (c+d x)+5 a (b B-a C)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (5 b B+4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a (b B-a C)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\right )\)

\(\Big \downarrow \) 3538

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {5 \left (a b (b B-a C)+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}+\frac {5 \int \frac {a b (b B-a C)+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {5 \int \frac {a b (b B-a C)+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {5 \int \frac {a b (b B-a C)+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\right )\)

\(\Big \downarrow \) 3481

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {5 \left (\frac {3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {\left (-3 a^3 C+3 a^2 b B-a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {5 \left (\frac {3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {\left (-3 a^3 C+3 a^2 b B-a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {5 \left (\frac {3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-3 a^3 C+3 a^2 b B-a b^2 (3 A+C)+b^3 B\right )}{b d}\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\right )\)

\(\Big \downarrow \) 3284

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d}+\frac {5 \left (\frac {6 a^2 \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-3 a^3 C+3 a^2 b B-a b^2 (3 A+C)+b^3 B\right )}{b d}\right )}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*C*Cos[c + d*x]^(3/2)*Sin[c + d*x 
])/(5*b*d) + (((6*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b^2*C)*EllipticE[(c + d 
*x)/2, 2])/(b*d) + (5*((2*(3*a^2*b*B + b^3*B - 3*a^3*C - a*b^2*(3*A + C))* 
EllipticF[(c + d*x)/2, 2])/(b*d) + (6*a^2*(A*b^2 - a*(b*B - a*C))*Elliptic 
Pi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d)))/b)/(3*b) + (10*(b*B - a 
*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b*d))/(5*b))
 

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 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 3528

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)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*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] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

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. \(802\) vs. \(2(251)=502\).

Time = 3.82 (sec) , antiderivative size = 803, normalized size of antiderivative = 2.97

method	result	size
default	\(\text {Expression too large to display}\)	\(803\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))/sec(d*x+c)^(3/2),x,me 
thod=_RETURNVERBOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*a^2*(A*b^2- 
B*a*b+C*a^2)/b^3/(-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+A*b^3-B*a^2*b 
-B*a*b^2-B*b^3+C*a^3+C*a^2*b+C*a*b^2+C*b^3)/b^4*(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))-4/5*C/b/(-2*sin(1/2* 
d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x 
+1/2*c)^6-14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+6*sin(1/2*d*x+1/2*c)^ 
2*cos(1/2*d*x+1/2*c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^ 
2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*(sin(1/2*d*x+1/2*c)^2)^ 
(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2 
)))+2/b^3*(A*b^2-B*a*b-2*B*b^2+C*a^2+2*C*a*b+3*C*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))+EllipticF(co 
s(1/2*d*x+1/2*c),2^(1/2)))+4/3/b^2*(B*b-C*a-3*C*b)*(2*sin(1/2*d*x+1/2*c)^4 
*cos(1/2*d*x+1/2*c)-sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+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)*EllipticF(cos(1/2*d*x+1/ 
2*c),2^(1/2))-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 
/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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