\(\int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [1084]

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 = 33, antiderivative size = 134 \[ \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (5 A+7 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 a (5 A+7 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d} \] Output:

2/5*a*(3*A+5*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/21*a*(5*A+7*C)*I 
nverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/21*a*(5*A+7*C)*cos(d*x+c)^(1/2)* 
sin(d*x+c)/d+2/5*a*A*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/7*a*A*cos(d*x+c)^(5/2 
)*sin(d*x+c)/d
 

Mathematica [C] (warning: unable to verify)

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

Time = 6.48 (sec) , antiderivative size = 872, normalized size of antiderivative = 6.51 \[ \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

a*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*(-1/5*((3*A 
+ 5*C)*Cot[c])/d + ((23*A + 28*C)*Cos[d*x]*Sin[c])/(84*d) + (A*Cos[2*d*x]* 
Sin[2*c])/(10*d) + (A*Cos[3*d*x]*Sin[3*c])/(28*d) + ((23*A + 28*C)*Cos[c]* 
Sin[d*x])/(84*d) + (A*Cos[2*c]*Sin[2*d*x])/(10*d) + (A*Cos[3*c]*Sin[3*d*x] 
)/(28*d)) - (5*A*(1 + Cos[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, { 
5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - ArcTan[C 
ot[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]]]])/(21*d* 
Sqrt[1 + Cot[c]^2]) - (C*(1 + Cos[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 
 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^2*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]) - (3*A*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x 
)/2]^2*((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]]]*Tan[ 
c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Ta 
n[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt 
[1 + Tan[c]^2]]))/(10*d) - (C*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)...
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4602, 3042, 3513, 27, 3042, 3502, 27, 3042, 3227, 3042, 3115, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4602

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3513

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (7 a A \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (5 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )+7 a C\right )dx+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {\cos (c+d x)} (7 a (3 A+5 C)+5 a (5 A+7 C) \cos (c+d x))dx+\frac {14 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \sqrt {\cos (c+d x)} (7 a (3 A+5 C)+5 a (5 A+7 C) \cos (c+d x))dx+\frac {14 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (7 a (3 A+5 C)+5 a (5 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {14 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 a (5 A+7 C) \int \cos ^{\frac {3}{2}}(c+d x)dx+7 a (3 A+5 C) \int \sqrt {\cos (c+d x)}dx\right )+\frac {14 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (7 a (3 A+5 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 a (5 A+7 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )+\frac {14 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (7 a (3 A+5 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 a (5 A+7 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (7 a (3 A+5 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 a (5 A+7 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 a (5 A+7 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {14 a (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {14 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {14 a (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+5 a (5 A+7 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

Input:

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

Output:

(2*a*A*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*d) + ((14*a*A*Cos[c + d*x]^(3/2 
)*Sin[c + d*x])/(5*d) + ((14*a*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/d + 
5*a*(5*A + 7*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x 
]]*Sin[c + d*x])/(3*d)))/5)/7
 

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 3115
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

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 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 3513
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ 
(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3) 
)), x] + Simp[1/(b*(m + 3))   Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c 
*(m + 3) + b*d*(C*(m + 2) + A*(m + 3))*Sin[e + f*x] - (2*a*C*d - b*c*C*(m + 
 3))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && 
 NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 4602
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[d^( 
m + 2)   Int[(b + a*Cos[e + f*x])^m*(d*Cos[e + f*x])^(n - m - 2)*(C + A*Cos 
[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !IntegerQ[n] 
 && IntegerQ[m]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(377\) vs. \(2(121)=242\).

Time = 31.45 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.82

method result size
default \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \left (240 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-528 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (448 A +140 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-122 A -70 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+25 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+35 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(378\)

Input:

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

Output:

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(240*A*co 
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-528*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x 
+1/2*c)^6+(448*A+140*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-122*A-70 
*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+25*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)) 
-63*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)*Ellipt 
icE(cos(1/2*d*x+1/2*c),2^(1/2))+35*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))-105*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) 
/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.38 \[ \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {-5 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (15 \, A a \cos \left (d x + c\right )^{2} + 21 \, A a \cos \left (d x + c\right ) + 5 \, {\left (5 \, A + 7 \, C\right )} a\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d} \] Input:

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

Output:

1/105*(-5*I*sqrt(2)*(5*A + 7*C)*a*weierstrassPInverse(-4, 0, cos(d*x + c) 
+ I*sin(d*x + c)) + 5*I*sqrt(2)*(5*A + 7*C)*a*weierstrassPInverse(-4, 0, c 
os(d*x + c) - I*sin(d*x + c)) + 21*I*sqrt(2)*(3*A + 5*C)*a*weierstrassZeta 
(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I* 
sqrt(2)*(3*A + 5*C)*a*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co 
s(d*x + c) - I*sin(d*x + c))) + 2*(15*A*a*cos(d*x + c)^2 + 21*A*a*cos(d*x 
+ c) + 5*(5*A + 7*C)*a)*sqrt(cos(d*x + c))*sin(d*x + c))/d
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.04 \[ \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,C\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,C\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,A\,a\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,a\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

(2*C*a*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3 
*d) + (2*C*a*ellipticE(c/2 + (d*x)/2, 2))/d - (2*A*a*cos(c + d*x)^(7/2)*si 
n(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x) 
^2)^(1/2)) - (2*A*a*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 
13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2))
 

Reduce [F]

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

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

Output:

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