\(\int (a+a \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [31]

Optimal result

Integrand size = 31, antiderivative size = 177 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {1}{2} a^4 (12 A+7 C) x+\frac {a^4 A \text {arctanh}(\sin (c+d x))}{d}+\frac {a^4 (10 A+7 C) \sin (c+d x)}{2 d}+\frac {a C (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(8 A+7 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d} \] Output:

1/2*a^4*(12*A+7*C)*x+a^4*A*arctanh(sin(d*x+c))/d+1/2*a^4*(10*A+7*C)*sin(d* 
x+c)/d+1/5*a*C*(a+a*cos(d*x+c))^3*sin(d*x+c)/d+1/5*C*(a+a*cos(d*x+c))^4*si 
n(d*x+c)/d+1/15*(5*A+7*C)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/d+1/6*(8*A+7*C 
)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 3.63 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.83 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {a^4 \left (1440 A d x+840 C d x-240 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+240 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+30 (54 A+49 C) \sin (c+d x)+240 (A+2 C) \sin (2 (c+d x))+20 A \sin (3 (c+d x))+145 C \sin (3 (c+d x))+30 C \sin (4 (c+d x))+3 C \sin (5 (c+d x))\right )}{240 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x],x]
 

Output:

(a^4*(1440*A*d*x + 840*C*d*x - 240*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/ 
2]] + 240*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 30*(54*A + 49*C)*Si 
n[c + d*x] + 240*(A + 2*C)*Sin[2*(c + d*x)] + 20*A*Sin[3*(c + d*x)] + 145* 
C*Sin[3*(c + d*x)] + 30*C*Sin[4*(c + d*x)] + 3*C*Sin[5*(c + d*x)]))/(240*d 
)
 

Rubi [A] (verified)

Time = 1.47 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.06, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.613, Rules used = {3042, 3525, 3042, 3455, 27, 3042, 3455, 27, 3042, 3455, 27, 3042, 3447, 3042, 3502, 3042, 3214, 3042, 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.

\(\displaystyle \int \sec (c+d x) (a \cos (c+d x)+a)^4 \left (A+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3525

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

\(\displaystyle \frac {\frac {1}{3} \int 5 (\cos (c+d x) a+a)^2 \left (3 A a^3+(8 A+7 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5}{3} \int (\cos (c+d x) a+a)^2 \left (3 A a^3+(8 A+7 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (3 A a^3+(8 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 3455

\(\displaystyle \frac {\frac {5}{3} \left (\frac {1}{2} \int 3 (\cos (c+d x) a+a) \left (2 A a^4+(10 A+7 C) \cos (c+d x) a^4\right ) \sec (c+d x)dx+\frac {(8 A+7 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5}{3} \left (\frac {3}{2} \int (\cos (c+d x) a+a) \left (2 A a^4+(10 A+7 C) \cos (c+d x) a^4\right ) \sec (c+d x)dx+\frac {(8 A+7 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{3} \left (\frac {3}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (2 A a^4+(10 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(8 A+7 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {5}{3} \left (\frac {3}{2} \int \left ((10 A+7 C) \cos ^2(c+d x) a^5+2 A a^5+\left (2 A a^5+(10 A+7 C) a^5\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {(8 A+7 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{3} \left (\frac {3}{2} \int \frac {(10 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^5+2 A a^5+\left (2 A a^5+(10 A+7 C) a^5\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(8 A+7 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {5}{3} \left (\frac {3}{2} \left (\int \left (2 A a^5+(12 A+7 C) \cos (c+d x) a^5\right ) \sec (c+d x)dx+\frac {a^5 (10 A+7 C) \sin (c+d x)}{d}\right )+\frac {(8 A+7 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{3} \left (\frac {3}{2} \left (\int \frac {2 A a^5+(12 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^5}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^5 (10 A+7 C) \sin (c+d x)}{d}\right )+\frac {(8 A+7 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {5}{3} \left (\frac {3}{2} \left (2 a^5 A \int \sec (c+d x)dx+\frac {a^5 (10 A+7 C) \sin (c+d x)}{d}+a^5 x (12 A+7 C)\right )+\frac {(8 A+7 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{3} \left (\frac {3}{2} \left (2 a^5 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {a^5 (10 A+7 C) \sin (c+d x)}{d}+a^5 x (12 A+7 C)\right )+\frac {(8 A+7 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {5}{3} \left (\frac {3}{2} \left (\frac {2 a^5 A \text {arctanh}(\sin (c+d x))}{d}+\frac {a^5 (10 A+7 C) \sin (c+d x)}{d}+a^5 x (12 A+7 C)\right )+\frac {(8 A+7 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (5 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {a^2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d}}{5 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}\)

Input:

Int[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x],x]
 

Output:

(C*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(5*d) + ((a^3*(5*A + 7*C)*(a + a*C 
os[c + d*x])^2*Sin[c + d*x])/(3*d) + (a^2*C*(a + a*Cos[c + d*x])^3*Sin[c + 
 d*x])/d + (5*(((8*A + 7*C)*(a^5 + a^5*Cos[c + d*x])*Sin[c + d*x])/(2*d) + 
 (3*(a^5*(12*A + 7*C)*x + (2*a^5*A*ArcTanh[Sin[c + d*x]])/d + (a^5*(10*A + 
 7*C)*Sin[c + d*x])/d))/2))/3)/(5*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 3042

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

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

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 3455

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + 
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 3502

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

rule 3525

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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1 
)/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2))   Int[(a + b*Sin[e + f* 
x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1 
)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
 

rule 4257

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

Maple [A] (verified)

Time = 12.55 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.68

method	result	size
parallelrisch	\(-\frac {\left (A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-A -2 C \right ) \sin \left (2 d x +2 c \right )+\left (-\frac {A}{12}-\frac {29 C}{48}\right ) \sin \left (3 d x +3 c \right )-\frac {\sin \left (4 d x +4 c \right ) C}{8}-\frac {C \sin \left (5 d x +5 c \right )}{80}+\left (-\frac {27 A}{4}-\frac {49 C}{8}\right ) \sin \left (d x +c \right )-6 d x \left (A +\frac {7 C}{12}\right )\right ) a^{4}}{d}\)	\(120\)
parts	\(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (a^{4} A +6 a^{4} C \right ) \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}+\frac {\left (4 a^{4} A +4 a^{4} C \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (6 a^{4} A +a^{4} C \right ) \sin \left (d x +c \right )}{d}+\frac {a^{4} C \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {4 a^{4} A \left (d x +c \right )}{d}+\frac {4 a^{4} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(209\)
derivativedivides	\(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {a^{4} C \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+6 a^{4} A \sin \left (d x +c \right )+2 a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+4 a^{4} A \left (d x +c \right )+4 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} C \sin \left (d x +c \right )}{d}\)	\(230\)
default	\(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {a^{4} C \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+6 a^{4} A \sin \left (d x +c \right )+2 a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+4 a^{4} A \left (d x +c \right )+4 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} C \sin \left (d x +c \right )}{d}\)	\(230\)
risch	\(6 a^{4} x A +\frac {7 a^{4} C x}{2}-\frac {27 i {\mathrm e}^{i \left (d x +c \right )} a^{4} A}{8 d}-\frac {49 i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{16 d}+\frac {27 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} A}{8 d}+\frac {49 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{16 d}+\frac {a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a^{4} C \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) a^{4} C}{8 d}+\frac {\sin \left (3 d x +3 c \right ) a^{4} A}{12 d}+\frac {29 \sin \left (3 d x +3 c \right ) a^{4} C}{48 d}+\frac {\sin \left (2 d x +2 c \right ) a^{4} A}{d}+\frac {2 \sin \left (2 d x +2 c \right ) a^{4} C}{d}\)	\(242\)
norman	\(\frac {\left (6 a^{4} A +\frac {7}{2} a^{4} C \right ) x +\left (6 a^{4} A +\frac {7}{2} a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (90 a^{4} A +\frac {105}{2} a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (90 a^{4} A +\frac {105}{2} a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (36 a^{4} A +21 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (36 a^{4} A +21 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (120 a^{4} A +70 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {a^{4} \left (10 A +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {a^{4} \left (18 A +25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{4} \left (166 A +119 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {a^{4} \left (238 A +233 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {2 a^{4} \left (310 A +231 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 d}+\frac {2 a^{4} \left (350 A +281 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}+\frac {a^{4} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{4} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(383\)

Input:

int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c),x,method=_RETURNVERBO 
SE)
 

Output:

-(A*ln(tan(1/2*d*x+1/2*c)-1)-A*ln(tan(1/2*d*x+1/2*c)+1)+(-A-2*C)*sin(2*d*x 
+2*c)+(-1/12*A-29/48*C)*sin(3*d*x+3*c)-1/8*sin(4*d*x+4*c)*C-1/80*C*sin(5*d 
*x+5*c)+(-27/4*A-49/8*C)*sin(d*x+c)-6*d*x*(A+7/12*C))*a^4/d
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.78 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {15 \, {\left (12 \, A + 7 \, C\right )} a^{4} d x + 15 \, A a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, A a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, C a^{4} \cos \left (d x + c\right )^{4} + 30 \, C a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (5 \, A + 34 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, A + 7 \, C\right )} a^{4} \cos \left (d x + c\right ) + 2 \, {\left (100 \, A + 83 \, C\right )} a^{4}\right )} \sin \left (d x + c\right )}{30 \, d} \] Input:

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="f 
ricas")
 

Output:

1/30*(15*(12*A + 7*C)*a^4*d*x + 15*A*a^4*log(sin(d*x + c) + 1) - 15*A*a^4* 
log(-sin(d*x + c) + 1) + (6*C*a^4*cos(d*x + c)^4 + 30*C*a^4*cos(d*x + c)^3 
 + 2*(5*A + 34*C)*a^4*cos(d*x + c)^2 + 15*(4*A + 7*C)*a^4*cos(d*x + c) + 2 
*(100*A + 83*C)*a^4)*sin(d*x + c))/d
 

Sympy [F]

\[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=a^{4} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 4 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 A \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 C \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 C \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 C \cos ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{6}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \] Input:

integrate((a+a*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c),x)
 

Output:

a**4*(Integral(A*sec(c + d*x), x) + Integral(4*A*cos(c + d*x)*sec(c + d*x) 
, x) + Integral(6*A*cos(c + d*x)**2*sec(c + d*x), x) + Integral(4*A*cos(c 
+ d*x)**3*sec(c + d*x), x) + Integral(A*cos(c + d*x)**4*sec(c + d*x), x) + 
 Integral(C*cos(c + d*x)**2*sec(c + d*x), x) + Integral(4*C*cos(c + d*x)** 
3*sec(c + d*x), x) + Integral(6*C*cos(c + d*x)**4*sec(c + d*x), x) + Integ 
ral(4*C*cos(c + d*x)**5*sec(c + d*x), x) + Integral(C*cos(c + d*x)**6*sec( 
c + d*x), x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.25 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=-\frac {40 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 480 \, {\left (d x + c\right )} A a^{4} - 8 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{4} + 240 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 120 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 720 \, A a^{4} \sin \left (d x + c\right ) - 120 \, C a^{4} \sin \left (d x + c\right )}{120 \, d} \] Input:

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="m 
axima")
 

Output:

-1/120*(40*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 120*(2*d*x + 2*c + si 
n(2*d*x + 2*c))*A*a^4 - 480*(d*x + c)*A*a^4 - 8*(3*sin(d*x + c)^5 - 10*sin 
(d*x + c)^3 + 15*sin(d*x + c))*C*a^4 + 240*(sin(d*x + c)^3 - 3*sin(d*x + c 
))*C*a^4 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^ 
4 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 - 120*A*a^4*log(sec(d*x + c 
) + tan(d*x + c)) - 720*A*a^4*sin(d*x + c) - 120*C*a^4*sin(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.40 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {30 \, A a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 30 \, A a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 15 \, {\left (12 \, A a^{4} + 7 \, C a^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (150 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 105 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 680 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 490 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1180 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 896 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 920 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 790 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 270 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 375 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{30 \, d} \] Input:

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="g 
iac")
 

Output:

1/30*(30*A*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 30*A*a^4*log(abs(tan(1 
/2*d*x + 1/2*c) - 1)) + 15*(12*A*a^4 + 7*C*a^4)*(d*x + c) + 2*(150*A*a^4*t 
an(1/2*d*x + 1/2*c)^9 + 105*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 680*A*a^4*tan(1 
/2*d*x + 1/2*c)^7 + 490*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 1180*A*a^4*tan(1/2* 
d*x + 1/2*c)^5 + 896*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 920*A*a^4*tan(1/2*d*x 
+ 1/2*c)^3 + 790*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 270*A*a^4*tan(1/2*d*x + 1/ 
2*c) + 375*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
 

Mupad [B] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.14 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {12\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+7\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+A\,a^4\,\sin \left (2\,c+2\,d\,x\right )+\frac {A\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{12}+2\,C\,a^4\,\sin \left (2\,c+2\,d\,x\right )+\frac {29\,C\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{48}+\frac {C\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{8}+\frac {C\,a^4\,\sin \left (5\,c+5\,d\,x\right )}{80}+\frac {27\,A\,a^4\,\sin \left (c+d\,x\right )}{4}+\frac {49\,C\,a^4\,\sin \left (c+d\,x\right )}{8}}{d} \] Input:

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^4)/cos(c + d*x),x)
 

Output:

(12*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 2*A*a^4*atanh(sin( 
c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 7*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c 
/2 + (d*x)/2)) + A*a^4*sin(2*c + 2*d*x) + (A*a^4*sin(3*c + 3*d*x))/12 + 2* 
C*a^4*sin(2*c + 2*d*x) + (29*C*a^4*sin(3*c + 3*d*x))/48 + (C*a^4*sin(4*c + 
 4*d*x))/8 + (C*a^4*sin(5*c + 5*d*x))/80 + (27*A*a^4*sin(c + d*x))/4 + (49 
*C*a^4*sin(c + d*x))/8)/d
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.83 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {a^{4} \left (-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c +60 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +135 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +6 \sin \left (d x +c \right )^{5} c -10 \sin \left (d x +c \right )^{3} a -80 \sin \left (d x +c \right )^{3} c +210 \sin \left (d x +c \right ) a +240 \sin \left (d x +c \right ) c +180 a d x +105 c d x \right )}{30 d} \] Input:

int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c),x)
 

Output:

(a**4*( - 30*cos(c + d*x)*sin(c + d*x)**3*c + 60*cos(c + d*x)*sin(c + d*x) 
*a + 135*cos(c + d*x)*sin(c + d*x)*c - 30*log(tan((c + d*x)/2) - 1)*a + 30 
*log(tan((c + d*x)/2) + 1)*a + 6*sin(c + d*x)**5*c - 10*sin(c + d*x)**3*a 
- 80*sin(c + d*x)**3*c + 210*sin(c + d*x)*a + 240*sin(c + d*x)*c + 180*a*d 
*x + 105*c*d*x))/(30*d)