3.2.43 \(\int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [143]

3.2.43.1 Optimal result
3.2.43.2 Mathematica [C] (verified)
3.2.43.3 Rubi [A] (verified)
3.2.43.4 Maple [A] (verified)
3.2.43.5 Fricas [A] (verification not implemented)
3.2.43.6 Sympy [F(-1)]
3.2.43.7 Maxima [F(-1)]
3.2.43.8 Giac [F]
3.2.43.9 Mupad [F(-1)]

3.2.43.1 Optimal result

Integrand size = 33, antiderivative size = 254 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (283 A+326 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{128 d}+\frac {a^3 (283 A+326 B) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (283 A+326 B) \cos (c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (157 A+170 B) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 A+10 B) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d} \]

output
1/128*a^(5/2)*(283*A+326*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/ 
2))/d+1/5*a*A*cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/128*a^3*( 
283*A+326*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/192*a^3*(283*A+326*B)*c 
os(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/240*a^3*(157*A+170*B)*cos( 
d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/40*a^2*(13*A+10*B)*cos(d*x+ 
c)^3*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
 
3.2.43.2 Mathematica [C] (verified)

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

Time = 1.25 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.64 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^2 \left (25935 A \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+28350 B \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+11651 A \sqrt {1-\sec (c+d x)}+9702 B \sqrt {1-\sec (c+d x)}+37029 A \cos (c+d x) \sqrt {1-\sec (c+d x)}+35658 B \cos (c+d x) \sqrt {1-\sec (c+d x)}+12653 A \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+9786 B \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+3818 A \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}+2436 B \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}+1002 A \cos (4 (c+d x)) \sqrt {1-\sec (c+d x)}+84 B \cos (4 (c+d x)) \sqrt {1-\sec (c+d x)}+72 A \cos (5 (c+d x)) \sqrt {1-\sec (c+d x)}+21504 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},5,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}+15360 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},6,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{13440 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]

input
Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x 
]
 
output
(a^2*(25935*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 28350*B*ArcTanh[Sqrt[1 - S 
ec[c + d*x]]] + 11651*A*Sqrt[1 - Sec[c + d*x]] + 9702*B*Sqrt[1 - Sec[c + d 
*x]] + 37029*A*Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] + 35658*B*Cos[c + d*x]* 
Sqrt[1 - Sec[c + d*x]] + 12653*A*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 
 9786*B*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 3818*A*Cos[3*(c + d*x)]* 
Sqrt[1 - Sec[c + d*x]] + 2436*B*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 
1002*A*Cos[4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 84*B*Cos[4*(c + d*x)]*Sqr 
t[1 - Sec[c + d*x]] + 72*A*Cos[5*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 21504 
*B*Hypergeometric2F1[1/2, 5, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]] 
 + 15360*A*Hypergeometric2F1[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])/(13440*d*(1 + Cos[c + d 
*x])*Sqrt[1 - Sec[c + d*x]])
 
3.2.43.3 Rubi [A] (verified)

Time = 1.50 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4505, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

\(\displaystyle \frac {1}{5} \int \frac {1}{2} \cos ^4(c+d x) (\sec (c+d x) a+a)^{3/2} (a (13 A+10 B)+5 a (A+2 B) \sec (c+d x))dx+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{10} \int \cos ^4(c+d x) (\sec (c+d x) a+a)^{3/2} (a (13 A+10 B)+5 a (A+2 B) \sec (c+d x))dx+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (13 A+10 B)+5 a (A+2 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4505

\(\displaystyle \frac {1}{10} \left (\frac {1}{4} \int \frac {1}{2} \cos ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left ((157 A+170 B) a^2+5 (21 A+26 B) \sec (c+d x) a^2\right )dx+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \int \cos ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left ((157 A+170 B) a^2+5 (21 A+26 B) \sec (c+d x) a^2\right )dx+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((157 A+170 B) a^2+5 (21 A+26 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (283 A+326 B) \int \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (157 A+170 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (283 A+326 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^3 (157 A+170 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4292

\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (283 A+326 B) \left (\frac {3}{4} \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (157 A+170 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (283 A+326 B) \left (\frac {3}{4} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (157 A+170 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4292

\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (283 A+326 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (157 A+170 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (283 A+326 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (157 A+170 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (283 A+326 B) \left (\frac {3}{4} \left (\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (157 A+170 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{10} \left (\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}+\frac {1}{8} \left (\frac {a^3 (157 A+170 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {5}{6} a^2 (283 A+326 B) \left (\frac {3}{4} \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\right )\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

input
Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]
 
output
(a*A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + ((a^2 
*(13*A + 10*B)*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d) 
 + ((a^3*(157*A + 170*B)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[ 
c + d*x]]) + (5*a^2*(283*A + 326*B)*((a*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sq 
rt[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
 

3.2.43.3.1 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]
 
3.2.43.4 Maple [A] (verified)

Time = 2.12 (sec) , antiderivative size = 437, normalized size of antiderivative = 1.72

\[\frac {a^{2} \left (384 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}+1392 A \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+480 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}+2264 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}+1840 B \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+4245 A \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+2830 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+4890 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+3260 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+4245 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+4245 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+4890 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+4890 B \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{1920 d \left (\cos \left (d x +c \right )+1\right )}\]

input
int(cos(d*x+c)^5*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)
 
output
1/1920*a^2/d*(384*A*sin(d*x+c)*cos(d*x+c)^5+1392*A*cos(d*x+c)^4*sin(d*x+c) 
+480*B*sin(d*x+c)*cos(d*x+c)^4+2264*A*sin(d*x+c)*cos(d*x+c)^3+1840*B*cos(d 
*x+c)^3*sin(d*x+c)+4245*A*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/( 
cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+2830*A 
*cos(d*x+c)^2*sin(d*x+c)+4890*B*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+ 
3260*B*sin(d*x+c)*cos(d*x+c)^2+4245*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a 
rctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+4245* 
A*cos(d*x+c)*sin(d*x+c)+4890*B*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x 
+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+4890*B*cos(d 
*x+c)*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
 
3.2.43.5 Fricas [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 460, normalized size of antiderivative = 1.81 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left [\frac {15 \, {\left ({\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (283 \, A + 326 \, B\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (384 \, A a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (29 \, A + 10 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (283 \, A + 230 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3840 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {15 \, {\left ({\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (283 \, A + 326 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (384 \, A a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (29 \, A + 10 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (283 \, A + 230 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{1920 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

input
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorith 
m="fricas")
 
output
[1/3840*(15*((283*A + 326*B)*a^2*cos(d*x + c) + (283*A + 326*B)*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*(384*A*a^2*cos(d*x + c)^5 + 48*(29*A + 10*B)*a^2*cos(d*x + c)^4 + 8*( 
283*A + 230*B)*a^2*cos(d*x + c)^3 + 10*(283*A + 326*B)*a^2*cos(d*x + c)^2 
+ 15*(283*A + 326*B)*a^2*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), -1/1920*(15*((283*A + 326*B)*a^2* 
cos(d*x + c) + (283*A + 326*B)*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))) - (384*A*a^2*cos(d*x 
 + c)^5 + 48*(29*A + 10*B)*a^2*cos(d*x + c)^4 + 8*(283*A + 230*B)*a^2*cos( 
d*x + c)^3 + 10*(283*A + 326*B)*a^2*cos(d*x + c)^2 + 15*(283*A + 326*B)*a^ 
2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*c 
os(d*x + c) + d)]
 
3.2.43.6 Sympy [F(-1)]

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

input
integrate(cos(d*x+c)**5*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)
 
output
Timed out
 
3.2.43.7 Maxima [F(-1)]

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

input
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorith 
m="maxima")
 
output
Timed out
 
3.2.43.8 Giac [F]

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

input
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorith 
m="giac")
 
output
sage0*x
 
3.2.43.9 Mupad [F(-1)]

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

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