\(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx\) [103]

Optimal result

Integrand size = 35, antiderivative size = 193 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 (35 A+37 C) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}-\frac {2 C \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (35 A+31 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 a d} \] 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-4/105*(35*A+37*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-2/35*C* 
cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/7*C*cos(d*x+c)^3*sin(d* 
x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/105*(35*A+31*C)*(a+a*cos(d*x+c))^(1/2)*sin 
(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.46 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (105 (A+C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 (70 A+101 C+24 C \cos (c+d x)+15 C \cos (2 (c+d x))) \sin ^3\left (\frac {1}{2} (c+d x)\right )\right )}{105 d \sqrt {a (1+\cos (c+d x))}} \] Input:

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

Output:

(2*Cos[(c + d*x)/2]*(105*(A + C)*ArcTanh[Sin[(c + d*x)/2]] - 2*(70*A + 101 
*C + 24*C*Cos[c + d*x] + 15*C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]^3))/(105* 
d*Sqrt[a*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.13, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 3525, 27, 3042, 3462, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3525

\(\displaystyle \frac {2 \int \frac {\cos ^2(c+d x) (a (7 A+6 C)-a C \cos (c+d x))}{2 \sqrt {\cos (c+d x) a+a}}dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (a (7 A+6 C)-a C \cos (c+d x))}{\sqrt {\cos (c+d x) a+a}}dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3462

\(\displaystyle \frac {\frac {2 \int -\frac {\cos (c+d x) \left (4 a^2 C-a^2 (35 A+31 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\cos (c+d x) \left (4 a^2 C-a^2 (35 A+31 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (4 a^2 C-a^2 (35 A+31 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {-\frac {\int \frac {4 a^2 C \cos (c+d x)-a^2 (35 A+31 C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {4 a^2 C \sin \left (c+d x+\frac {\pi }{2}\right )-a^2 (35 A+31 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {-\frac {\frac {2 \int -\frac {a^3 (35 A+31 C)-2 a^3 (35 A+37 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {2 a (35 A+31 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {\int \frac {a^3 (35 A+31 C)-2 a^3 (35 A+37 C) \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {2 a (35 A+31 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {\int \frac {a^3 (35 A+31 C)-2 a^3 (35 A+37 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a (35 A+31 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {-\frac {-\frac {105 a^3 (A+C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx-\frac {4 a^3 (35 A+37 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 a (35 A+31 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {105 a^3 (A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^3 (35 A+37 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 a (35 A+31 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {-\frac {-\frac {-\frac {210 a^3 (A+C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {4 a^3 (35 A+37 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 a (35 A+31 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {-\frac {\frac {105 \sqrt {2} a^{5/2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {4 a^3 (35 A+37 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 a (35 A+31 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}\)

Input:

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

Output:

(2*C*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*Sqrt[a + a*Cos[c + d*x]]) + ((-2*a* 
C*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]]) - ((-2*a*(35 
*A + 31*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) - ((105*Sqrt[2]*a^ 
(5/2)*(A + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d 
*x]])])/d - (4*a^3*(35*A + 37*C)*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]] 
))/(3*a))/(5*a))/(7*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 3462

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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Sin[e + f*x])^m*(c + d*S 
in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m 
 + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 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 = 0.37 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.31

method	result	size
default	\(\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (-240 C \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+336 C \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-140 \sqrt {2}\, \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (A +2 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+105 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a A +105 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a C \right )}{105 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(253\)
parts	\(\frac {A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (-4 \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{3 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (-240 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+336 \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-280 \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+105 \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{105 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(331\)

Input:

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

Output:

1/105*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(-240*C*2^(1/2)*(s 
in(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^6+336*C*2^(1/2)*(s 
in(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^4-140*2^(1/2)*a^(1 
/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(A+2*C)*sin(1/2*d*x+1/2*c)^2+105*2^(1/2 
)*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+ 
105*2^(1/2)*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*C)/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 = 173, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {4 \, {\left (15 \, C \cos \left (d x + c\right )^{3} - 3 \, C \cos \left (d x + c\right )^{2} + {\left (35 \, A + 31 \, C\right )} \cos \left (d x + c\right ) - 35 \, A - 43 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac {105 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right ) + {\left (A + C\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{210 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/210*(4*(15*C*cos(d*x + c)^3 - 3*C*cos(d*x + c)^2 + (35*A + 31*C)*cos(d*x 
 + c) - 35*A - 43*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) + 105*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)/sqrt(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(-1)]

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

integrate(cos(d*x+c)**2*(A+C*cos(d*x+c)**2)/(a+a*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. 715643 vs. \(2 (168) = 336\).

Time = 24.89 (sec) , antiderivative size = 715643, normalized size of antiderivative = 3707.99 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/5040*(84*(20*(cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c)^3 + 8*(cos(d*x + c) 
^2 + sin(d*x + c)^2 + 2*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c)^3 - 20*cos( 
5/2*d*x + 5/2*c)^3*sin(d*x + c) + 2*(15*(log(cos(1/2*d*x + 1/2*c)^2 + sin( 
1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c 
)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c)^2 
 + 15*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x 
 + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*s 
in(1/2*d*x + 1/2*c) + 1))*sin(d*x + c)^2 + 30*(log(cos(1/2*d*x + 1/2*c)^2 
+ sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 
 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x 
+ c) + 4*(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*cos(d*x + c) + 1)*sin(3/2*d* 
x + 3/2*c) - 20*cos(3/2*d*x + 3/2*c)*sin(d*x + c) + 15*log(cos(1/2*d*x + 1 
/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 15*log(co 
s(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1 
))*cos(5/2*d*x + 5/2*c)^2 + 30*((log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x 
+ 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + si 
n(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c)^2 + (log( 
cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 
 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x 
+ 1/2*c) + 1))*sin(d*x + c)^2 + 2*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2...
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4}}{\cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )+1}d x \right ) a \right )}{a} \] Input:

int(cos(d*x+c)^2*(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)**4)/(cos(c + d*x) + 1), 
x)*c + int((sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2)/(cos(c + d*x) + 1),x)* 
a))/a