\(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx\) [120]

Optimal result

Integrand size = 35, antiderivative size = 212 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(19 A+163 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(A+17 C) \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(21 A+197 C) \sin (c+d x)}{24 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {5 (3 A+19 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d} \] Output:

1/32*(19*A+163*C)*arctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^ 
(1/2))*2^(1/2)/a^(5/2)/d-1/4*(A+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+ 
c))^(5/2)-1/16*(A+17*C)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2) 
-1/24*(21*A+197*C)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)+5/48*(3*A+19*C) 
*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/a^3/d
 

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.53 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {6 (19 A+163 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )-(27 A+379 C+(39 A+479 C) \cos (c+d x)+80 C \cos (2 (c+d x))-8 C \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{48 a d (a (1+\cos (c+d x)))^{3/2}} \] Input:

Integrate[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^(5/ 
2),x]
 

Output:

(6*(19*A + 163*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 - (27*A + 3 
79*C + (39*A + 479*C)*Cos[c + d*x] + 80*C*Cos[2*(c + d*x)] - 8*C*Cos[3*(c 
+ d*x)])*Tan[(c + d*x)/2])/(48*a*d*(a*(1 + Cos[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 3521, 27, 3042, 3456, 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.

\(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

\(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (2 a (A-3 C)+a (3 A+11 C) \cos (c+d x))}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (2 a (A-3 C)+a (3 A+11 C) \cos (c+d x))}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a (A-3 C)+a (3 A+11 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\frac {\int -\frac {\cos (c+d x) \left (4 a^2 (A+17 C)-5 a^2 (3 A+19 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\cos (c+d x) \left (4 a^2 (A+17 C)-5 a^2 (3 A+19 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (4 a^2 (A+17 C)-5 a^2 (3 A+19 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {-\frac {\int \frac {4 a^2 (A+17 C) \cos (c+d x)-5 a^2 (3 A+19 C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {4 a^2 (A+17 C) \sin \left (c+d x+\frac {\pi }{2}\right )-5 a^2 (3 A+19 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {-\frac {\frac {2 \int -\frac {5 a^3 (3 A+19 C)-2 a^3 (21 A+197 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {10 a (3 A+19 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {\int \frac {5 a^3 (3 A+19 C)-2 a^3 (21 A+197 C) \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {10 a (3 A+19 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {\int \frac {5 a^3 (3 A+19 C)-2 a^3 (21 A+197 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {10 a (3 A+19 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {-\frac {-\frac {3 a^3 (19 A+163 C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx-\frac {4 a^3 (21 A+197 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {10 a (3 A+19 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {3 a^3 (19 A+163 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^3 (21 A+197 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {10 a (3 A+19 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {-\frac {-\frac {-\frac {6 a^3 (19 A+163 C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {4 a^3 (21 A+197 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {10 a (3 A+19 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {-\frac {\frac {3 \sqrt {2} a^{5/2} (19 A+163 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {4 a^3 (21 A+197 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {10 a (3 A+19 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A+17 C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

Input:

Int[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^(5/2),x]
 

Output:

-1/4*((A + C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(5/2)) 
+ (-1/2*(a*(A + 17*C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]) 
^(3/2)) - ((-10*a*(3*A + 19*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d 
) - ((3*Sqrt[2]*a^(5/2)*(19*A + 163*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqr 
t[2]*Sqrt[a + a*Cos[c + d*x]])])/d - (4*a^3*(21*A + 197*C)*Sin[c + d*x])/( 
d*Sqrt[a + a*Cos[c + d*x]]))/(3*a))/(4*a^2))/(8*a^2)
 

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 3456

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[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 
a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + 
 b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & 
& NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In 
tegerQ[2*n] || EqQ[c, 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 3521

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*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
 + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1))   I 
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) 
- b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* 
c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
 - d^2, 0] && LtQ[m, -2^(-1)]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.71

method	result	size
default	\(\frac {\sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (128 C \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+57 A \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +489 C \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -512 C \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}-39 A \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-87 C \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}+6 A \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}+6 C \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\right )}{96 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(362\)
parts	\(\frac {A \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (19 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -13 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {C \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (128 \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+489 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -512 \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-87 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\right )}{96 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(418\)

Input:

int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),x,method=_RETUR 
NVERBOSE)
 

Output:

1/96*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(128*C*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a 
)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6+57*A*2^(1/2)*ln(2*(2*a^(1/2)*(sin(1/2 
*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a+489 
*C*2^(1/2)*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x 
+1/2*c))*cos(1/2*d*x+1/2*c)^4*a-512*C*2^(1/2)*cos(1/2*d*x+1/2*c)^4*(sin(1/ 
2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)-39*A*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2 
)*a^(1/2)*cos(1/2*d*x+1/2*c)^2-87*C*2^(1/2)*cos(1/2*d*x+1/2*c)^2*(sin(1/2* 
d*x+1/2*c)^2*a)^(1/2)*a^(1/2)+6*A*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a 
^(1/2)+6*C*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2))/cos(1/2*d*x+1/2 
*c)^3/a^(7/2)/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 = 0.10 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {3 \, \sqrt {2} {\left ({\left (19 \, A + 163 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A + 163 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A + 163 \, C\right )} \cos \left (d x + c\right ) + 19 \, A + 163 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (32 \, C \cos \left (d x + c\right )^{3} - 160 \, C \cos \left (d x + c\right )^{2} - {\left (39 \, A + 503 \, C\right )} \cos \left (d x + c\right ) - 27 \, A - 299 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{192 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),x, algori 
thm="fricas")
 

Output:

1/192*(3*sqrt(2)*((19*A + 163*C)*cos(d*x + c)^3 + 3*(19*A + 163*C)*cos(d*x 
 + c)^2 + 3*(19*A + 163*C)*cos(d*x + c) + 19*A + 163*C)*sqrt(a)*log(-(a*co 
s(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2 
*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(32*C*co 
s(d*x + c)^3 - 160*C*cos(d*x + c)^2 - (39*A + 503*C)*cos(d*x + c) - 27*A - 
 299*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a 
^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**2*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),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. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:

int((cos(c + d*x)^2*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^(5/2),x)
 

Output:

int((cos(c + d*x)^2*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) a \right )}{a^{3}} \] Input:

int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),x)
 

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*cos(c + d*x)**4)/(cos(c + d*x)**3 + 
3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*c + int((sqrt(cos(c + d*x) + 1) 
*cos(c + d*x)**2)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 
1),x)*a))/a**3