3.4.80 \(\int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\) [380]

3.4.80.1 Optimal result
3.4.80.2 Mathematica [C] (verified)
3.4.80.3 Rubi [A] (verified)
3.4.80.4 Maple [B] (verified)
3.4.80.5 Fricas [C] (verification not implemented)
3.4.80.6 Sympy [F(-1)]
3.4.80.7 Maxima [F]
3.4.80.8 Giac [F]
3.4.80.9 Mupad [F(-1)]

3.4.80.1 Optimal result

Integrand size = 23, antiderivative size = 124 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {5 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {5 \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {3 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \]

output
3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/ 
2*c),2^(1/2))/a/d+5/3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli 
pticF(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+5/3*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)- 
sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))-3*sin(d*x+c)/a/d/cos(d*x+c) 
^(1/2)
 
3.4.80.2 Mathematica [C] (verified)

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

Time = 3.09 (sec) , antiderivative size = 338, normalized size of antiderivative = 2.73 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {\left (10 \cos \left (\frac {1}{2} (c-d x)\right )+8 \cos \left (\frac {1}{2} (3 c+d x)\right )+4 \cos \left (\frac {1}{2} (c+3 d x)\right )+5 \cos \left (\frac {1}{2} (5 c+3 d x)\right )+9 \cos \left (\frac {1}{2} (3 c+5 d x)\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )}{4 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 i \sqrt {2} e^{-i (c+d x)} \left (9 \left (1+e^{2 i (c+d x)}\right )+9 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )-5 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right ) \sec (c+d x)}{d \left (-1+e^{2 i c}\right ) \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{3 a (1+\sec (c+d x))} \]

input
Integrate[1/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])),x]
 
output
(Cos[(c + d*x)/2]^2*(-1/4*((10*Cos[(c - d*x)/2] + 8*Cos[(3*c + d*x)/2] + 4 
*Cos[(c + 3*d*x)/2] + 5*Cos[(5*c + 3*d*x)/2] + 9*Cos[(3*c + 5*d*x)/2])*Csc 
[c/2]*Sec[c/2]*Sec[(c + d*x)/2])/(d*Cos[c + d*x]^(5/2)) + ((2*I)*Sqrt[2]*( 
9*(1 + E^((2*I)*(c + d*x))) + 9*(-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*E^(I*(c 
 + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F 
1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c + d*x])/(d*E^(I*(c + d*x))*( 
-1 + E^((2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))])))/(3*a* 
(1 + Sec[c + d*x]))
 
3.4.80.3 Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.52, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 4752, 3042, 4305, 27, 3042, 4274, 3042, 4255, 3042, 4258, 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.

\(\displaystyle \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\)

\(\Big \downarrow \) 4752

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4305

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

input
Int[1/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])),x]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-((Sec[c + d*x]^(5/2)*Sin[c + d*x]) 
/(d*(a + a*Sec[c + d*x]))) - (3*a*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d 
*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d) - 
 5*a*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/ 
(3*d) + (2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d)))/(2*a^2))
 

3.4.80.3.1 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 4255
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4258
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4305
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_)), x_Symbol] :> Simp[d^2*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 2)/(f*(a + 
b*Csc[e + f*x]))), x] - Simp[d^2/(a*b)   Int[(d*Csc[e + f*x])^(n - 2)*(b*(n 
 - 2) - a*(n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ 
[a^2 - b^2, 0] && GtQ[n, 1]
 

rule 4752
Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Csc[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 
3.4.80.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(412\) vs. \(2(166)=332\).

Time = 9.17 (sec) , antiderivative size = 413, normalized size of antiderivative = 3.33

method result size
default \(\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-18 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-36 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}+9 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}+44 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-11 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{3 a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(413\)

input
int(1/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a/sin(1/2*d* 
x+1/2*c)^3/cos(1/2*d*x+1/2*c)/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c) 
^2+1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(10*cos(1/2*d*x 
+1/2*c)*EllipticF(cos(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)^(1/2)*sin(1/2*d*x+1/2*c)^2-18*cos(1/2*d*x+1/2* 
c)*EllipticE(cos(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)^(1/2)*sin(1/2*d*x+1/2*c)^2-36*sin(1/2*d*x+1/2*c)^6- 
5*cos(1/2*d*x+1/2*c)*EllipticF(cos(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)^(1/2)+9*cos(1/2*d*x+1/2*c)*Ellipt 
icE(cos(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)^(1/2)+44*sin(1/2*d*x+1/2*c)^4-11*sin(1/2*d*x+1/2*c)^2)/(2*co 
s(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 
3.4.80.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.08 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=-\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]

input
integrate(1/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")
 
output
-1/6*(2*(9*cos(d*x + c)^2 + 4*cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x 
 + c) + 5*(I*sqrt(2)*cos(d*x + c)^3 + I*sqrt(2)*cos(d*x + c)^2)*weierstras 
sPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(-I*sqrt(2)*cos(d*x + 
c)^3 - I*sqrt(2)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - 
 I*sin(d*x + c)) + 9*(-I*sqrt(2)*cos(d*x + c)^3 - I*sqrt(2)*cos(d*x + c)^2 
)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d 
*x + c))) + 9*(I*sqrt(2)*cos(d*x + c)^3 + I*sqrt(2)*cos(d*x + c)^2)*weiers 
trassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) 
))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)
 
3.4.80.6 Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\text {Timed out} \]

input
integrate(1/cos(d*x+c)**(7/2)/(a+a*sec(d*x+c)),x)
 
output
Timed out
 
3.4.80.7 Maxima [F]

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

input
integrate(1/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")
 
output
integrate(1/((a*sec(d*x + c) + a)*cos(d*x + c)^(7/2)), x)
 
3.4.80.8 Giac [F]

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

input
integrate(1/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c)),x, algorithm="giac")
 
output
integrate(1/((a*sec(d*x + c) + a)*cos(d*x + c)^(7/2)), x)
 
3.4.80.9 Mupad [F(-1)]

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

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