\(\int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [506]

Optimal result

Integrand size = 43, antiderivative size = 207 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (25 A+38 B+40 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (3 A+2 B-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a (5 A+6 B) \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d} \] Output:

1/8*a^(5/2)*(25*A+38*B+40*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1 
/2))/d+1/24*a^3*(49*A+54*B-24*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/4*a 
^2*(3*A+2*B-8*C)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d+1/12*a*(5*A+6*B)*cos( 
d*x+c)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+a*sec(d*x 
+c))^(5/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.78 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.50 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {a^2 \cos (c+d x) \sqrt {a (1+\sec (c+d x))} \left (6 B \sqrt {1-\sec (c+d x)} \sin (c+d x)-3 C \sqrt {1-\sec (c+d x)} \sin (c+d x)+4 A \cos ^2(c+d x) \sqrt {1-\sec (c+d x)} \sin (c+d x)+A \sqrt {1-\sec (c+d x)} \sin (2 (c+d x))+4 B \sqrt {1-\sec (c+d x)} \sin (2 (c+d x))-15 C \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)-6 C \sqrt {1-\sec (c+d x)} \tan (c+d x)-38 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)} \tan (c+d x)-30 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)} \tan (c+d x)\right )}{3 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \] Input:

Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + 
C*Sec[c + d*x]^2),x]
 

Output:

-1/3*(a^2*Cos[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(6*B*Sqrt[1 - Sec[c + d* 
x]]*Sin[c + d*x] - 3*C*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 4*A*Cos[c + d 
*x]^2*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + A*Sqrt[1 - Sec[c + d*x]]*Sin[2 
*(c + d*x)] + 4*B*Sqrt[1 - Sec[c + d*x]]*Sin[2*(c + d*x)] - 15*C*ArcTanh[S 
qrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 6*C*Sqrt[1 - Sec[c + d*x]]*Tan[c + d 
*x] - 38*B*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c 
 + d*x]]*Tan[c + d*x] - 30*A*Hypergeometric2F1[1/2, 4, 3/2, 1 - Sec[c + d* 
x]]*Sqrt[1 - Sec[c + d*x]]*Tan[c + d*x]))/(d*(1 + Cos[c + d*x])*Sqrt[1 - S 
ec[c + d*x]])
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 215, 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, 4505, 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.

\(\displaystyle \int \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

\(\displaystyle \frac {\int \frac {1}{2} \cos ^2(c+d x) (\sec (c+d x) a+a)^{5/2} (a (5 A+6 B)-a (A-6 C) \sec (c+d x))dx}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \cos ^2(c+d x) (\sec (c+d x) a+a)^{5/2} (a (5 A+6 B)-a (A-6 C) \sec (c+d x))dx}{6 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

\(\displaystyle \frac {\frac {1}{2} \int \frac {1}{2} \cos (c+d x) (\sec (c+d x) a+a)^{3/2} \left (a^2 (31 A+42 B+24 C)-3 a^2 (3 A+2 B-8 C) \sec (c+d x)\right )dx+\frac {a^2 (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{4} \int \cos (c+d x) (\sec (c+d x) a+a)^{3/2} \left (a^2 (31 A+42 B+24 C)-3 a^2 (3 A+2 B-8 C) \sec (c+d x)\right )dx+\frac {a^2 (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a^2 (31 A+42 B+24 C)-3 a^2 (3 A+2 B-8 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\frac {1}{4} \left (2 \int \frac {1}{2} \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left ((49 A+54 B-24 C) a^3+(13 A+30 B+72 C) \sec (c+d x) a^3\right )dx-\frac {6 a^3 (3 A+2 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{4} \left (\int \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left ((49 A+54 B-24 C) a^3+(13 A+30 B+72 C) \sec (c+d x) a^3\right )dx-\frac {6 a^3 (3 A+2 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((49 A+54 B-24 C) a^3+(13 A+30 B+72 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {6 a^3 (3 A+2 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {\frac {1}{4} \left (\frac {3}{2} a^3 (25 A+38 B+40 C) \int \sqrt {\sec (c+d x) a+a}dx+\frac {a^4 (49 A+54 B-24 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^3 (3 A+2 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (\frac {3}{2} a^3 (25 A+38 B+40 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^4 (49 A+54 B-24 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^3 (3 A+2 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {1}{4} \left (-\frac {3 a^4 (25 A+38 B+40 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {a^4 (49 A+54 B-24 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^3 (3 A+2 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {a^2 (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}+\frac {1}{4} \left (\frac {3 a^{7/2} (25 A+38 B+40 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^4 (49 A+54 B-24 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^3 (3 A+2 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )}{6 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}\)

Input:

Int[Cos[c + d*x]^3*(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]^2*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(3*d) + ((a^2*( 
5*A + 6*B)*Cos[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d) + ( 
(3*a^(7/2)*(25*A + 38*B + 40*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*S 
ec[c + d*x]]])/d + (a^4*(49*A + 54*B - 24*C)*Sin[c + d*x])/(d*Sqrt[a + a*S 
ec[c + d*x]]) - (6*a^3*(3*A + 2*B - 8*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + 
d*x])/d)/4)/(6*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 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]
 

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 [A] (verified)

Time = 268.51 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.77

method	result	size
default	\(\frac {a^{2} \left (\left (75 \cos \left (d x +c \right )+75\right ) A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (114 \cos \left (d x +c \right )+114\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (120 \cos \left (d x +c \right )+120\right ) C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (8 \cos \left (d x +c \right )^{2}+34 \cos \left (d x +c \right )+75\right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (12 \cos \left (d x +c \right )+66\right ) B +\left (24 \cos \left (d x +c \right )+48\right ) \sin \left (d x +c \right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{24 d \left (\cos \left (d x +c \right )+1\right )}\)	\(366\)

Input:

int(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, 
method=_RETURNVERBOSE)
 

Output:

1/24/d*a^2*((75*cos(d*x+c)+75)*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan 
h(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)))+(114*cos(d*x+c)+114)*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)))+(120*cos(d*x+c)+120)*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)*(8*cos(d* 
x+c)^2+34*cos(d*x+c)+75)*A+sin(d*x+c)*cos(d*x+c)*(12*cos(d*x+c)+66)*B+(24* 
cos(d*x+c)+48)*sin(d*x+c)*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.98 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left ({\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (25 \, A + 22 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (25 \, A + 22 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="fricas")
 

Output:

[1/48*(3*((25*A + 38*B + 40*C)*a^2*cos(d*x + c) + (25*A + 38*B + 40*C)*a^2 
)*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*(8*A*a^2*cos(d*x + c)^3 + 2*(17*A + 6*B)*a^2*cos(d*x + c)^2 + 
3*(25*A + 22*B + 8*C)*a^2*cos(d*x + c) + 48*C*a^2)*sqrt((a*cos(d*x + c) + 
a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/24*(3*((25*A + 38* 
B + 40*C)*a^2*cos(d*x + c) + (25*A + 38*B + 40*C)*a^2)*sqrt(a)*arctan(sqrt 
((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 
 (8*A*a^2*cos(d*x + c)^3 + 2*(17*A + 6*B)*a^2*cos(d*x + c)^2 + 3*(25*A + 2 
2*B + 8*C)*a^2*cos(d*x + c) + 48*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
+ c))*sin(d*x + c))/(d*cos(d*x + c) + d)]
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**3*(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. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^3*(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. 1319 vs. \(2 (183) = 366\).

Time = 1.42 (sec) , antiderivative size = 1319, normalized size of antiderivative = 6.37 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="giac")
 

Output:

-1/48*(96*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*C*a^3*sgn(cos(d*x + 
c))*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 - a) + 3*(25*A*sqrt(-a) 
*a^2*sgn(cos(d*x + c)) + 38*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 40*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*(25*A*sqrt(-a)*a 
^2*sgn(cos(d*x + c)) + 38*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 40*C*sqrt(-a) 
*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*t 
an(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*(75*sqrt(2)*(sqrt( 
-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*sqrt( 
-a)*a^3*sgn(cos(d*x + c)) + 114*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - s 
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 
72*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 
 + a))^10*C*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 1125*sqrt(2)*(sqrt(-a)*tan(1/ 
2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*sqrt(-a)*a^4*sgn 
(cos(d*x + c)) - 1710*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan 
(1/2*d*x + 1/2*c)^2 + a))^8*B*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 888*sqrt(2) 
*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*C 
*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 6174*sqrt(2)*(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^5*sgn(cos(d*x + 
c)) + 6804*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x...
 

Mupad [F(-1)]

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

int(cos(c + d*x)^3*(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)^3*(a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos( 
c + d*x)^2), x)
 

Reduce [F]

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

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