\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [1442]

Optimal result

Integrand size = 37, antiderivative size = 348 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \left (2 A b^2-a^2 (A-C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a^3 \sqrt {a+b} d \sqrt {\sec (c+d x)}}-\frac {2 (2 A b+a (A-C)) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a^2 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^2+a^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \] Output:

-2*(2*A*b^2-a^2*(A-C))*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticE((a+b*cos(d*x+ 
c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x 
+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/(a+b)^(1/2)/d/sec(d*x 
+c)^(1/2)-2*(2*A*b+a*(A-C))*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticF((a+b*cos 
(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-se 
c(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/(a+b)^(1/2)/d/se 
c(d*x+c)^(1/2)+2*(A*b^2+C*a^2)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/(a^2-b^2)/d/( 
a+b*cos(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 14.99 (sec) , antiderivative size = 456, normalized size of antiderivative = 1.31 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 A-2 A b^2-a^2 C\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right )}+\frac {2 \left (A b^2 \sin (c+d x)+a^2 C \sin (c+d x)\right )}{a \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )}{d}+\frac {2 \sqrt {2} \sqrt {\frac {\cos (c+d x)}{(1+\cos (c+d x))^2}} \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (-\left ((a+b) \left (\left (-2 A b^2+a^2 (A-C)\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+a (2 A b+a (-A+C)) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )\right ) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {(a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}} \sec (c+d x)\right )+\left (2 A b^2+a^2 (-A+C)\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right ) d \sqrt {\frac {1}{1+\cos (c+d x)}} \sqrt {a+b \cos (c+d x)} \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2}} \] Input:

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

Output:

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(a^2*A - 2*A*b^2 - a^2*C) 
*Sin[c + d*x])/(a^2*(a^2 - b^2)) + (2*(A*b^2*Sin[c + d*x] + a^2*C*Sin[c + 
d*x]))/(a*(a^2 - b^2)*(a + b*Cos[c + d*x]))))/d + (2*Sqrt[2]*Sqrt[Cos[c + 
d*x]/(1 + Cos[c + d*x])^2]*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*(Cos[(c + 
 d*x)/2]^2*Sec[c + d*x])^(3/2)*(-((a + b)*((-2*A*b^2 + a^2*(A - C))*Ellipt 
icE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + a*(2*A*b + a*(-A + C))*E 
llipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)])*(Cos[c + d*x]*Sec[(c 
 + d*x)/2]^2)^(3/2)*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b) 
]*Sec[c + d*x]) + (2*A*b^2 + a^2*(-A + C))*Cos[c + d*x]*(a + b*Cos[c + d*x 
])*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2]))/(a^2*(a^2 - b^2)*d*Sqrt[(1 + Cos[ 
c + d*x])^(-1)]*Sqrt[a + b*Cos[c + d*x]]*(Sec[(c + d*x)/2]^2)^(3/2))
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {3042, 4709, 3042, 3535, 27, 3042, 3477, 3042, 3295, 3473}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {\left (2 A b^2-a^2 (A-C)\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} (a (A-C)+2 A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3473

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 (a-b) \sqrt {a+b} \left (2 A b^2-a^2 (A-C)\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 (a-b) \sqrt {a+b} (a (A-C)+2 A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-(((2*(a - b)*Sqrt[a + b]*(2*A*b^2 
- a^2*(A - C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqr 
t[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d* 
x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a^2*d) + (2*(a - b)*S 
qrt[a + b]*(2*A*b + a*(A - C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Co 
s[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a 
*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d)) 
/(a*(a^2 - b^2))) + (2*(A*b^2 + a^2*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt 
[Cos[c + d*x]]*Sqrt[a + b*Cos[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 3295

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 3473

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) 
^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* 
(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] 
)/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + 
d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], 
x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && 
PosQ[(c + d)/b]
 

rule 3477

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S 
imp[(A - B)/(a - b)   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* 
x]]), x], x] - Simp[(A*b - a*B)/(a - b)   Int[(1 + Sin[e + f*x])/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e 
, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && NeQ[A, B]
 

rule 3535

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[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S 
in[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[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n 
+ 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d 
*(A*b^2 + a^2*C)*(m + n + 3)*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) || 
 !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 
 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. \(1297\) vs. \(2(314)=628\).

Time = 21.64 (sec) , antiderivative size = 1298, normalized size of antiderivative = 3.73

method	result	size
default	\(\text {Expression too large to display}\)	\(1298\)
parts	\(\text {Expression too large to display}\)	\(1352\)

Input:

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

Output:

-2/d/(a+b)/(a-b)/a^2*sec(d*x+c)^(3/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(( 
(1-cos(d*x+c))^3*csc(d*x+c)^3+csc(d*x+c)-cot(d*x+c))*a^3*A+(2*(1-cos(d*x+c 
))^3*csc(d*x+c)^3-2*csc(d*x+c)+2*cot(d*x+c))*b^3*A+(-(1-cos(d*x+c))^3*csc( 
d*x+c)^3+csc(d*x+c)-cot(d*x+c))*a^3*C+(-(1-cos(d*x+c))^3*csc(d*x+c)^3+csc( 
d*x+c)-cot(d*x+c))*b*a^2*A-2*A*a*b^2*(1-cos(d*x+c))^3*csc(d*x+c)^3+((1-cos 
(d*x+c))^3*csc(d*x+c)^3-csc(d*x+c)+cot(d*x+c))*b*a^2*C-2*A*EllipticF(-csc( 
d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1 
/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)+2*A*EllipticF(-csc(d*x 
+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b*(cos(d*x+c)/(1+cos(d*x+c)))^(1/ 
2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)+4*A*EllipticF(-csc(d*x+ 
c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2 
)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)+2*A*EllipticE(-csc(d*x+c 
)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*( 
1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)+2*A*EllipticE(-csc(d*x+c)+c 
ot(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1 
/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)-4*A*EllipticE(-csc(d*x+c)+co 
t(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/ 
(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)-4*A*EllipticE(-csc(d*x+c)+cot 
(d*x+c),(-(a-b)/(a+b))^(1/2))*b^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+ 
b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)+2*C*EllipticF(-csc(d*x+c)+cot...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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