\(\int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [1396]

Optimal result
Mathematica [B] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 241 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \left (5 a^2 C+b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^3 d}-\frac {2 a \left (3 A b^2+\left (3 a^2+b^2\right ) C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^4 d}+\frac {2 a^2 \left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^4 (a+b) d}+\frac {2 C \sin (c+d x)}{5 b d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 a C \sin (c+d x)}{3 b^2 d \sqrt {\sec (c+d x)}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(597\) vs. \(2(241)=482\).

Time = 7.97 (sec) , antiderivative size = 597, normalized size of antiderivative = 2.48 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\frac {2 \left (15 A b^2+5 a^2 C+9 b^2 C\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {16 a C \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{(a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (15 A b^2+15 a^2 C+9 b^2 C\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{30 b^2 d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {C \sin (c+d x)}{10 b}-\frac {a C \sin (2 (c+d x))}{3 b^2}+\frac {C \sin (3 (c+d x))}{10 b}\right )}{d} \] Input:

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

Output:

((2*(15*A*b^2 + 5*a^2*C + 9*b^2*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[S 
ec[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)) + (16*a*C*Cos[c + d*x]^2*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 + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((15*A*b^ 
2 + 15*a^2*C + 9*b^2*C)*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[Sq 
rt[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]]*Sqr 
t[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)))/(30*b^2*d) + (Sqrt[Sec[c + d*x]]*((C*Sin[c + d*x])/(10*b) - (a*C* 
Sin[2*(c + d*x)])/(3*b^2) + (C*Sin[3*(c + d*x)])/(10*b)))/d
 

Rubi [A] (verified)

Time = 1.74 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.91, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4709, 3042, 3529, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3529

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

Int[(A + 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) + (-1/3*((-6*(5*a^2*C + b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 
 2])/(b*d) + (5*((2*a*(3*A*b^2 + (3*a^2 + b^2)*C)*EllipticF[(c + d*x)/2, 2 
])/(b*d) - (6*a^2*(A*b^2 + a^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2 
])/(b*(a + b)*d)))/b)/b - (10*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 3529
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* 
(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, 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. \(947\) vs. \(2(222)=444\).

Time = 5.75 (sec) , antiderivative size = 948, normalized size of antiderivative = 3.93

method result size
default \(\text {Expression too large to display}\) \(948\)

Input:

int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))/sec(d*x+c)^(3/2),x,method=_RETURNV 
ERBOSE)
                                                                                    
                                                                                    
 

Output:

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((-24*C*a*b^ 
3+24*C*b^4)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(-20*C*a^2*b^2+44*C*a* 
b^3-24*C*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(10*C*a^2*b^2-16*C*a 
*b^3+6*C*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+15*A*(sin(1/2*d*x+1/ 
2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2* 
c),-2*b/(a-b),2^(1/2))*a^2*b^2-15*A*a^2*b^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))+15*a 
*A*b^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)*Ellip 
ticF(cos(1/2*d*x+1/2*c),2^(1/2))-15*A*(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))*a*b^3+15*A 
*(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(c 
os(1/2*d*x+1/2*c),2^(1/2))*b^4+15*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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))* 
a^4-15*a^4*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) 
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+15*a^3*b*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) 
)-5*C*a^2*b^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))+5*C*a*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) 
)-15*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)*El...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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