\(\int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [499]

Optimal result

Integrand size = 43, antiderivative size = 263 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3/2} (133 A+150 B+176 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{128 d}+\frac {a^2 (133 A+150 B+176 C) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (133 A+150 B+176 C) \cos (c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (67 A+90 B+80 C) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt {a+a \sec (c+d x)}}+\frac {a (3 A+10 B) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d} \] Output:

1/128*a^(3/2)*(133*A+150*B+176*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c 
))^(1/2))/d+1/128*a^2*(133*A+150*B+176*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1 
/2)+1/192*a^2*(133*A+150*B+176*C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c)) 
^(1/2)+1/240*a^2*(67*A+90*B+80*C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c 
))^(1/2)+1/40*a*(3*A+10*B)*cos(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/ 
d+1/5*A*cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 2.22 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (693 C \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+693 C \cos (c+d x) \sqrt {1-\sec (c+d x)}+462 C \cos ^2(c+d x) \sqrt {1-\sec (c+d x)}+168 C \cos ^3(c+d x) \sqrt {1-\sec (c+d x)}-144 B \cos ^4(c+d x) \sqrt {1-\sec (c+d x)}-112 A \cos ^5(c+d x) \sqrt {1-\sec (c+d x)}+2160 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},5,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}+2128 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},6,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{504 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \] Input:

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

Output:

(a*(693*C*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 693*C*Cos[c + d*x]*Sqrt[1 - Se 
c[c + d*x]] + 462*C*Cos[c + d*x]^2*Sqrt[1 - Sec[c + d*x]] + 168*C*Cos[c + 
d*x]^3*Sqrt[1 - Sec[c + d*x]] - 144*B*Cos[c + d*x]^4*Sqrt[1 - Sec[c + d*x] 
] - 112*A*Cos[c + d*x]^5*Sqrt[1 - Sec[c + d*x]] + 2160*B*Hypergeometric2F1 
[1/2, 5, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]] + 2128*A*Hypergeome 
tric2F1[1/2, 6, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*(1 + 
 Sec[c + d*x])]*Sin[c + d*x])/(504*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d 
*x]])
 

Rubi [A] (verified)

Time = 1.56 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.99, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 4574, 27, 3042, 4505, 27, 3042, 4503, 3042, 4292, 3042, 4292, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

\(\displaystyle \frac {\int \frac {1}{2} \cos ^4(c+d x) (\sec (c+d x) a+a)^{3/2} (a (3 A+10 B)+5 a (A+2 C) \sec (c+d x))dx}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \cos ^4(c+d x) (\sec (c+d x) a+a)^{3/2} (a (3 A+10 B)+5 a (A+2 C) \sec (c+d x))dx}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (3 A+10 B)+5 a (A+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4505

\(\displaystyle \frac {\frac {1}{4} \int \frac {1}{2} \cos ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left ((67 A+90 B+80 C) a^2+5 (11 A+10 B+16 C) \sec (c+d x) a^2\right )dx+\frac {a^2 (3 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{8} \int \cos ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left ((67 A+90 B+80 C) a^2+5 (11 A+10 B+16 C) \sec (c+d x) a^2\right )dx+\frac {a^2 (3 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((67 A+90 B+80 C) a^2+5 (11 A+10 B+16 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 (3 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \int \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (67 A+90 B+80 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^3 (67 A+90 B+80 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4292

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (67 A+90 B+80 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (67 A+90 B+80 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4292

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (67 A+90 B+80 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (67 A+90 B+80 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \left (\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (67 A+90 B+80 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (3 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {a^2 (3 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}+\frac {1}{8} \left (\frac {a^3 (67 A+90 B+80 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {5}{6} a^2 (133 A+150 B+176 C) \left (\frac {3}{4} \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\right )}{10 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

Input:

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

Output:

(A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + ((a^2*( 
3*A + 10*B)*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d) + 
((a^3*(67*A + 90*B + 80*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Se 
c[c + d*x]]) + (5*a^2*(133*A + 150*B + 176*C)*((a*Cos[c + d*x]*Sin[c + d*x 
])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (3*((Sqrt[a]*ArcTan[(Sqrt[a]*Tan[c + d 
*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + 
 d*x]])))/4))/6)/8)/(10*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 4292

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a 
+ b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*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}, x] && 
 EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
 

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 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] (warning: unable to verify)

Time = 25.32 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.60

method	result	size
default	\(-\frac {a \left (\left (1995 \cos \left (d x +c \right )+1995\right ) A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (2250 \cos \left (d x +c \right )+2250\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (2640 \cos \left (d x +c \right )+2640\right ) C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-384 \cos \left (d x +c \right )^{4}-912 \cos \left (d x +c \right )^{3}-1064 \cos \left (d x +c \right )^{2}-1330 \cos \left (d x +c \right )-1995\right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-480 \cos \left (d x +c \right )^{3}-1200 \cos \left (d x +c \right )^{2}-1500 \cos \left (d x +c \right )-2250\right ) B +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-640 \cos \left (d x +c \right )^{2}-1760 \cos \left (d x +c \right )-2640\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{1920 d \left (\cos \left (d x +c \right )+1\right )}\)	\(420\)

Input:

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

Output:

-1/1920/d*a*((1995*cos(d*x+c)+1995)*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a 
rctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x 
+c)+cot(d*x+c)^2-1)^(1/2))+(2250*cos(d*x+c)+2250)*B*(-cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc( 
d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2))+(2640*cos(d*x+c)+2640)*C*(-cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d 
*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2))+sin(d*x+c)*cos(d*x+ 
c)*(-384*cos(d*x+c)^4-912*cos(d*x+c)^3-1064*cos(d*x+c)^2-1330*cos(d*x+c)-1 
995)*A+sin(d*x+c)*cos(d*x+c)*(-480*cos(d*x+c)^3-1200*cos(d*x+c)^2-1500*cos 
(d*x+c)-2250)*B+sin(d*x+c)*cos(d*x+c)*(-640*cos(d*x+c)^2-1760*cos(d*x+c)-2 
640)*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 462, normalized size of antiderivative = 1.76 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {15 \, {\left ({\left (133 \, A + 150 \, B + 176 \, C\right )} a \cos \left (d x + c\right ) + {\left (133 \, A + 150 \, B + 176 \, C\right )} a\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 (384 \, A a \cos \left (d x + c\right )^{5} + 48 \, {\left (19 \, A + 10 \, B\right )} a \cos \left (d x + c\right )^{4} + 8 \, {\left (133 \, A + 150 \, B + 80 \, C\right )} a \cos \left (d x + c\right )^{3} + 10 \, {\left (133 \, A + 150 \, B + 176 \, C\right )} a \cos \left (d x + c\right )^{2} + 15 \, {\left (133 \, A + 150 \, B + 176 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3840 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {15 \, {\left ({\left (133 \, A + 150 \, B + 176 \, C\right )} a \cos \left (d x + c\right ) + {\left (133 \, A + 150 \, B + 176 \, C\right )} a\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 (384 \, A a \cos \left (d x + c\right )^{5} + 48 \, {\left (19 \, A + 10 \, B\right )} a \cos \left (d x + c\right )^{4} + 8 \, {\left (133 \, A + 150 \, B + 80 \, C\right )} a \cos \left (d x + c\right )^{3} + 10 \, {\left (133 \, A + 150 \, B + 176 \, C\right )} a \cos \left (d x + c\right )^{2} + 15 \, {\left (133 \, A + 150 \, B + 176 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{1920 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

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

Output:

[1/3840*(15*((133*A + 150*B + 176*C)*a*cos(d*x + c) + (133*A + 150*B + 176 
*C)*a)*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*(384*A*a*cos(d*x + c)^5 + 48*(19*A + 10*B)*a*cos(d*x + c) 
^4 + 8*(133*A + 150*B + 80*C)*a*cos(d*x + c)^3 + 10*(133*A + 150*B + 176*C 
)*a*cos(d*x + c)^2 + 15*(133*A + 150*B + 176*C)*a*cos(d*x + c))*sqrt((a*co 
s(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/1920* 
(15*((133*A + 150*B + 176*C)*a*cos(d*x + c) + (133*A + 150*B + 176*C)*a)*s 
qrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a 
)*sin(d*x + c))) - (384*A*a*cos(d*x + c)^5 + 48*(19*A + 10*B)*a*cos(d*x + 
c)^4 + 8*(133*A + 150*B + 80*C)*a*cos(d*x + c)^3 + 10*(133*A + 150*B + 176 
*C)*a*cos(d*x + c)^2 + 15*(133*A + 150*B + 176*C)*a*cos(d*x + c))*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. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-1)]

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

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^(3/2)*(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. 1953 vs. \(2 (235) = 470\).

Time = 1.32 (sec) , antiderivative size = 1953, normalized size of antiderivative = 7.43 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/3840*(15*(133*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 150*B*sqrt(-a)*a*sgn(cos 
(d*x + c)) + 176*C*sqrt(-a)*a*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) 
)) - 15*(133*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 150*B*sqrt(-a)*a*sgn(cos(d*x 
 + c)) + 176*C*sqrt(-a)*a*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))) + 
 4*(1995*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/ 
2*c)^2 + a))^18*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 2250*sqrt(2)*(sqrt(-a)* 
tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^18*B*sqrt(-a)* 
a^2*sgn(cos(d*x + c)) + 2640*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt 
(-a*tan(1/2*d*x + 1/2*c)^2 + a))^18*C*sqrt(-a)*a^2*sgn(cos(d*x + c)) - 385 
05*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 
 + a))^16*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 76110*sqrt(2)*(sqrt(-a)*tan(1 
/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^16*B*sqrt(-a)*a^3*s 
gn(cos(d*x + c)) - 55920*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a* 
tan(1/2*d*x + 1/2*c)^2 + a))^16*C*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 561660* 
sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + 
a))^14*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 737160*sqrt(2)*(sqrt(-a)*tan(1/2 
*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*B*sqrt(-a)*a^4*sgn 
(cos(d*x + c)) + 582720*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a*(int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**5*sec(c + d*x)**3,x)*c 
 + int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**5*sec(c + d*x)**2,x)*b + int(s 
qrt(sec(c + d*x) + 1)*cos(c + d*x)**5*sec(c + d*x)**2,x)*c + int(sqrt(sec( 
c + d*x) + 1)*cos(c + d*x)**5*sec(c + d*x),x)*a + int(sqrt(sec(c + d*x) + 
1)*cos(c + d*x)**5*sec(c + d*x),x)*b + int(sqrt(sec(c + d*x) + 1)*cos(c + 
d*x)**5,x)*a)