\(\int \frac {(a+a \cos (c+d x))^2 (A+C \cos ^2(c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [139]

Optimal result

Integrand size = 35, antiderivative size = 156 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {16 a^2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (17 A+15 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \] Output:

-16/5*a^2*A*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/3*a^2*(A+3*C)*Invers 
eJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/15*a^2*(17*A+15*C)*sin(d*x+c)/d/cos(d 
*x+c)^(1/2)+2/5*A*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(5/2)+8/15*A* 
(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(3/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 8.84 (sec) , antiderivative size = 656, normalized size of antiderivative = 4.21 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {(-16 A-5 C+5 C \cos (2 c)) \csc (c) \sec (c)}{20 d}+\frac {A \sec (c) \sec ^3(c+d x) \sin (d x)}{10 d}+\frac {\sec (c) \sec ^2(c+d x) (3 A \sin (c)+10 A \sin (d x))}{30 d}+\frac {\sec (c) \sec (c+d x) (10 A \sin (c)+24 A \sin (d x)+15 C \sin (d x))}{30 d}\right )-\frac {A (a+a \cos (c+d x))^2 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d \sqrt {1+\cot ^2(c)}}-\frac {C (a+a \cos (c+d x))^2 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d \sqrt {1+\cot ^2(c)}}+\frac {2 A (a+a \cos (c+d x))^2 \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d} \] Input:

Integrate[((a + a*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/ 
2),x]
 

Output:

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(-1/20*((-1 
6*A - 5*C + 5*C*Cos[2*c])*Csc[c]*Sec[c])/d + (A*Sec[c]*Sec[c + d*x]^3*Sin[ 
d*x])/(10*d) + (Sec[c]*Sec[c + d*x]^2*(3*A*Sin[c] + 10*A*Sin[d*x]))/(30*d) 
 + (Sec[c]*Sec[c + d*x]*(10*A*Sin[c] + 24*A*Sin[d*x] + 15*C*Sin[d*x]))/(30 
*d)) - (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[Cot[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]) - (C*(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[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c] 
^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]] 
)/(d*Sqrt[1 + Cot[c]^2]) + (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[Tan[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]]]*Ta 
n[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]]]*Sq 
rt[1 + Tan[c]^2]]))/(5*d)
 

Rubi [A] (verified)

Time = 1.16 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.06, 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, 3500, 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.

\(\displaystyle \int \frac {(a \cos (c+d x)+a)^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\)

\(\Big \downarrow \) 3523

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {2}{3} \int \frac {(\cos (c+d x) a+a) \left (a^2 (17 A+15 C)-a^2 (7 A-15 C) \cos (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \frac {(\cos (c+d x) a+a) \left (a^2 (17 A+15 C)-a^2 (7 A-15 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (17 A+15 C)-a^2 (7 A-15 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {1}{3} \int \frac {-\left ((7 A-15 C) \cos ^2(c+d x) a^3\right )+(17 A+15 C) a^3+\left (a^3 (17 A+15 C)-a^3 (7 A-15 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {-\left ((7 A-15 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3\right )+(17 A+15 C) a^3+\left (a^3 (17 A+15 C)-a^3 (7 A-15 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {\frac {1}{3} \left (2 \int \frac {5 a^3 (A+3 C)-12 a^3 A \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^3 (17 A+15 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (2 \int \frac {5 a^3 (A+3 C)-12 a^3 A \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^3 (17 A+15 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {1}{3} \left (2 \left (5 a^3 (A+3 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-12 a^3 A \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 a^3 (17 A+15 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (2 \left (5 a^3 (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-12 a^3 A \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 a^3 (17 A+15 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {1}{3} \left (2 \left (5 a^3 (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {24 a^3 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a^3 (17 A+15 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (17 A+15 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+2 \left (\frac {10 a^3 (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {24 a^3 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {8 A \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

Input:

Int[((a + a*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]
 

Output:

(2*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((8*A 
*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*((-2 
4*a^3*A*EllipticE[(c + d*x)/2, 2])/d + (10*a^3*(A + 3*C)*EllipticF[(c + d* 
x)/2, 2])/d) + (2*a^3*(17*A + 15*C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))/ 
3)/(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 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 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

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. \(755\) vs. \(2(143)=286\).

Time = 3.38 (sec) , antiderivative size = 756, normalized size of antiderivative = 4.85

method	result	size
default	\(\text {Expression too large to display}\)	\(756\)
parts	\(\text {Expression too large to display}\)	\(928\)

Input:

int((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-4/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2/(8*sin 
(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 
2*d*x+1/2*c)^3*(96*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-20*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)*EllipticF(cos(1/2*d* 
x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-48*A*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)^4+60*C*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-60*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))*sin(1/2*d*x+1/2*c)^4-116*A*cos(1/2*d*x+1/2*c)*sin(1/2*d 
*x+1/2*c)^4+20*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+48*A*Ellip 
ticE(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)^2-60*C*sin(1/2*d*x+1/2*c)^4*cos(1/ 
2*d*x+1/2*c)+60*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))*sin(1/2*d*x+1/2*c)^2+37*A*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-5*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*A*(si 
n(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))+15*C*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-15*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(...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.43 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 12 i \, \sqrt {2} A a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 i \, \sqrt {2} A a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, {\left (8 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 10 \, A a^{2} \cos \left (d x + c\right ) + 3 \, A a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algori 
thm="fricas")
 

Output:

-2/15*(5*I*sqrt(2)*(A + 3*C)*a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, 
 cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(A + 3*C)*a^2*cos(d*x + c)^3 
*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 12*I*sqrt(2)* 
A*a^2*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos 
(d*x + c) + I*sin(d*x + c))) - 12*I*sqrt(2)*A*a^2*cos(d*x + c)^3*weierstra 
ssZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 
 (3*(8*A + 5*C)*a^2*cos(d*x + c)^2 + 10*A*a^2*cos(d*x + c) + 3*A*a^2)*sqrt 
(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**2*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algori 
thm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^2/cos(d*x + c)^(7/2) 
, x)
 

Giac [F]

\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algori 
thm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^2/cos(d*x + c)^(7/2) 
, x)
 

Mupad [B] (verification not implemented)

Time = 1.12 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.29 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {6\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,A\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^2)/cos(c + d*x)^(7/2),x)
 

Output:

(6*A*a^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 20*A* 
a^2*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2) 
+ 30*A*a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + 
 d*x)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (2*C*a^2* 
ellipticE(c/2 + (d*x)/2, 2))/d + (4*C*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + 
 (2*C*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))
 

Reduce [F]

\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=a^{2} \left (2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a +2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) c \right ) \] Input:

int((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)
 

Output:

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