\(\int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx\) [758]

Optimal result

Integrand size = 37, antiderivative size = 308 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \left (2 A b^2-a^2 (A-C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a^3 \sqrt {a+b} d}-\frac {2 (2 A b+a (A-C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a^2 \sqrt {a+b} d}+\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \] Output:

-2*(2*A*b^2-a^2*(A-C))*cot(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-2*(2*A*b+a*(A-C))*cot(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-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/ 
2)/a^2/(a+b)^(1/2)/d+2*(A*b^2+C*a^2)*sin(d*x+c)/a/(a^2-b^2)/d/cos(d*x+c)^( 
1/2)/(a+b*cos(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.05 (sec) , antiderivative size = 1269, normalized size of antiderivative = 4.12 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

((-4*a*(2*a^2*A*b - 2*A*b^3)*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*S 
qrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d 
*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos 
[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x) 
/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - 4*a*(a^3*A 
- 2*a*A*b^2 - a^3*C)*((Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-( 
((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*C 
sc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d 
*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4) 
/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Co 
t[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^ 
2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*Elli 
pticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sq 
rt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a 
+ b*Cos[c + d*x]])) + 2*(a^2*A*b - 2*A*b^3 - a^2*b*C)*((I*Cos[(c + d*x)/2] 
*Sqrt[a + b*Cos[c + d*x]]*EllipticE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos[c 
+ d*x]]], (-2*a)/(-a - b)]*Sec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c 
+ d*x]]*Sqrt[((a + b*Cos[c + d*x])*Sec[c + d*x])/(a + b)]) + (2*a*((a*Sqrt 
[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[( 
c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Cs...
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

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

\(\Big \downarrow \) 3473

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

Input:

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

Output:

-(((2*(a - b)*Sqrt[a + b]*(2*A*b^2 - a^2*(A - C))*Cot[c + d*x]*EllipticE[A 
rcSin[Sqrt[a + b*Cos[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^2*d) + (2*(a - b)*Sqrt[a + b]*(2*A*b + a*(A - C))*Cot[c + 
d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[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])))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1291\) vs. \(2(282)=564\).

Time = 13.18 (sec) , antiderivative size = 1292, normalized size of antiderivative = 4.19

method	result	size
default	\(\text {Expression too large to display}\)	\(1292\)
parts	\(\text {Expression too large to display}\)	\(1340\)

Input:

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

Output:

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

Fricas [F]

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

integrate((A+C*cos(d*x+c)^2)/cos(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)*sqrt(cos(d*x + c) 
)/(b^2*cos(d*x + c)^4 + 2*a*b*cos(d*x + c)^3 + a^2*cos(d*x + c)^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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