\(\int \frac {\cos ^{\frac {7}{2}}(c+d x) (A+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [1107]

Optimal result

Integrand size = 35, antiderivative size = 192 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {3 (7 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}+\frac {5 (9 A+7 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 a d}+\frac {5 (9 A+7 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 a d}-\frac {(7 A+5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {(9 A+7 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 a d}-\frac {(A+C) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \] Output:

-3/5*(7*A+5*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+5/21*(9*A+7*C)*In 
verseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a/d+5/21*(9*A+7*C)*cos(d*x+c)^(1/2)*s 
in(d*x+c)/a/d-1/5*(7*A+5*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d+1/7*(9*A+7*C)* 
cos(d*x+c)^(5/2)*sin(d*x+c)/a/d-(A+C)*cos(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*c 
os(d*x+c))
 

Mathematica [C] (warning: unable to verify)

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

Time = 8.14 (sec) , antiderivative size = 1155, normalized size of antiderivative = 6.02 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2)*((4*(5*A + 
 5*C + 16*A*Cos[c] + 10*C*Cos[c])*Csc[c])/(5*d) + (2*(51*A + 28*C)*Cos[d*x 
]*Sin[c])/(21*d) - (4*A*Cos[2*d*x]*Sin[2*c])/(5*d) + (2*A*Cos[3*d*x]*Sin[3 
*c])/(7*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + C*Sin[(d*x)/ 
2]))/d + (2*(51*A + 28*C)*Cos[c]*Sin[d*x])/(21*d) - (4*A*Cos[2*c]*Sin[2*d* 
x])/(5*d) + (2*A*Cos[3*c]*Sin[3*d*x])/(7*d)))/((A + 2*C + A*Cos[2*c + 2*d* 
x])*(a + a*Sec[c + d*x])) - (30*A*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/ 
2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c 
/2]*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - Ar 
cTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]]) 
]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(7*d*(A + 2*C + A*Cos[2*c + 2*d*x]) 
*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) - (10*C*Cos[c/2 + (d*x)/2]^2*Cos 
[c + d*x]*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[C 
ot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d*x]^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*(A + 2*C + A*C 
os[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) + (21*A*Cos[c/2 
+ (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*((Hyper 
geometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + A 
rcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Co...
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.92, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4602, 3042, 3521, 27, 3042, 3227, 3042, 3115, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4602

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

\(\displaystyle \frac {\int -\frac {1}{2} \cos ^{\frac {5}{2}}(c+d x) (a (7 A+5 C)-a (9 A+7 C) \cos (c+d x))dx}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \cos ^{\frac {5}{2}}(c+d x) (a (7 A+5 C)-a (9 A+7 C) \cos (c+d x))dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (7 A+5 C)-a (9 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3227

\(\displaystyle -\frac {a (7 A+5 C) \int \cos ^{\frac {5}{2}}(c+d x)dx-a (9 A+7 C) \int \cos ^{\frac {7}{2}}(c+d x)dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a (7 A+5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx-a (9 A+7 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {a (7 A+5 C) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-a (9 A+7 C) \left (\frac {5}{7} \int \cos ^{\frac {3}{2}}(c+d x)dx+\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a (7 A+5 C) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-a (9 A+7 C) \left (\frac {5}{7} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {a (7 A+5 C) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-a (9 A+7 C) \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a (7 A+5 C) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-a (9 A+7 C) \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {a (7 A+5 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-a (9 A+7 C) \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {a (7 A+5 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-a (9 A+7 C) \left (\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {5}{7} \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

Input:

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

Output:

-(((A + C)*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))) - (a 
*(7*A + 5*C)*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)* 
Sin[c + d*x])/(5*d)) - a*(9*A + 7*C)*((2*Cos[c + d*x]^(5/2)*Sin[c + d*x])/ 
(7*d) + (5*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Si 
n[c + d*x])/(3*d)))/7))/(2*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 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 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)]
 

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 [A] (verified)

Time = 14.13 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.54

method	result	size
default	\(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (225 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+441 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+175 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+315 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-480 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+864 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-888 A -280 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (930 A +630 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-321 A -245 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{105 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(295\)

Input:

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

Output:

-1/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d* 
x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(22 
5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+441*A*EllipticE(cos(1/2*d*x+1/2* 
c),2^(1/2))+175*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+315*C*EllipticE(co 
s(1/2*d*x+1/2*c),2^(1/2)))-480*A*sin(1/2*d*x+1/2*c)^10+864*A*sin(1/2*d*x+1 
/2*c)^8+(-888*A-280*C)*sin(1/2*d*x+1/2*c)^6+(930*A+630*C)*sin(1/2*d*x+1/2* 
c)^4+(-321*A-245*C)*sin(1/2*d*x+1/2*c)^2)/a/cos(1/2*d*x+1/2*c)/(-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 = 280, normalized size of antiderivative = 1.46 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {2 \, {\left (30 \, A \cos \left (d x + c\right )^{3} - 12 \, A \cos \left (d x + c\right )^{2} + 2 \, {\left (39 \, A + 35 \, C\right )} \cos \left (d x + c\right ) + 225 \, A + 175 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 25 \, {\left (\sqrt {2} {\left (9 i \, A + 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (9 i \, A + 7 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 25 \, {\left (\sqrt {2} {\left (-9 i \, A - 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-9 i \, A - 7 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 63 \, {\left (\sqrt {2} {\left (7 i \, A + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (7 i \, A + 5 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 63 \, {\left (\sqrt {2} {\left (-7 i \, A - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-7 i \, A - 5 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{210 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/210*(2*(30*A*cos(d*x + c)^3 - 12*A*cos(d*x + c)^2 + 2*(39*A + 35*C)*cos( 
d*x + c) + 225*A + 175*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 25*(sqrt(2)*(9 
*I*A + 7*I*C)*cos(d*x + c) + sqrt(2)*(9*I*A + 7*I*C))*weierstrassPInverse( 
-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 25*(sqrt(2)*(-9*I*A - 7*I*C)*cos(d 
*x + c) + sqrt(2)*(-9*I*A - 7*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c 
) - I*sin(d*x + c)) - 63*(sqrt(2)*(7*I*A + 5*I*C)*cos(d*x + c) + sqrt(2)*( 
7*I*A + 5*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c))) - 63*(sqrt(2)*(-7*I*A - 5*I*C)*cos(d*x + c) + sqrt 
(2)*(-7*I*A - 5*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co 
s(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d*x + c) + a*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}}{\sec \left (d x +c \right )+1}d x \right ) a}{a} \] Input:

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

Output:

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