\(\int (a+a \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx\) [1169]

Optimal result

Integrand size = 35, antiderivative size = 196 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {16 a^2 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^2 (17 A+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:

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

Mathematica [C] (verified)

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

Time = 5.50 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.49 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {4 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (12 A \left (1+e^{2 i (c+d x)}\right )+12 A \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+5 (A+3 C) e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\sqrt {\sec (c+d x)} (3 (16 A+5 C-5 C \cos (2 c)) \cos (d x) \csc (c)+30 C \cos (c) \sin (d x)+2 A (10+3 \sec (c+d x)) \tan (c+d x))\right )}{60 d} \] Input:

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

Output:

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(((-4*I)*Sqrt[2]*Sqrt[E^(I*(c 
 + d*x))/(1 + E^((2*I)*(c + d*x)))]*(12*A*(1 + E^((2*I)*(c + d*x))) + 12*A 
*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 
1/2, 3/4, -E^((2*I)*(c + d*x))] + 5*(A + 3*C)*E^(I*(c + d*x))*(-1 + E^((2* 
I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^( 
(2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + Sqrt[Sec[c + d* 
x]]*(3*(16*A + 5*C - 5*C*Cos[2*c])*Cos[d*x]*Csc[c] + 30*C*Cos[c]*Sin[d*x] 
+ 2*A*(10 + 3*Sec[c + d*x])*Tan[c + d*x])))/(60*d)
 

Rubi [A] (verified)

Time = 1.35 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.95, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4709, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((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*((-24*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])
 

rule 4709

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(755\) vs. \(2(175)=350\).

Time = 209.74 (sec) , antiderivative size = 756, normalized size of antiderivative = 3.86

method	result	size
default	\(\text {Expression too large to display}\)	\(756\)
parts	\(\text {Expression too large to display}\)	\(1043\)

Input:

int((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(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*(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))*sin( 
1/2*d*x+1/2*c)^4+60*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-60*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*(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))*sin(1/2*d*x+1/2*c)^2-60*C*cos(1/2*d*x+1/2*c)*sin(1/2* 
d*x+1/2*c)^4+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*sin(1/ 
2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A-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*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*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 = 1.14 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 12 i \, \sqrt {2} A a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 i \, \sqrt {2} A a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, {\left (8 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 10 \, A a^{2} \cos \left (d x + c\right ) + 3 \, A a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(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)^2*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)^2 
*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 12*I*sqrt(2)* 
A*a^2*cos(d*x + c)^2*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)^2*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)*sin( 
d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2 \,d x \] Input:

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^2,x 
)
 

Output:

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^2, 
x)
 

Reduce [F]

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

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

Output:

a**2*(2*int(sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x)**3,x)*a + int(sqr 
t(sec(c + d*x))*cos(c + d*x)**4*sec(c + d*x)**3,x)*c + 2*int(sqrt(sec(c + 
d*x))*cos(c + d*x)**3*sec(c + d*x)**3,x)*c + int(sqrt(sec(c + d*x))*cos(c 
+ d*x)**2*sec(c + d*x)**3,x)*a + int(sqrt(sec(c + d*x))*cos(c + d*x)**2*se 
c(c + d*x)**3,x)*c + int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x)*a)