\(\int \frac {\cos (c+d x) (A+C \cos ^2(c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx\) [104]

Optimal result

Integrand size = 33, antiderivative size = 152 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 (15 A+14 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}-\frac {2 C \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 a d} \] Output:

-2^(1/2)*(A+C)*arctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^(1/ 
2))/a^(1/2)/d+2/15*(15*A+14*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/5*C*c 
os(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-2/15*C*(a+a*cos(d*x+c))^(1 
/2)*sin(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.57 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (-30 (A+C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 (30 A+29 C-2 C \cos (c+d x)+3 C \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{15 d \sqrt {a (1+\cos (c+d x))}} \] Input:

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

Output:

(Cos[(c + d*x)/2]*(-30*(A + C)*ArcTanh[Sin[(c + d*x)/2]] + 2*(30*A + 29*C 
- 2*C*Cos[c + d*x] + 3*C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(15*d*Sqrt[a 
*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.09, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3525, 27, 3042, 3447, 3042, 3502, 27, 3042, 3230, 3042, 3128, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3525

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {2 \int -\frac {a^2 C-a^2 (15 A+14 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {a^2 C-a^2 (15 A+14 C) \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {a^2 C-a^2 (15 A+14 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {-\frac {15 a^2 (A+C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx-\frac {2 a^2 (15 A+14 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {15 a^2 (A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^2 (15 A+14 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {-\frac {-\frac {30 a^2 (A+C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {2 a^2 (15 A+14 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {\frac {15 \sqrt {2} a^{3/2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 a^2 (15 A+14 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\)

Input:

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

Output:

(2*C*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]]) + ((-2*C* 
Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) - ((15*Sqrt[2]*a^(3/2)*(A + C 
)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d - 
(2*a^2*(15*A + 14*C)*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(3*a))/(5 
*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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3230

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( 
f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1))   Int[(a + b*Sin[e 
 + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] 
&& EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]
 

rule 3447

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

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -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 = 0.27 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.62

method	result	size
default	\(\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (24 C \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-20 C \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+30 A \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}-15 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a +30 C \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}-15 C \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{15 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(247\)
parts	\(\frac {A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (2 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}-\ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (24 \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-20 \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+30 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}-15 \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{15 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(305\)

Input:

int(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

1/15*cos(1/2*d*x+1/2*c)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(24*C*(sin( 
1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^4-20*C*a^(1/2)*(sin(1 
/2*d*x+1/2*c)^2*a)^(1/2)*sin(1/2*d*x+1/2*c)^2+30*A*a^(1/2)*(sin(1/2*d*x+1/ 
2*c)^2*a)^(1/2)-15*A*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^ 
2*a)^(1/2)+a))*a+30*C*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-15*C*ln(4/cos 
(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a)/a^(3/2)/sin 
(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.03 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {4 \, {\left (3 \, C \cos \left (d x + c\right )^{2} - C \cos \left (d x + c\right ) + 15 \, A + 13 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac {15 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right ) + {\left (A + C\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/30*(4*(3*C*cos(d*x + c)^2 - C*cos(d*x + c) + 15*A + 13*C)*sqrt(a*cos(d*x 
 + c) + a)*sin(d*x + c) + 15*sqrt(2)*((A + C)*a*cos(d*x + c) + (A + C)*a)* 
log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqr 
t(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a)) 
/(a*d*cos(d*x + c) + a*d)
 

Sympy [F]

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

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

Output:

Integral((A + C*cos(c + d*x)**2)*cos(c + d*x)/sqrt(a*(cos(c + d*x) + 1)), 
x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 927467 vs. \(2 (131) = 262\).

Time = 26.42 (sec) , antiderivative size = 927467, normalized size of antiderivative = 6101.76 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/1680*((84*(sqrt(2)*cos(3/2*d*x + 3/2*c)^2*sin(d*x + c) + 2*sqrt(2)*cos( 
3/2*d*x + 3/2*c)*cos(1/2*d*x + 1/2*c)*sin(d*x + c) + sqrt(2)*sin(3/2*d*x + 
 3/2*c)^2*sin(d*x + c) + 2*sqrt(2)*sin(3/2*d*x + 3/2*c)*sin(d*x + c)*sin(1 
/2*d*x + 1/2*c) + (sqrt(2)*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*sin(1/2*d*x + 
1/2*c)^2)*sin(d*x + c))*cos(7/2*d*x + 7/2*c)^3 - 84*((sqrt(2)*cos(d*x + c) 
 + sqrt(2))*cos(3/2*d*x + 3/2*c)^2 + (sqrt(2)*cos(d*x + c) + sqrt(2))*sin( 
3/2*d*x + 3/2*c)^2 + sqrt(2)*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*sin(1/2*d*x 
+ 1/2*c)^2 + 2*(sqrt(2)*cos(d*x + c)*cos(1/2*d*x + 1/2*c) + sqrt(2)*cos(1/ 
2*d*x + 1/2*c))*cos(3/2*d*x + 3/2*c) + (sqrt(2)*cos(1/2*d*x + 1/2*c)^2 + s 
qrt(2)*sin(1/2*d*x + 1/2*c)^2)*cos(d*x + c) + 2*(sqrt(2)*cos(d*x + c)*sin( 
1/2*d*x + 1/2*c) + sqrt(2)*sin(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c))*sin 
(7/2*d*x + 7/2*c)^3 - 24*((sqrt(2)*cos(d*x + c)^2 + sqrt(2)*sin(d*x + c)^2 
 + 2*sqrt(2)*cos(d*x + c) + sqrt(2))*cos(3/2*d*x + 3/2*c)^2 + (sqrt(2)*cos 
(1/2*d*x + 1/2*c)^2 + sqrt(2)*sin(1/2*d*x + 1/2*c)^2)*cos(d*x + c)^2 + (sq 
rt(2)*cos(d*x + c)^2 + sqrt(2)*sin(d*x + c)^2 + 2*sqrt(2)*cos(d*x + c) + s 
qrt(2))*sin(3/2*d*x + 3/2*c)^2 + (sqrt(2)*cos(1/2*d*x + 1/2*c)^2 + sqrt(2) 
*sin(1/2*d*x + 1/2*c)^2)*sin(d*x + c)^2 + sqrt(2)*cos(1/2*d*x + 1/2*c)^2 + 
 sqrt(2)*sin(1/2*d*x + 1/2*c)^2 + 2*(sqrt(2)*cos(d*x + c)^2*cos(1/2*d*x + 
1/2*c) + sqrt(2)*cos(1/2*d*x + 1/2*c)*sin(d*x + c)^2 + 2*sqrt(2)*cos(d*x + 
 c)*cos(1/2*d*x + 1/2*c) + sqrt(2)*cos(1/2*d*x + 1/2*c))*cos(3/2*d*x + ...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*cos(c + d*x))/(cos(c + d*x) + 1),x)* 
a + int((sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3)/(cos(c + d*x) + 1),x)*c)) 
/a