3.2.92 \(\int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\) [192]

3.2.92.1 Optimal result
3.2.92.2 Mathematica [C] (verified)
3.2.92.3 Rubi [A] (verified)
3.2.92.4 Maple [A] (verified)
3.2.92.5 Fricas [C] (verification not implemented)
3.2.92.6 Sympy [F(-1)]
3.2.92.7 Maxima [F]
3.2.92.8 Giac [F]
3.2.92.9 Mupad [F(-1)]

3.2.92.1 Optimal result

Integrand size = 33, antiderivative size = 234 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {4 a^2 (8 A+9 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (5 A+6 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (8 A+9 B) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (5 A+6 B) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

output
2/63*a^2*(11*A+9*B)*sin(d*x+c)/d/sec(d*x+c)^(5/2)+4/45*a^2*(8*A+9*B)*sin(d 
*x+c)/d/sec(d*x+c)^(3/2)+2/9*A*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d/sec(d*x+c 
)^(7/2)+4/21*a^2*(5*A+6*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+4/15*a^2*(8*A+9*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)/d+4/21*a^2*(5*A+6*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)/d
 
3.2.92.2 Mathematica [C] (verified)

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

Time = 3.74 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.93 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {a^2 e^{-i d x} \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (240 (5 A+6 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-112 i (8 A+9 B) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+\cos (c+d x) (2688 i A+3024 i B+30 (46 A+51 B) \sin (c+d x)+14 (37 A+36 B) \sin (2 (c+d x))+180 A \sin (3 (c+d x))+90 B \sin (3 (c+d x))+35 A \sin (4 (c+d x)))\right )}{1260 d} \]

input
Integrate[((a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(9/2) 
,x]
 
output
(a^2*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(240*(5*A + 6*B)*Sqrt[Cos[ 
c + d*x]]*EllipticF[(c + d*x)/2, 2] - (112*I)*(8*A + 9*B)*E^(I*(c + d*x))* 
Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*( 
c + d*x))] + Cos[c + d*x]*((2688*I)*A + (3024*I)*B + 30*(46*A + 51*B)*Sin[ 
c + d*x] + 14*(37*A + 36*B)*Sin[2*(c + d*x)] + 180*A*Sin[3*(c + d*x)] + 90 
*B*Sin[3*(c + d*x)] + 35*A*Sin[4*(c + d*x)])))/(1260*d*E^(I*d*x))
 
3.2.92.3 Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4505, 27, 3042, 4484, 25, 3042, 4274, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

\(\displaystyle \frac {2}{9} \int \frac {(\sec (c+d x) a+a) (a (11 A+9 B)+a (5 A+9 B) \sec (c+d x))}{2 \sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \int \frac {(\sec (c+d x) a+a) (a (11 A+9 B)+a (5 A+9 B) \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (11 A+9 B)+a (5 A+9 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 4484

\(\displaystyle \frac {1}{9} \left (\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {2}{7} \int -\frac {7 (8 A+9 B) a^2+9 (5 A+6 B) \sec (c+d x) a^2}{\sec ^{\frac {5}{2}}(c+d x)}dx\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \int \frac {7 (8 A+9 B) a^2+9 (5 A+6 B) \sec (c+d x) a^2}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \int \frac {7 (8 A+9 B) a^2+9 (5 A+6 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (7 a^2 (8 A+9 B) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)}dx+9 a^2 (5 A+6 B) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (7 a^2 (8 A+9 B) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+9 a^2 (5 A+6 B) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (7 a^2 (8 A+9 B) \left (\frac {3}{5} \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+9 a^2 (5 A+6 B) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (7 a^2 (8 A+9 B) \left (\frac {3}{5} \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+9 a^2 (5 A+6 B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (7 a^2 (8 A+9 B) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+9 a^2 (5 A+6 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (7 a^2 (8 A+9 B) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+9 a^2 (5 A+6 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (9 a^2 (5 A+6 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+7 a^2 (8 A+9 B) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{9} \left (\frac {2 a^2 (11 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2}{7} \left (9 a^2 (5 A+6 B) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )+7 a^2 (8 A+9 B) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )\right )\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{9 d \sec ^{\frac {7}{2}}(c+d x)}\)

input
Int[((a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(9/2),x]
 
output
(2*A*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + ((2 
*a^2*(11*A + 9*B)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*(7*a^2*(8*A 
+ 9*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))) + 9*a^2*(5*A + 6*B)*( 
(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) 
+ (2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]))))/7)/9
 

3.2.92.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 4484
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*a*Cot[e + 
f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Simp[1/(d*n)   Int[(d*Csc[e + f*x])^( 
n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] 
/; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]
 

rule 4505
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot 
[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim 
p[b/(a*d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim 
p[a*A*(m - n - 1) - b*B*n - (a*B*n + 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 
] && GtQ[m, 1/2] && LtQ[n, -1]
 
3.2.92.4 Maple [A] (verified)

Time = 24.32 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.76

method result size
default \(-\frac {4 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{2} \left (-560 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (1840 A +360 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-2368 A -1044 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (1568 A +1134 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-387 A -351 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+75 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-168 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+90 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-189 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{315 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(413\)
parts \(\text {Expression too large to display}\) \(822\)

input
int((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x,method=_RETURNV 
ERBOSE)
 
output
-4/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-560*A 
*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+(1840*A+360*B)*sin(1/2*d*x+1/2*c 
)^8*cos(1/2*d*x+1/2*c)+(-2368*A-1044*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1 
/2*c)+(1568*A+1134*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-387*A-351* 
B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+75*A*(2*sin(1/2*d*x+1/2*c)^2-1) 
^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))- 
168*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt 
icE(cos(1/2*d*x+1/2*c),2^(1/2))+90*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin 
(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-189*B*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(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
 
3.2.92.5 Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.99 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (15 i \, \sqrt {2} {\left (5 \, A + 6 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (5 \, A + 6 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 i \, \sqrt {2} {\left (8 \, A + 9 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} {\left (8 \, A + 9 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (35 \, A a^{2} \cos \left (d x + c\right )^{4} + 45 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{3} + 14 \, {\left (8 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, {\left (5 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{315 \, d} \]

input
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algorith 
m="fricas")
 
output
-2/315*(15*I*sqrt(2)*(5*A + 6*B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + 
c) + I*sin(d*x + c)) - 15*I*sqrt(2)*(5*A + 6*B)*a^2*weierstrassPInverse(-4 
, 0, cos(d*x + c) - I*sin(d*x + c)) - 21*I*sqrt(2)*(8*A + 9*B)*a^2*weierst 
rassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) 
 + 21*I*sqrt(2)*(8*A + 9*B)*a^2*weierstrassZeta(-4, 0, weierstrassPInverse 
(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (35*A*a^2*cos(d*x + c)^4 + 45*(2 
*A + B)*a^2*cos(d*x + c)^3 + 14*(8*A + 9*B)*a^2*cos(d*x + c)^2 + 30*(5*A + 
 6*B)*a^2*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d
 
3.2.92.6 Sympy [F(-1)]

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

input
integrate((a+a*sec(d*x+c))**2*(A+B*sec(d*x+c))/sec(d*x+c)**(9/2),x)
 
output
Timed out
 
3.2.92.7 Maxima [F]

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

input
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algorith 
m="maxima")
 
output
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2/sec(d*x + c)^(9/2), 
x)
 
3.2.92.8 Giac [F]

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

input
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algorith 
m="giac")
 
output
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2/sec(d*x + c)^(9/2), 
x)
 
3.2.92.9 Mupad [F(-1)]

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

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