3.3.8 \(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\) [208]

3.3.8.1 Optimal result
3.3.8.2 Mathematica [C] (verified)
3.3.8.3 Rubi [A] (verified)
3.3.8.4 Maple [A] (verified)
3.3.8.5 Fricas [C] (verification not implemented)
3.3.8.6 Sympy [F]
3.3.8.7 Maxima [F]
3.3.8.8 Giac [F]
3.3.8.9 Mupad [F(-1)]

3.3.8.1 Optimal result

Integrand size = 33, antiderivative size = 230 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=-\frac {21 (A-B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}+\frac {5 (9 A-7 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 a d}+\frac {(9 A-7 B) \sin (c+d x)}{7 a d \sec ^{\frac {5}{2}}(c+d x)}-\frac {7 (A-B) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 A-7 B) \sin (c+d x)}{21 a d \sqrt {\sec (c+d x)}}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \]

output
1/7*(9*A-7*B)*sin(d*x+c)/a/d/sec(d*x+c)^(5/2)-7/5*(A-B)*sin(d*x+c)/a/d/sec 
(d*x+c)^(3/2)-(A-B)*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))+5/21*(9 
*A-7*B)*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)-21/5*(A-B)*(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))*cos(d*x+c)^ 
(1/2)*sec(d*x+c)^(1/2)/a/d+5/21*(9*A-7*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos 
(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec 
(d*x+c)^(1/2)/a/d
 
3.3.8.2 Mathematica [C] (verified)

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

Time = 4.85 (sec) , antiderivative size = 568, normalized size of antiderivative = 2.47 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) (A+B \sec (c+d x)) \left (588 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-588 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+1800 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-1400 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+\sqrt {\sec (c+d x)} \left (63 (A-B) (17+11 \cos (2 c)) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )+20 (27 A-14 B) \cos (2 d x) \sin (2 c)-84 (A-B) \cos (3 d x) \sin (3 c)+30 A \cos (4 d x) \sin (4 c)-840 (A-B) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )-2772 (A-B) \cos (c) \sin (d x)+20 (27 A-14 B) \cos (2 c) \sin (2 d x)-84 (A-B) \cos (3 c) \sin (3 d x)+30 A \cos (4 c) \sin (4 d x)-840 (A-B) \tan \left (\frac {c}{2}\right )\right )\right )}{420 a d (B+A \cos (c+d x)) (1+\sec (c+d x))} \]

input
Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(7/2)*(a + a*Sec[c + d*x])),x 
]
 
output
(Cos[(c + d*x)/2]^2*(A + B*Sec[c + d*x])*((588*Sqrt[2]*A*Sqrt[E^(I*(c + d* 
x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sq 
rt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeomet 
ric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) - (588*Sqrt[2]*B*S 
qrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x) 
)]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I) 
*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + 
1800*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 1 
400*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + Sq 
rt[Sec[c + d*x]]*(63*(A - B)*(17 + 11*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2] 
 + 20*(27*A - 14*B)*Cos[2*d*x]*Sin[2*c] - 84*(A - B)*Cos[3*d*x]*Sin[3*c] + 
 30*A*Cos[4*d*x]*Sin[4*c] - 840*(A - B)*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d*x 
)/2] - 2772*(A - B)*Cos[c]*Sin[d*x] + 20*(27*A - 14*B)*Cos[2*c]*Sin[2*d*x] 
 - 84*(A - B)*Cos[3*c]*Sin[3*d*x] + 30*A*Cos[4*c]*Sin[4*d*x] - 840*(A - B) 
*Tan[c/2])))/(420*a*d*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x]))
 
3.3.8.3 Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.94, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4508, 27, 3042, 4274, 3042, 4256, 3042, 4256, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (9 A-7 B) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx-7 a (A-B) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

input
Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(7/2)*(a + a*Sec[c + d*x])),x]
 
output
-(((A - B)*Sin[c + d*x])/(d*Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x]))) + (- 
7*a*(A - B)*((6*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + 
d*x]])/(5*d) + (2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))) + a*(9*A - 7*B)* 
((2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (5*((2*Sqrt[Cos[c + d*x]]*Ell 
ipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*Sin[c + d*x])/(3*d*S 
qrt[Sec[c + d*x]])))/7))/(2*a^2)
 

3.3.8.3.1 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 4256
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 4508
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b 
- a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 
 1))), x] - Simp[1/(a^2*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs 
c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ 
e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B 
, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]
 
3.3.8.4 Maple [A] (verified)

Time = 9.12 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.30

method result size
default \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (225 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+441 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-175 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-441 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-480 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (864 A +336 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-888 A -392 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (930 A -210 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-321 A +161 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{105 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(300\)

input
int((A+B*sec(d*x+c))/sec(d*x+c)^(7/2)/(a+a*sec(d*x+c)),x,method=_RETURNVER 
BOSE)
 
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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(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*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-441*B*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+336*B)*sin(1 
/2*d*x+1/2*c)^8+(-888*A-392*B)*sin(1/2*d*x+1/2*c)^6+(930*A-210*B)*sin(1/2* 
d*x+1/2*c)^4+(-321*A+161*B)*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*co 
s(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 
3.3.8.5 Fricas [C] (verification not implemented)

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

Time = 0.12 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.29 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=-\frac {25 \, {\left (\sqrt {2} {\left (9 i \, A - 7 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (9 i \, A - 7 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 \, {\left (\sqrt {2} {\left (-9 i \, A + 7 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-9 i \, A + 7 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 441 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 441 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (30 \, A \cos \left (d x + c\right )^{4} - 6 \, {\left (2 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (39 \, A - 14 \, B\right )} \cos \left (d x + c\right )^{2} + 25 \, {\left (9 \, A - 7 \, B\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{210 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

input
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(7/2)/(a+a*sec(d*x+c)),x, algorithm= 
"fricas")
 
output
-1/210*(25*(sqrt(2)*(9*I*A - 7*I*B)*cos(d*x + c) + sqrt(2)*(9*I*A - 7*I*B) 
)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 25*(sqrt(2)* 
(-9*I*A + 7*I*B)*cos(d*x + c) + sqrt(2)*(-9*I*A + 7*I*B))*weierstrassPInve 
rse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 441*(sqrt(2)*(I*A - I*B)*cos(d 
*x + c) + sqrt(2)*(I*A - I*B))*weierstrassZeta(-4, 0, weierstrassPInverse( 
-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 441*(sqrt(2)*(-I*A + I*B)*cos(d*x 
 + c) + sqrt(2)*(-I*A + I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(- 
4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(30*A*cos(d*x + c)^4 - 6*(2*A - 
7*B)*cos(d*x + c)^3 + 2*(39*A - 14*B)*cos(d*x + c)^2 + 25*(9*A - 7*B)*cos( 
d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d)
 
3.3.8.6 Sympy [F]

\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {\int \frac {A}{\sec ^{\frac {9}{2}}{\left (c + d x \right )} + \sec ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{\frac {9}{2}}{\left (c + d x \right )} + \sec ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx}{a} \]

input
integrate((A+B*sec(d*x+c))/sec(d*x+c)**(7/2)/(a+a*sec(d*x+c)),x)
 
output
(Integral(A/(sec(c + d*x)**(9/2) + sec(c + d*x)**(7/2)), x) + Integral(B*s 
ec(c + d*x)/(sec(c + d*x)**(9/2) + sec(c + d*x)**(7/2)), x))/a
 
3.3.8.7 Maxima [F]

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

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

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

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

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

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