\(\int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [201]

Optimal result

Integrand size = 23, antiderivative size = 202 \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {7 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {7 \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {10 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac {7 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \] Output:

7*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/2)/ 
a^2/d+10/3*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x 
+c)^(1/2)/a^2/d-7*sec(d*x+c)^(1/2)*sin(d*x+c)/a^2/d+10/3*sec(d*x+c)^(3/2)* 
sin(d*x+c)/a^2/d-7/3*sec(d*x+c)^(5/2)*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3* 
sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2
 

Mathematica [C] (verified)

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

Time = 3.67 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {e^{-\frac {1}{2} i (4 c+3 d x)} \left (-1+e^{i c}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \csc \left (\frac {c}{2}\right ) \left (-10-37 e^{i (c+d x)}-65 e^{2 i (c+d x)}-82 e^{3 i (c+d x)}-68 e^{4 i (c+d x)}-53 e^{5 i (c+d x)}-21 e^{6 i (c+d x)}+10 i \left (1+e^{i (c+d x)}\right )^3 \left (1+e^{2 i (c+d x)}\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+7 e^{i (c+d x)} \left (1+e^{i (c+d x)}\right )^3 \left (1+e^{2 i (c+d x)}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec ^{\frac {5}{2}}(c+d x)}{12 a^2 d \left (1+e^{2 i (c+d x)}\right ) (1+\sec (c+d x))^2} \] Input:

Integrate[Sec[c + d*x]^(9/2)/(a + a*Sec[c + d*x])^2,x]
 

Output:

-1/12*((-1 + E^(I*c))*Cos[(c + d*x)/2]*Csc[c/2]*(-10 - 37*E^(I*(c + d*x)) 
- 65*E^((2*I)*(c + d*x)) - 82*E^((3*I)*(c + d*x)) - 68*E^((4*I)*(c + d*x)) 
 - 53*E^((5*I)*(c + d*x)) - 21*E^((6*I)*(c + d*x)) + (10*I)*(1 + E^(I*(c + 
 d*x)))^3*(1 + E^((2*I)*(c + d*x)))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x) 
/2, 2] + 7*E^(I*(c + d*x))*(1 + E^(I*(c + d*x)))^3*(1 + E^((2*I)*(c + d*x) 
))^(3/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c + d 
*x]^(5/2))/(a^2*d*E^((I/2)*(4*c + 3*d*x))*(1 + E^((2*I)*(c + d*x)))*(1 + S 
ec[c + d*x])^2)
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 4303, 27, 3042, 4507, 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.

\(\displaystyle \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{9/2}}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\)

\(\Big \downarrow \) 4303

\(\displaystyle -\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (5 a-9 a \sec (c+d x))}{2 (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (5 a-9 a \sec (c+d x))}{\sec (c+d x) a+a}dx}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (5 a-9 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4507

\(\displaystyle -\frac {\frac {\int 3 \sec ^{\frac {3}{2}}(c+d x) \left (7 a^2-10 a^2 \sec (c+d x)\right )dx}{a^2}+\frac {14 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {3 \int \sec ^{\frac {3}{2}}(c+d x) \left (7 a^2-10 a^2 \sec (c+d x)\right )dx}{a^2}+\frac {14 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (7 a^2-10 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {14 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4274

\(\displaystyle -\frac {\frac {3 \left (7 a^2 \int \sec ^{\frac {3}{2}}(c+d x)dx-10 a^2 \int \sec ^{\frac {5}{2}}(c+d x)dx\right )}{a^2}+\frac {14 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3 \left (7 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-10 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )}{a^2}+\frac {14 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4255

\(\displaystyle -\frac {\frac {3 \left (7 a^2 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )-10 a^2 \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}+\frac {14 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3 \left (7 a^2 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-10 a^2 \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}+\frac {14 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4258

\(\displaystyle -\frac {\frac {3 \left (7 a^2 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )-10 a^2 \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}+\frac {14 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3 \left (7 a^2 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-10 a^2 \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}+\frac {14 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {\frac {3 \left (7 a^2 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-10 a^2 \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}+\frac {14 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {\frac {3 \left (7 a^2 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-10 a^2 \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )\right )}{a^2}+\frac {14 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

Input:

Int[Sec[c + d*x]^(9/2)/(a + a*Sec[c + d*x])^2,x]
 

Output:

-1/3*(Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^2) - ((14*S 
ec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])) + (3*(7*a^2*((-2*Sq 
rt[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) - 10*a^2*((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))))/a^2)/(6*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 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 4303

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

rule 4507

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b 
- a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 
2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)* 
(d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m 
 - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, 
A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G 
tQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(412\) vs. \(2(181)=362\).

Time = 8.14 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.04

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {6 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {22 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {14 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {16 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{2 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(413\)

Input:

int(sec(d*x+c)^(9/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*(1/3*(- 
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)^3+6* 
(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)-22 
/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin( 
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c), 
2^(1/2))+14*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2) 
/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d 
*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2/3*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)/(cos(1/2*d*x+1 
/2*c)^2-1/2)^2+16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d* 
x+1/2*c)^2+1)*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 = 328, normalized size of antiderivative = 1.62 \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {10 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 10 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (21 \, \cos \left (d x + c\right )^{3} + 32 \, \cos \left (d x + c\right )^{2} + 8 \, \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \] Input:

integrate(sec(d*x+c)^(9/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
 

Output:

-1/6*(10*(I*sqrt(2)*cos(d*x + c)^3 + 2*I*sqrt(2)*cos(d*x + c)^2 + I*sqrt(2 
)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) 
+ 10*(-I*sqrt(2)*cos(d*x + c)^3 - 2*I*sqrt(2)*cos(d*x + c)^2 - I*sqrt(2)*c 
os(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2 
1*(-I*sqrt(2)*cos(d*x + c)^3 - 2*I*sqrt(2)*cos(d*x + c)^2 - I*sqrt(2)*cos( 
d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + 
 I*sin(d*x + c))) + 21*(I*sqrt(2)*cos(d*x + c)^3 + 2*I*sqrt(2)*cos(d*x + c 
)^2 + I*sqrt(2)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(- 
4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(21*cos(d*x + c)^3 + 32*cos(d*x 
+ c)^2 + 8*cos(d*x + c) - 2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d 
*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**(9/2)/(a+a*sec(d*x+c))**2,x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)^(9/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {9}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate(sec(d*x+c)^(9/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate(sec(d*x + c)^(9/2)/(a*sec(d*x + c) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{9/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \] Input:

int((1/cos(c + d*x))^(9/2)/(a + a/cos(c + d*x))^2,x)
 

Output:

int((1/cos(c + d*x))^(9/2)/(a + a/cos(c + d*x))^2, x)
 

Reduce [F]

\[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x}{a^{2}} \] Input:

int(sec(d*x+c)^(9/2)/(a+a*sec(d*x+c))^2,x)
 

Output:

int((sqrt(sec(c + d*x))*sec(c + d*x)**4)/(sec(c + d*x)**2 + 2*sec(c + d*x) 
 + 1),x)/a**2