\(\int \frac {(a+a \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [539]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 41, antiderivative size = 182 \[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a (3 (A+B)+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (5 A+7 (B+C)) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a (A+B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a (5 A+7 (B+C)) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \] Output:

2/5*a*(3*A+3*B+5*C)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)) 
*sec(d*x+c)^(1/2)/d+2/21*a*(5*A+7*B+7*C)*cos(d*x+c)^(1/2)*InverseJacobiAM( 
1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+2/7*a*A*sin(d*x+c)/d/sec(d*x+c)^ 
(5/2)+2/5*a*(A+B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/21*a*(5*A+7*B+7*C)*sin(d 
*x+c)/d/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 4.16 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.10 \[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a e^{-i d x} \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (40 (5 A+7 (B+C)) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-56 i (3 A+3 B+5 C) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+2 \cos (c+d x) (84 i (3 A+3 B+5 C)+5 (23 A+28 (B+C)) \sin (c+d x)+42 (A+B) \sin (2 (c+d x))+15 A \sin (3 (c+d x)))\right )}{420 d} \] Input:

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

Output:

(a*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(40*(5*A + 7*(B + C))*Sqrt[C 
os[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (56*I)*(3*A + 3*B + 5*C)*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))] + 2*Cos[c + d*x]*((84*I)*(3*A + 3*B + 5*C) + 5*(23*A + 
28*(B + C))*Sin[c + d*x] + 42*(A + B)*Sin[2*(c + d*x)] + 15*A*Sin[3*(c + d 
*x)])))/(420*d*E^(I*d*x))
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.98, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.390, Rules used = {3042, 4562, 27, 3042, 4535, 3042, 4256, 3042, 4258, 3042, 3120, 4533, 3042, 4258, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4562

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 4533

\(\displaystyle \frac {1}{7} \left (\frac {7}{5} a (3 (A+B)+5 C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+a (5 A+7 (B+C)) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )+\frac {14 a (A+B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {7}{5} a (3 (A+B)+5 C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+a (5 A+7 (B+C)) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )+\frac {14 a (A+B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {1}{7} \left (\frac {7}{5} a (3 (A+B)+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+a (5 A+7 (B+C)) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )+\frac {14 a (A+B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {7}{5} a (3 (A+B)+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+a (5 A+7 (B+C)) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )+\frac {14 a (A+B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

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

Input:

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

Output:

(2*a*A*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + ((14*a*(3*(A + B) + 5*C)*S 
qrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (1 
4*a*(A + B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + a*(5*A + 7*(B + C))*( 
(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
 

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 4533
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + 
Simp[(C*m + A*(m + 1))/(b^2*m)   Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr 
eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
 

rule 4535
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

rule 4562
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si 
mp[1/(d*n)   Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* 
b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ 
{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(480\) vs. \(2(164)=328\).

Time = 17.88 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.64

method result size
default \(-\frac {2 \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 \left (240 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-528 A -168 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (448 A +308 B +140 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-122 A -112 B -70 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+25 A \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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 A \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}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+35 B \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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 B \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}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+35 C \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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 C \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}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \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}\) \(481\)
parts \(\text {Expression too large to display}\) \(729\)

Input:

int((a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x,me 
thod=_RETURNVERBOSE)
 

Output:

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(240*A*co 
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+(-528*A-168*B)*sin(1/2*d*x+1/2*c)^6* 
cos(1/2*d*x+1/2*c)+(448*A+308*B+140*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/ 
2*c)+(-122*A-112*B-70*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+25*A*(sin 
(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/ 
2*d*x+1/2*c),2^(1/2))-63*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2 
*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+35*B*(sin(1/2*d*x+1/2 
*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c) 
,2^(1/2))-63*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/ 
2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+35*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-10 
5*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)*Elliptic 
E(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
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.15 \[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {-5 i \, \sqrt {2} {\left (5 \, A + 7 \, B + 7 \, C\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (5 \, A + 7 \, B + 7 \, C\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (3 \, A + 3 \, B + 5 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (3 \, A + 3 \, B + 5 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (15 \, A a \cos \left (d x + c\right )^{3} + 21 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{2} + 5 \, {\left (5 \, A + 7 \, B + 7 \, C\right )} a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d} \] Input:

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

Output:

1/105*(-5*I*sqrt(2)*(5*A + 7*B + 7*C)*a*weierstrassPInverse(-4, 0, cos(d*x 
 + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*(5*A + 7*B + 7*C)*a*weierstrassPInve 
rse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*I*sqrt(2)*(3*A + 3*B + 5*C) 
*a*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin( 
d*x + c))) - 21*I*sqrt(2)*(3*A + 3*B + 5*C)*a*weierstrassZeta(-4, 0, weier 
strassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(15*A*a*cos(d*x 
+ c)^3 + 21*(A + B)*a*cos(d*x + c)^2 + 5*(5*A + 7*B + 7*C)*a*cos(d*x + c)) 
*sin(d*x + c)/sqrt(cos(d*x + c)))/d
 

Sympy [F]

\[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=a \left (\int \frac {A}{\sec ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx + \int \frac {A}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C}{\sqrt {\sec {\left (c + d x \right )}}}\, dx\right ) \] Input:

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

Output:

a*(Integral(A/sec(c + d*x)**(7/2), x) + Integral(A/sec(c + d*x)**(5/2), x) 
 + Integral(B/sec(c + d*x)**(5/2), x) + Integral(B/sec(c + d*x)**(3/2), x) 
 + Integral(C/sec(c + d*x)**(3/2), x) + Integral(C/sqrt(sec(c + d*x)), x))
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=a \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) c \right ) \] Input:

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

Output:

a*(int(sqrt(sec(c + d*x))/sec(c + d*x)**4,x)*a + int(sqrt(sec(c + d*x))/se 
c(c + d*x)**3,x)*a + int(sqrt(sec(c + d*x))/sec(c + d*x)**3,x)*b + int(sqr 
t(sec(c + d*x))/sec(c + d*x)**2,x)*b + int(sqrt(sec(c + d*x))/sec(c + d*x) 
**2,x)*c + int(sqrt(sec(c + d*x))/sec(c + d*x),x)*c)