\(\int \frac {(a+a \cos (c+d x))^2 (A+C \cos ^2(c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [138]

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

Optimal result

Integrand size = 35, antiderivative size = 156 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {4 a^2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {8 a^2 (A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {2 a^2 (5 A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}} \] Output:

-4*a^2*(A-C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+8/3*a^2*(A+C)*Inverse 
JacobiAM(1/2*d*x+1/2*c,2^(1/2))/d-2/3*a^2*(5*A-C)*cos(d*x+c)^(1/2)*sin(d*x 
+c)/d+2/3*A*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(3/2)+8/3*A*(a^2+a^ 
2*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 8.27 (sec) , antiderivative size = 865, normalized size of antiderivative = 5.54 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(-1/2*((-2* 
A + C + C*Cos[2*c])*Csc[c]*Sec[c])/d + (C*Cos[d*x]*Sin[c])/(6*d) + (C*Cos[ 
c]*Sin[d*x])/(6*d) + (A*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(6*d) + (Sec[c]*Se 
c[c + d*x]*(A*Sin[c] + 6*A*Sin[d*x]))/(6*d)) - (2*A*(a + a*Cos[c + d*x])^2 
*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]* 
Sec[c/2 + (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[C 
ot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt 
[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (2*C*(a + a*Co 
s[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan 
[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d 
*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Co 
t[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) + ( 
A*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((HypergeometricPFQ[{ 
-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]] 
*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Ta 
n[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + 
 Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2* 
Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^ 
2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d) - (C 
*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((HypergeometricPFQ...
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3523, 27, 3042, 3454, 27, 3042, 3447, 3042, 3502, 3042, 3227, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle \frac {3 \left (\frac {2}{3} \left (2 a^3 (A+C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^3 (A-C) \int \sqrt {\cos (c+d x)}dx\right )-\frac {2 a^3 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {2}{3} \left (2 a^3 (A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^3 (A-C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 a^3 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {3 \left (\frac {2}{3} \left (2 a^3 (A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a^3 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^3 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {3 \left (\frac {2}{3} \left (\frac {4 a^3 (A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^3 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^3 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

Input:

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

Output:

(2*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((8*A 
*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + 3*((2*((- 
6*a^3*(A - C)*EllipticE[(c + d*x)/2, 2])/d + (4*a^3*(A + C)*EllipticF[(c + 
 d*x)/2, 2])/d))/3 - (2*a^3*(5*A - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3* 
d)))/(3*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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

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 3454
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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{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] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[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 3523
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^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + 
 f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - 
d^2))   Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a 
*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* 
(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - 
d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(651\) vs. \(2(145)=290\).

Time = 2.72 (sec) , antiderivative size = 652, normalized size of antiderivative = 4.18

method result size
default \(-\frac {4 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{2} \left (-4 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-6 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-4 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) \(652\)
parts \(\text {Expression too large to display}\) \(749\)

Input:

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

Output:

-4/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2/(4*sin( 
1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)^3*(-4*C*sin( 
1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+12*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/ 
2*c)^4-4*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E 
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-6*A*EllipticE(co 
s(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2* 
c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+4*C*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/ 
2*c)-4*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ell 
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+6*C*EllipticE(cos( 
1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c) 
^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-7*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c) 
^2+2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellip 
ticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1 
/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-C*sin(1/2*d 
*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2 
*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*C*(sin(1/2* 
d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x 
+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1 
+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.40 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (2 i \, \sqrt {2} {\left (A + C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 2 i \, \sqrt {2} {\left (A + C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (C a^{2} \cos \left (d x + c\right )^{2} + 6 \, A a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{3 \, d \cos \left (d x + c\right )^{2}} \] Input:

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

Output:

-2/3*(2*I*sqrt(2)*(A + C)*a^2*cos(d*x + c)^2*weierstrassPInverse(-4, 0, co 
s(d*x + c) + I*sin(d*x + c)) - 2*I*sqrt(2)*(A + C)*a^2*cos(d*x + c)^2*weie 
rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)*(A - C 
)*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos 
(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(A - C)*a^2*cos(d*x + c)^2*weie 
rstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c 
))) - (C*a^2*cos(d*x + c)^2 + 6*A*a^2*cos(d*x + c) + A*a^2)*sqrt(cos(d*x + 
 c))*sin(d*x + c))/(d*cos(d*x + c)^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2\,C\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+6\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

(2*C*a^2*(cos(c + d*x)^(1/2)*sin(c + d*x) + 6*ellipticE(c/2 + (d*x)/2, 2) 
+ 4*ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*A*a^2*ellipticF(c/2 + (d*x)/2 
, 2))/d + (4*A*a^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2 
))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*A*a^2*sin(c + d*x)*h 
ypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c 
 + d*x)^2)^(1/2))
 

Reduce [F]

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

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

Output:

a**2*(int(sqrt(cos(c + d*x))/cos(c + d*x),x)*a + int(sqrt(cos(c + d*x))/co 
s(c + d*x),x)*c + int(sqrt(cos(c + d*x))/cos(c + d*x)**3,x)*a + 2*int(sqrt 
(cos(c + d*x))/cos(c + d*x)**2,x)*a + 2*int(sqrt(cos(c + d*x)),x)*c + int( 
sqrt(cos(c + d*x))*cos(c + d*x),x)*c)