\(\int \frac {(a+a \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [181]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [B] (verification not implemented)
Giac [F(-1)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 37, antiderivative size = 171 \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {a^{3/2} (24 A+11 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}+\frac {a^2 (24 A+19 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a C \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {C \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d} \] Output:

1/8*a^(3/2)*(24*A+11*C)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/ 
d+1/24*a^2*(24*A+19*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2 
)+1/4*a*C*cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d+1/3*C*cos(d 
*x+c)^(1/2)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66 \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 \sqrt {2} (24 A+11 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (24 A+37 C+22 C \cos (c+d x)+4 C \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \] Input:

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

Output:

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sqrt[2]*(24*A + 11*C)*Ar 
cSin[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(24*A + 37*C + 22*C* 
Cos[c + d*x] + 4*C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 3525, 27, 3042, 3455, 27, 3042, 3460, 3042, 3253, 223}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3525

\(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^{3/2} (a (6 A+C)+3 a C \cos (c+d x))}{2 \sqrt {\cos (c+d x)}}dx}{3 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^{3/2} (a (6 A+C)+3 a C \cos (c+d x))}{\sqrt {\cos (c+d x)}}dx}{6 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (6 A+C)+3 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{6 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3455

\(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((24 A+7 C) a^2+(24 A+19 C) \cos (c+d x) a^2\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {3 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{4} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((24 A+7 C) a^2+(24 A+19 C) \cos (c+d x) a^2\right )}{\sqrt {\cos (c+d x)}}dx+\frac {3 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((24 A+7 C) a^2+(24 A+19 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {3 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {\frac {1}{4} \left (\frac {3}{2} a^2 (24 A+11 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a^3 (24 A+19 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (\frac {3}{2} a^2 (24 A+11 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^3 (24 A+19 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3253

\(\displaystyle \frac {\frac {1}{4} \left (\frac {a^3 (24 A+19 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 a^2 (24 A+11 C) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {3 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\frac {3 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {3 a^{5/2} (24 A+11 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a^3 (24 A+19 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )}{6 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}\)

Input:

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

Output:

(C*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + ((3 
*a^2*C*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(2*d) + ( 
(3*a^(5/2)*(24*A + 11*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + 
d*x]]])/d + (a^3*(24*A + 19*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a 
+ a*Cos[c + d*x]]))/4)/(6*a)
 

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 223
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3253
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-2/f   Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co 
s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E 
qQ[a^2 - b^2, 0] && EqQ[d, a/b]
 

rule 3455
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + 
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 
 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1 
] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 3460
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] 
 /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - 
 b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]
 

rule 3525
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/(b*d*(m + n + 2))   Int[(a + b*Sin[e + f* 
x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1 
)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
 
Maple [A] (verified)

Time = 8.71 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.16

method result size
default \(\frac {a \sqrt {2}\, \left (72 A \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+33 C \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+24 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\left (8 \cos \left (d x +c \right )^{2}+22 \cos \left (d x +c \right )+33\right ) \sin \left (d x +c \right ) C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{24 d \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )}\) \(198\)
parts \(\frac {A a \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\frac {\sin \left (2 d x +2 c \right )}{2}+3 \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )\right )}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}}+\frac {C a \sqrt {2}\, \left (33 \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\left (8 \cos \left (d x +c \right )^{2}+22 \cos \left (d x +c \right )+33\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{24 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) \(244\)

Input:

int((a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x,method=_R 
ETURNVERBOSE)
 

Output:

1/24/d*a*2^(1/2)*(72*A*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c) 
)+33*C*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))+24*A*(cos(d*x+ 
c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+(8*cos(d*x+c)^2+22*cos(d*x+c)+33)*sin( 
d*x+c)*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^(1/2)*(a*cos(1/2*d* 
x+1/2*c)^2)^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.92 \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {{\left (8 \, C a \cos \left (d x + c\right )^{2} + 22 \, C a \cos \left (d x + c\right ) + 3 \, {\left (8 \, A + 11 \, C\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (24 \, A + 11 \, C\right )} a \cos \left (d x + c\right ) + {\left (24 \, A + 11 \, C\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, al 
gorithm="fricas")
 

Output:

1/24*((8*C*a*cos(d*x + c)^2 + 22*C*a*cos(d*x + c) + 3*(8*A + 11*C)*a)*sqrt 
(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((24*A + 11*C)*a* 
cos(d*x + c) + (24*A + 11*C)*a)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sq 
rt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) 
)/(d*cos(d*x + c) + d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2746 vs. \(2 (145) = 290\).

Time = 0.51 (sec) , antiderivative size = 2746, normalized size of antiderivative = 16.06 \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, al 
gorithm="maxima")
 

Output:

1/96*(24*(2*(a*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*si 
n(d*x + c) - (a*cos(d*x + c) - a)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2* 
d*x + 2*c) + 1)))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 
 2*c) + 1)^(1/4)*sqrt(a) + 3*(a*arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 
 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), 
 cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2* 
d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c 
)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d* 
x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x 
+ 2*c), cos(2*d*x + 2*c) + 1))) + 1) - a*arctan2(-(cos(2*d*x + 2*c)^2 + si 
n(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d* 
x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arcta 
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2* 
d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2 
(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) - a*arctan2((cos(2*d*x + 2*c 
)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 
 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), 
cos(2*d*x + 2*c) + 1)) + 1) + a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x...
 

Giac [F(-1)]

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

integrate((a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, al 
gorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a*(int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/cos(c + d*x),x) 
*a + int(sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x),x)*c + int 
(sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*c + int(sqrt 
(cos(c + d*x) + 1)*sqrt(cos(c + d*x)),x)*a)