\(\int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [505]

Optimal result

Integrand size = 43, antiderivative size = 197 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (19 A+20 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-4 B-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d} \] Output:

1/4*a^(5/2)*(19*A+20*B+8*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/ 
2))/d+1/12*a^3*(27*A-12*B-56*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/2*a^ 
2*(A-4*B-8*C)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d-1/6*a*(3*A-4*C)*(a+a*sec 
(d*x+c))^(3/2)*sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a+a*sec(d*x+c))^(5/2)*sin(d* 
x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 6.70 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.77 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {a (1+\sec (c+d x))} \left (6 \sqrt {2} (19 A+20 B+8 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)+2 (33 A+12 B+16 C+(9 A+48 B+128 C) \cos (c+d x)+3 (11 A+4 B) \cos (2 (c+d x))+3 A \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \] Input:

Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + 
C*Sec[c + d*x]^2),x]
 

Output:

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(6*Sqrt[2]*( 
19*A + 20*B + 8*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(3/2) + 2 
*(33*A + 12*B + 16*C + (9*A + 48*B + 128*C)*Cos[c + d*x] + 3*(11*A + 4*B)* 
Cos[2*(c + d*x)] + 3*A*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)
 

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4574, 27, 3042, 4506, 27, 3042, 4506, 27, 3042, 4503, 3042, 4261, 216}

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.

\(\displaystyle \int \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

\(\displaystyle \frac {\int \frac {1}{2} \cos (c+d x) (\sec (c+d x) a+a)^{5/2} (a (5 A+4 B)-a (3 A-4 C) \sec (c+d x))dx}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \cos (c+d x) (\sec (c+d x) a+a)^{5/2} (a (5 A+4 B)-a (3 A-4 C) \sec (c+d x))dx}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+4 B)-a (3 A-4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\frac {2}{3} \int \frac {1}{2} \cos (c+d x) (\sec (c+d x) a+a)^{3/2} \left (a^2 (21 A+12 B-8 C)-3 a^2 (A-4 B-8 C) \sec (c+d x)\right )dx-\frac {2 a^2 (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \cos (c+d x) (\sec (c+d x) a+a)^{3/2} \left (a^2 (21 A+12 B-8 C)-3 a^2 (A-4 B-8 C) \sec (c+d x)\right )dx-\frac {2 a^2 (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a^2 (21 A+12 B-8 C)-3 a^2 (A-4 B-8 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^2 (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\frac {1}{3} \left (2 \int \frac {1}{2} \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left ((27 A-12 B-56 C) a^3+(15 A+36 B+40 C) \sec (c+d x) a^3\right )dx-\frac {6 a^3 (A-4 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {2 a^2 (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \left (\int \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left ((27 A-12 B-56 C) a^3+(15 A+36 B+40 C) \sec (c+d x) a^3\right )dx-\frac {6 a^3 (A-4 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {2 a^2 (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((27 A-12 B-56 C) a^3+(15 A+36 B+40 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {6 a^3 (A-4 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {2 a^2 (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} a^3 (19 A+20 B+8 C) \int \sqrt {\sec (c+d x) a+a}dx+\frac {a^4 (27 A-12 B-56 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^3 (A-4 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {2 a^2 (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} a^3 (19 A+20 B+8 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^4 (27 A-12 B-56 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^3 (A-4 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {2 a^2 (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {1}{3} \left (-\frac {3 a^4 (19 A+20 B+8 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {a^4 (27 A-12 B-56 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^3 (A-4 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {2 a^2 (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 a^{7/2} (19 A+20 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^4 (27 A-12 B-56 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^3 (A-4 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {2 a^2 (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d}\)

Input:

Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[ 
c + d*x]^2),x]
 

Output:

(A*Cos[c + d*x]*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(2*d) + ((-2*a^2* 
(3*A - 4*C)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + ((3*a^(7/2)*( 
19*A + 20*B + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]] 
)/d + (a^4*(27*A - 12*B - 56*C)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) 
 - (6*a^3*(A - 4*B - 8*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d)/3)/(4* 
a)
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4261

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 4503

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

rule 4506

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

rule 4574

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 
_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e 
 + f*x])^n/(f*n)), x] - Simp[1/(b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[ 
e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] 
, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] 
&&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(350\) vs. \(2(173)=346\).

Time = 83.43 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.78

method	result	size
default	\(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (57 \cos \left (d x +c \right )+57\right ) A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (60 \cos \left (d x +c \right )+60\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (24 \cos \left (d x +c \right )+24\right ) C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (6 \cos \left (d x +c \right )+33\right ) A +\sin \left (d x +c \right ) \left (12 \cos \left (d x +c \right )+24\right ) B +C \left (64 \sin \left (d x +c \right )+8 \tan \left (d x +c \right )\right )\right )}{12 d \left (\cos \left (d x +c \right )+1\right )}\)	\(351\)

Input:

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, 
method=_RETURNVERBOSE)
 

Output:

1/12/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*((57*cos(d*x+c)+57)*A*( 
-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)/(csc(d*x+c)^2-2*csc(d*x+ 
c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c)))+(60*cos(d*x+c 
)+60)*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)/(csc(d*x+c)^2-2 
*csc(d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c)))+(24* 
cos(d*x+c)+24)*C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)/(csc(d 
*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+ 
c)))+sin(d*x+c)*cos(d*x+c)*(6*cos(d*x+c)+33)*A+sin(d*x+c)*(12*cos(d*x+c)+2 
4)*B+C*(64*sin(d*x+c)+8*tan(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 438, normalized size of antiderivative = 2.22 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left ({\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (11 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (3 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, -\frac {3 \, {\left ({\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (11 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (3 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \] Input:

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="fricas")
 

Output:

[1/24*(3*((19*A + 20*B + 8*C)*a^2*cos(d*x + c)^2 + (19*A + 20*B + 8*C)*a^2 
*cos(d*x + c))*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d 
*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a) 
/(cos(d*x + c) + 1)) + 2*(6*A*a^2*cos(d*x + c)^3 + 3*(11*A + 4*B)*a^2*cos( 
d*x + c)^2 + 8*(3*B + 8*C)*a^2*cos(d*x + c) + 8*C*a^2)*sqrt((a*cos(d*x + c 
) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), -1 
/12*(3*((19*A + 20*B + 8*C)*a^2*cos(d*x + c)^2 + (19*A + 20*B + 8*C)*a^2*c 
os(d*x + c))*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d* 
x + c)/(sqrt(a)*sin(d*x + c))) - (6*A*a^2*cos(d*x + c)^3 + 3*(11*A + 4*B)* 
a^2*cos(d*x + c)^2 + 8*(3*B + 8*C)*a^2*cos(d*x + c) + 8*C*a^2)*sqrt((a*cos 
(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + 
 c))]
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+ 
c)**2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 811 vs. \(2 (173) = 346\).

Time = 1.27 (sec) , antiderivative size = 811, normalized size of antiderivative = 4.12 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="giac")
 

Output:

-1/24*(3*(19*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 20*B*sqrt(-a)*a^2*sgn(cos( 
d*x + c)) + 8*C*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2* 
d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3)) 
) - 3*(19*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 20*B*sqrt(-a)*a^2*sgn(cos(d*x 
 + c)) + 8*C*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x 
 + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 
 16*(3*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 9*sqrt(2)*C*a^4*sgn(cos(d*x + c)) 
 - (3*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 7*sqrt(2)*C*a^4*sgn(cos(d*x + c))) 
*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - 
 a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) + 12*sqrt(2)*(19*(sqrt(-a)*tan(1/ 
2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^3*sgn 
(cos(d*x + c)) + 12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 
 1/2*c)^2 + a))^6*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 171*(sqrt(-a)*tan(1/2 
*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^4*sgn( 
cos(d*x + c)) - 76*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 
1/2*c)^2 + a))^4*B*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 89*(sqrt(-a)*tan(1/2*d 
*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^5*sgn(co 
s(d*x + c)) + 36*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/ 
2*c)^2 + a))^2*B*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 9*A*sqrt(-a)*a^6*sgn(cos 
(d*x + c)) - 4*B*sqrt(-a)*a^6*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x...
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \] Input:

int(cos(c + d*x)^2*(a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos( 
c + d*x)^2),x)
                                                                                    
                                                                                    
 

Output:

int(cos(c + d*x)^2*(a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos( 
c + d*x)^2), x)
 

Reduce [F]

\[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) b +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**4,x 
)*c + int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**3,x)*b + 2* 
int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**3,x)*c + int(sqrt 
(sec(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**2,x)*a + 2*int(sqrt(sec(c 
 + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**2,x)*b + int(sqrt(sec(c + d*x) 
+ 1)*cos(c + d*x)**2*sec(c + d*x)**2,x)*c + 2*int(sqrt(sec(c + d*x) + 1)*c 
os(c + d*x)**2*sec(c + d*x),x)*a + int(sqrt(sec(c + d*x) + 1)*cos(c + d*x) 
**2*sec(c + d*x),x)*b + int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**2,x)*a)