3.1.26 \(\int (a+a \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\) [26]

3.1.26.1 Optimal result
3.1.26.2 Mathematica [A] (verified)
3.1.26.3 Rubi [A] (verified)
3.1.26.4 Maple [A] (verified)
3.1.26.5 Fricas [A] (verification not implemented)
3.1.26.6 Sympy [F(-1)]
3.1.26.7 Maxima [A] (verification not implemented)
3.1.26.8 Giac [A] (verification not implemented)
3.1.26.9 Mupad [B] (verification not implemented)

3.1.26.1 Optimal result

Integrand size = 33, antiderivative size = 194 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a^3 (13 A+20 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (38 A+55 C) \tan (c+d x)}{15 d}+\frac {a^3 (109 A+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(11 A+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d} \]

output
1/8*a^3*(13*A+20*C)*arctanh(sin(d*x+c))/d+1/15*a^3*(38*A+55*C)*tan(d*x+c)/ 
d+1/120*a^3*(109*A+140*C)*sec(d*x+c)*tan(d*x+c)/d+1/30*(11*A+10*C)*(a^3+a^ 
3*cos(d*x+c))*sec(d*x+c)^2*tan(d*x+c)/d+3/20*A*(a^2+a^2*cos(d*x+c))^2*sec( 
d*x+c)^3*tan(d*x+c)/a/d+1/5*A*(a+a*cos(d*x+c))^3*sec(d*x+c)^4*tan(d*x+c)/d
 
3.1.26.2 Mathematica [A] (verified)

Time = 1.71 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.01 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {13 a^3 A \text {arctanh}(\sin (c+d x))}{8 d}+\frac {5 a^3 C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {4 a^3 A \tan (c+d x)}{d}+\frac {4 a^3 C \tan (c+d x)}{d}+\frac {13 a^3 A \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 a^3 C \sec (c+d x) \tan (c+d x)}{2 d}+\frac {3 a^3 A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 a^3 A \tan ^3(c+d x)}{3 d}+\frac {a^3 C \tan ^3(c+d x)}{3 d}+\frac {a^3 A \tan ^5(c+d x)}{5 d} \]

input
Integrate[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]
 
output
(13*a^3*A*ArcTanh[Sin[c + d*x]])/(8*d) + (5*a^3*C*ArcTanh[Sin[c + d*x]])/( 
2*d) + (4*a^3*A*Tan[c + d*x])/d + (4*a^3*C*Tan[c + d*x])/d + (13*a^3*A*Sec 
[c + d*x]*Tan[c + d*x])/(8*d) + (3*a^3*C*Sec[c + d*x]*Tan[c + d*x])/(2*d) 
+ (3*a^3*A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a^3*A*Tan[c + d*x]^3)/( 
3*d) + (a^3*C*Tan[c + d*x]^3)/(3*d) + (a^3*A*Tan[c + d*x]^5)/(5*d)
 
3.1.26.3 Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 3523, 3042, 3454, 3042, 3454, 3042, 3447, 3042, 3500, 3042, 3227, 3042, 4254, 24, 4257}

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.

\(\displaystyle \int \sec ^6(c+d x) (a \cos (c+d x)+a)^3 \left (A+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\)

\(\Big \downarrow \) 3523

\(\displaystyle \frac {\int (\cos (c+d x) a+a)^3 (3 a A+a (A+5 C) \cos (c+d x)) \sec ^5(c+d x)dx}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {1}{4} \int (\cos (c+d x) a+a)^2 \left (2 (11 A+10 C) a^2+(7 A+20 C) \cos (c+d x) a^2\right ) \sec ^4(c+d x)dx+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 (11 A+10 C) a^2+(7 A+20 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int (\cos (c+d x) a+a) \left ((109 A+140 C) a^3+(43 A+80 C) \cos (c+d x) a^3\right ) \sec ^3(c+d x)dx+\frac {2 (11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}\right )+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((109 A+140 C) a^3+(43 A+80 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 (11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}\right )+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int \left ((43 A+80 C) \cos ^2(c+d x) a^4+(109 A+140 C) a^4+\left ((43 A+80 C) a^4+(109 A+140 C) a^4\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {2 (11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}\right )+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int \frac {(43 A+80 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+(109 A+140 C) a^4+\left ((43 A+80 C) a^4+(109 A+140 C) a^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 (11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}\right )+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (8 (38 A+55 C) a^4+15 (13 A+20 C) \cos (c+d x) a^4\right ) \sec ^2(c+d x)dx+\frac {a^4 (109 A+140 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}\right )+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {8 (38 A+55 C) a^4+15 (13 A+20 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^4 (109 A+140 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}\right )+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (8 a^4 (38 A+55 C) \int \sec ^2(c+d x)dx+15 a^4 (13 A+20 C) \int \sec (c+d x)dx\right )+\frac {a^4 (109 A+140 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}\right )+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a^4 (13 A+20 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+8 a^4 (38 A+55 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {a^4 (109 A+140 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}\right )+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a^4 (13 A+20 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {8 a^4 (38 A+55 C) \int 1d(-\tan (c+d x))}{d}\right )+\frac {a^4 (109 A+140 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}\right )+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a^4 (13 A+20 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {8 a^4 (38 A+55 C) \tan (c+d x)}{d}\right )+\frac {a^4 (109 A+140 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}\right )+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {15 a^4 (13 A+20 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {8 a^4 (38 A+55 C) \tan (c+d x)}{d}\right )+\frac {a^4 (109 A+140 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}\right )+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d}\)

input
Int[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]
 
output
(A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((3*A*(a^2 
+ a^2*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((2*(11*A + 10* 
C)*(a^4 + a^4*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a^4*(10 
9*A + 140*C)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((15*a^4*(13*A + 20*C)*Arc 
Tanh[Sin[c + d*x]])/d + (8*a^4*(38*A + 55*C)*Tan[c + d*x])/d)/2)/3)/4)/(5* 
a)
 

3.1.26.3.1 Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3447
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3454
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 
])
 

rule 3500
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

rule 3523
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + 
 f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - 
d^2))   Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a 
*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* 
(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - 
d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
 

rule 4254
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
3.1.26.4 Maple [A] (verified)

Time = 10.40 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.09

method result size
parts \(-\frac {A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,a^{3}+3 C \,a^{3}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (3 A \,a^{3}+C \,a^{3}\right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{3}}{d}+\frac {3 A \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {3 C \,a^{3} \tan \left (d x +c \right )}{d}\) \(211\)
parallelrisch \(\frac {40 \left (-\frac {39 \left (A +\frac {20 C}{13}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{32}+\frac {39 \left (A +\frac {20 C}{13}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{32}+\left (\frac {15 A}{16}+\frac {9 C}{20}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {19 A}{20}+\frac {37 C}{40}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {39 A}{160}+\frac {9 C}{40}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {19 A}{100}+\frac {11 C}{40}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +\frac {13 C}{20}\right )\right ) a^{3}}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) \(217\)
derivativedivides \(\frac {A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-3 A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 C \,a^{3} \tan \left (d x +c \right )+3 A \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 C \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-C \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) \(242\)
default \(\frac {A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-3 A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 C \,a^{3} \tan \left (d x +c \right )+3 A \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 C \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-C \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) \(242\)
risch \(-\frac {i a^{3} \left (195 A \,{\mathrm e}^{9 i \left (d x +c \right )}+180 C \,{\mathrm e}^{9 i \left (d x +c \right )}-360 C \,{\mathrm e}^{8 i \left (d x +c \right )}+750 A \,{\mathrm e}^{7 i \left (d x +c \right )}+360 C \,{\mathrm e}^{7 i \left (d x +c \right )}-720 A \,{\mathrm e}^{6 i \left (d x +c \right )}-1680 C \,{\mathrm e}^{6 i \left (d x +c \right )}-2320 A \,{\mathrm e}^{4 i \left (d x +c \right )}-2720 C \,{\mathrm e}^{4 i \left (d x +c \right )}-750 A \,{\mathrm e}^{3 i \left (d x +c \right )}-360 C \,{\mathrm e}^{3 i \left (d x +c \right )}-1520 A \,{\mathrm e}^{2 i \left (d x +c \right )}-1840 C \,{\mathrm e}^{2 i \left (d x +c \right )}-195 A \,{\mathrm e}^{i \left (d x +c \right )}-180 C \,{\mathrm e}^{i \left (d x +c \right )}-304 A -440 C \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {13 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {13 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\) \(299\)

input
int((a+cos(d*x+c)*a)^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x,method=_RETURNVER 
BOSE)
 
output
-A*a^3/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+(A*a^3+3*C* 
a^3)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-(3*A*a^3+ 
C*a^3)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+1/d*C*ln(sec(d*x+c)+tan(d*x+c) 
)*a^3+3*A*a^3/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec 
(d*x+c)+tan(d*x+c)))+3*C*a^3/d*tan(d*x+c)
 
3.1.26.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.83 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (13 \, A + 20 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (13 \, A + 20 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (38 \, A + 55 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \, {\left (13 \, A + 12 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (19 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 90 \, A a^{3} \cos \left (d x + c\right ) + 24 \, A a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

input
integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm= 
"fricas")
 
output
1/240*(15*(13*A + 20*C)*a^3*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(13* 
A + 20*C)*a^3*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8*(38*A + 55*C)*a 
^3*cos(d*x + c)^4 + 15*(13*A + 12*C)*a^3*cos(d*x + c)^3 + 8*(19*A + 5*C)*a 
^3*cos(d*x + c)^2 + 90*A*a^3*cos(d*x + c) + 24*A*a^3)*sin(d*x + c))/(d*cos 
(d*x + c)^5)
 
3.1.26.6 Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]

input
integrate((a+a*cos(d*x+c))**3*(A+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)
 
output
Timed out
 
3.1.26.7 Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.51 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 45 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 720 \, C a^{3} \tan \left (d x + c\right )}{240 \, d} \]

input
integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm= 
"maxima")
 
output
1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 
 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 80*(tan(d*x + c)^3 + 3*tan( 
d*x + c))*C*a^3 - 45*A*a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x 
 + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x 
+ c) - 1)) - 60*A*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + 
 c) + 1) + log(sin(d*x + c) - 1)) - 180*C*a^3*(2*sin(d*x + c)/(sin(d*x + c 
)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*C*a^3*(log 
(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 720*C*a^3*tan(d*x + c))/d
 
3.1.26.8 Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.27 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (13 \, A a^{3} + 20 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (13 \, A a^{3} + 20 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (195 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 300 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 910 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1400 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1664 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2560 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1330 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 765 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 660 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

input
integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm= 
"giac")
 
output
1/120*(15*(13*A*a^3 + 20*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(1 
3*A*a^3 + 20*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(195*A*a^3*tan( 
1/2*d*x + 1/2*c)^9 + 300*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 910*A*a^3*tan(1/2* 
d*x + 1/2*c)^7 - 1400*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 1664*A*a^3*tan(1/2*d* 
x + 1/2*c)^5 + 2560*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 1330*A*a^3*tan(1/2*d*x 
+ 1/2*c)^3 - 2120*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 765*A*a^3*tan(1/2*d*x + 1 
/2*c) + 660*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
 
3.1.26.9 Mupad [B] (verification not implemented)

Time = 3.66 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.15 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (13\,A+20\,C\right )}{4\,d}-\frac {\left (\frac {13\,A\,a^3}{4}+5\,C\,a^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {91\,A\,a^3}{6}-\frac {70\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {416\,A\,a^3}{15}+\frac {128\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {133\,A\,a^3}{6}-\frac {106\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {51\,A\,a^3}{4}+11\,C\,a^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

input
int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^3)/cos(c + d*x)^6,x)
 
output
(a^3*atanh(tan(c/2 + (d*x)/2))*(13*A + 20*C))/(4*d) - (tan(c/2 + (d*x)/2)* 
((51*A*a^3)/4 + 11*C*a^3) + tan(c/2 + (d*x)/2)^9*((13*A*a^3)/4 + 5*C*a^3) 
- tan(c/2 + (d*x)/2)^7*((91*A*a^3)/6 + (70*C*a^3)/3) - tan(c/2 + (d*x)/2)^ 
3*((133*A*a^3)/6 + (106*C*a^3)/3) + tan(c/2 + (d*x)/2)^5*((416*A*a^3)/15 + 
 (128*C*a^3)/3))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10 
*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1 
))