\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\) [1107]

Optimal result

Integrand size = 43, antiderivative size = 306 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {\left (3 A b^2-a b B-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2-a (b B-a C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a b \left (a^2-b^2\right ) d}+\frac {\left (3 A b^4+3 a^3 b B-a b^3 B-a^4 C-a^2 b^2 (5 A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 (a-b) b (a+b)^2 d}-\frac {\left (3 A b^2-a b B-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 5.52 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.15 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 \left (2 a A \left (a^2-b^2\right )+b \left (-3 A b^2+a b B+a^2 (2 A-C)\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (9 A b^3+4 a^3 B-3 a b^2 B-a^2 b (10 A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}-\frac {8 a \left (-2 A b^2+a b B+a^2 (A-C)\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{b (a+b)}-\frac {2 \left (-3 A b^2+a b B+a^2 (2 A-C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(-a+b) (a+b)}}{4 a^2 d} \] Input:

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

Output:

((4*(2*a*A*(a^2 - b^2) + b*(-3*A*b^2 + a*b*B + a^2*(2*A - C))*Cos[c + d*x] 
)*Sin[c + d*x])/((a^2 - b^2)*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])) - (( 
2*(9*A*b^3 + 4*a^3*B - 3*a*b^2*B - a^2*b*(10*A + C))*EllipticPi[(2*b)/(a + 
 b), (c + d*x)/2, 2])/(a + b) - (8*a*(-2*A*b^2 + a*b*B + a^2*(A - C))*((a 
+ b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 
2]))/(b*(a + b)) - (2*(-3*A*b^2 + a*b*B + a^2*(2*A - C))*(-2*a*b*EllipticE 
[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c 
 + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x 
]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2]))/((-a + b)*(a + b)))/(4 
*a^2*d)
 

Rubi [A] (verified)

Time = 2.20 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.93, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 3534, 27, 3042, 3534, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]] 
*(a + b*Cos[c + d*x])) - (-(((2*(3*A*b^2 - a*b*B - a^2*(2*A - C))*Elliptic 
E[(c + d*x)/2, 2])/d + ((2*a*(A*b^2 - a*(b*B - a*C))*EllipticF[(c + d*x)/2 
, 2])/d + (2*(3*A*b^4 + 3*a^3*b*B - a*b^3*B - a^4*C - a^2*b^2*(5*A + C))*E 
llipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d))/b)/a) + (2*(3*A*b^2 
 - a*b*B - a^2*(2*A - C))*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]]))/(2*a*(a^ 
2 - b^2))
 

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 3534

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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(874\) vs. \(2(307)=614\).

Time = 5.99 (sec) , antiderivative size = 875, normalized size of antiderivative = 2.86

method	result	size
default	\(\text {Expression too large to display}\)	\(875\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^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*A/a^2/sin(1/ 
2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-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*cos(1/2*d*x+1/2*c)-(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)))+4*(A*b^2-C*a^2)/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 
*b^2-B*a*b+C*a^2)/b/a*(-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)*EllipticPi(c 
os(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*(si 
n(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...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4} b^{2}+2 \cos \left (d x +c \right )^{3} a b +\cos \left (d x +c \right )^{2} a^{2}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3} b^{2}+2 \cos \left (d x +c \right )^{2} a b +\cos \left (d x +c \right ) a^{2}}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) c \] Input:

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

Output:

int(sqrt(cos(c + d*x))/(cos(c + d*x)**4*b**2 + 2*cos(c + d*x)**3*a*b + cos 
(c + d*x)**2*a**2),x)*a + int(sqrt(cos(c + d*x))/(cos(c + d*x)**3*b**2 + 2 
*cos(c + d*x)**2*a*b + cos(c + d*x)*a**2),x)*b + int(sqrt(cos(c + d*x))/(c 
os(c + d*x)**2*b**2 + 2*cos(c + d*x)*a*b + a**2),x)*c