\(\int (a+a \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\) [25]

Optimal result

Integrand size = 33, antiderivative size = 169 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=a^3 C x+\frac {a^3 (15 A+28 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {5 a^3 (3 A+4 C) \tan (c+d x)}{8 d}+\frac {(5 A+4 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:

a^3*C*x+1/8*a^3*(15*A+28*C)*arctanh(sin(d*x+c))/d+5/8*a^3*(3*A+4*C)*tan(d* 
x+c)/d+1/8*(5*A+4*C)*(a^3+a^3*cos(d*x+c))*sec(d*x+c)*tan(d*x+c)/d+1/4*A*(a 
^2+a^2*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/a/d+1/4*A*(a+a*cos(d*x+c))^3* 
sec(d*x+c)^3*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 2.42 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.04 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=a^3 C x+\frac {3 a^3 C \coth ^{-1}(\sin (c+d x))}{d}+\frac {15 a^3 A \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {4 a^3 A \tan (c+d x)}{d}+\frac {3 a^3 C \tan (c+d x)}{d}+\frac {15 a^3 A \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 C \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^3 A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 A \tan ^3(c+d x)}{d} \] Input:

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

Output:

a^3*C*x + (3*a^3*C*ArcCoth[Sin[c + d*x]])/d + (15*a^3*A*ArcTanh[Sin[c + d* 
x]])/(8*d) + (a^3*C*ArcTanh[Sin[c + d*x]])/(2*d) + (4*a^3*A*Tan[c + d*x])/ 
d + (3*a^3*C*Tan[c + d*x])/d + (15*a^3*A*Sec[c + d*x]*Tan[c + d*x])/(8*d) 
+ (a^3*C*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^3*A*Sec[c + d*x]^3*Tan[c + 
d*x])/(4*d) + (a^3*A*Tan[c + d*x]^3)/d
 

Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3523, 3042, 3454, 27, 3042, 3454, 3042, 3447, 3042, 3500, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

\(\displaystyle \frac {\frac {1}{2} \left (\int \left ((15 A+28 C) a^4+8 C \cos (c+d x) a^4\right ) \sec (c+d x)dx+\frac {5 a^4 (3 A+4 C) \tan (c+d x)}{d}\right )+\frac {(5 A+4 C) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (\int \frac {(15 A+28 C) a^4+8 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 a^4 (3 A+4 C) \tan (c+d x)}{d}\right )+\frac {(5 A+4 C) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {1}{2} \left (a^4 (15 A+28 C) \int \sec (c+d x)dx+\frac {5 a^4 (3 A+4 C) \tan (c+d x)}{d}+8 a^4 C x\right )+\frac {(5 A+4 C) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (a^4 (15 A+28 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {5 a^4 (3 A+4 C) \tan (c+d x)}{d}+8 a^4 C x\right )+\frac {(5 A+4 C) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {1}{2} \left (\frac {a^4 (15 A+28 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (3 A+4 C) \tan (c+d x)}{d}+8 a^4 C x\right )+\frac {(5 A+4 C) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\)

Input:

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

Output:

(A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (((5*A + 4* 
C)*(a^4 + a^4*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (A*(a^2 + a 
^2*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/d + (8*a^4*C*x + (a^4*(15* 
A + 28*C)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(3*A + 4*C)*Tan[c + d*x])/d)/2 
)/(4*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 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 4257

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

Maple [A] (verified)

Time = 1.85 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.10

method	result	size
parts	\(\frac {a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (a^{3} A +3 C \,a^{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 a^{3} A +C \,a^{3}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {C \,a^{3} \left (d x +c \right )}{d}+\frac {3 C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{3}}{d}-\frac {3 a^{3} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(186\)
derivativedivides	\(\frac {a^{3} A \tan \left (d x +c \right )+C \,a^{3} \left (d x +c \right )+3 a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-3 a^{3} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 C \,a^{3} \tan \left (d x +c \right )+a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(205\)
default	\(\frac {a^{3} A \tan \left (d x +c \right )+C \,a^{3} \left (d x +c \right )+3 a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-3 a^{3} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 C \,a^{3} \tan \left (d x +c \right )+a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(205\)
parallelrisch	\(\frac {10 \left (-\frac {3 \left (A +\frac {28 C}{15}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}+\frac {3 \left (A +\frac {28 C}{15}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+\frac {2 d x C \cos \left (2 d x +2 c \right )}{5}+\frac {d x C \cos \left (4 d x +4 c \right )}{10}+\left (A +\frac {3 C}{5}\right ) \sin \left (2 d x +2 c \right )+\frac {\left (\frac {3 A}{4}+\frac {C}{5}\right ) \sin \left (3 d x +3 c \right )}{2}+\frac {3 \left (A +C \right ) \sin \left (4 d x +4 c \right )}{10}+\frac {\left (\frac {23 A}{4}+C \right ) \sin \left (d x +c \right )}{10}+\frac {3 d x C}{10}\right ) a^{3}}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(209\)
risch	\(a^{3} C x -\frac {i a^{3} \left (15 A \,{\mathrm e}^{7 i \left (d x +c \right )}+4 C \,{\mathrm e}^{7 i \left (d x +c \right )}-8 A \,{\mathrm e}^{6 i \left (d x +c \right )}-24 C \,{\mathrm e}^{6 i \left (d x +c \right )}+23 A \,{\mathrm e}^{5 i \left (d x +c \right )}+4 C \,{\mathrm e}^{5 i \left (d x +c \right )}-72 A \,{\mathrm e}^{4 i \left (d x +c \right )}-72 C \,{\mathrm e}^{4 i \left (d x +c \right )}-23 A \,{\mathrm e}^{3 i \left (d x +c \right )}-4 C \,{\mathrm e}^{3 i \left (d x +c \right )}-88 A \,{\mathrm e}^{2 i \left (d x +c \right )}-72 C \,{\mathrm e}^{2 i \left (d x +c \right )}-15 A \,{\mathrm e}^{i \left (d x +c \right )}-4 C \,{\mathrm e}^{i \left (d x +c \right )}-24 A -24 C \right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {15 a^{3} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {15 a^{3} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\)	\(293\)

Input:

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

Output:

a^3*A/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)))+(A*a^3+3*C*a^3)/d*tan(d*x+c)+(3*A*a^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)))+C*a^3/d*(d*x+c)+3/d*C*ln(sec(d 
*x+c)+tan(d*x+c))*a^3-3*a^3*A/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.89 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {16 \, C a^{3} d x \cos \left (d x + c\right )^{4} + {\left (15 \, A + 28 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (15 \, A + 28 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, {\left (A + C\right )} a^{3} \cos \left (d x + c\right )^{3} + {\left (15 \, A + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, A a^{3} \cos \left (d x + c\right ) + 2 \, A a^{3}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:

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

Output:

1/16*(16*C*a^3*d*x*cos(d*x + c)^4 + (15*A + 28*C)*a^3*cos(d*x + c)^4*log(s 
in(d*x + c) + 1) - (15*A + 28*C)*a^3*cos(d*x + c)^4*log(-sin(d*x + c) + 1) 
 + 2*(24*(A + C)*a^3*cos(d*x + c)^3 + (15*A + 4*C)*a^3*cos(d*x + c)^2 + 8* 
A*a^3*cos(d*x + c) + 2*A*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.52 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 16 \, {\left (d x + c\right )} C a^{3} - A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, A a^{3} \tan \left (d x + c\right ) + 48 \, C a^{3} \tan \left (d x + c\right )}{16 \, d} \] Input:

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

Output:

1/16*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 16*(d*x + c)*C*a^3 - 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)) - 12*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)) - 4*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)) + 24*C*a^3*(log(sin(d*x + c) + 1) - log(s 
in(d*x + c) - 1)) + 16*A*a^3*tan(d*x + c) + 48*C*a^3*tan(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.31 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {8 \, {\left (d x + c\right )} C a^{3} + {\left (15 \, A a^{3} + 28 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (15 \, A a^{3} + 28 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 20 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 55 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 68 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 73 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 76 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 49 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 28 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{8 \, d} \] Input:

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

Output:

1/8*(8*(d*x + c)*C*a^3 + (15*A*a^3 + 28*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c 
) + 1)) - (15*A*a^3 + 28*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15 
*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 20*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 55*A*a^3 
*tan(1/2*d*x + 1/2*c)^5 - 68*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 73*A*a^3*tan(1 
/2*d*x + 1/2*c)^3 + 76*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 49*A*a^3*tan(1/2*d*x 
 + 1/2*c) - 28*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4) 
/d
 

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.37 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {15\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {2\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {15\,A\,a^3\,\sin \left (c+d\,x\right )}{8\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^3}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{4\,d\,{\cos \left (c+d\,x\right )}^4}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2} \] Input:

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

Output:

(15*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*d) + (2*C*a^3*a 
tan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (7*C*a^3*atanh(sin(c/2 + ( 
d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*A*a^3*sin(c + d*x))/(d*cos(c + d*x)) + 
 (15*A*a^3*sin(c + d*x))/(8*d*cos(c + d*x)^2) + (A*a^3*sin(c + d*x))/(d*co 
s(c + d*x)^3) + (A*a^3*sin(c + d*x))/(4*d*cos(c + d*x)^4) + (3*C*a^3*sin(c 
 + d*x))/(d*cos(c + d*x)) + (C*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 530, normalized size of antiderivative = 3.14 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx =\text {Too large to display} \] Input:

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

Output:

(a**3*( - 15*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 28 
*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*c + 30*cos(c + d*x 
)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 56*cos(c + d*x)*log(tan((c 
 + d*x)/2) - 1)*sin(c + d*x)**2*c - 15*cos(c + d*x)*log(tan((c + d*x)/2) - 
 1)*a - 28*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*c + 15*cos(c + d*x)*log( 
tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a + 28*cos(c + d*x)*log(tan((c + d*x 
)/2) + 1)*sin(c + d*x)**4*c - 30*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*si 
n(c + d*x)**2*a - 56*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)** 
2*c + 15*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a + 28*cos(c + d*x)*log(ta 
n((c + d*x)/2) + 1)*c + 8*cos(c + d*x)*sin(c + d*x)**4*c*d*x - 15*cos(c + 
d*x)*sin(c + d*x)**3*a - 4*cos(c + d*x)*sin(c + d*x)**3*c - 16*cos(c + d*x 
)*sin(c + d*x)**2*c*d*x + 17*cos(c + d*x)*sin(c + d*x)*a + 4*cos(c + d*x)* 
sin(c + d*x)*c + 8*cos(c + d*x)*c*d*x + 24*sin(c + d*x)**5*a + 24*sin(c + 
d*x)**5*c - 56*sin(c + d*x)**3*a - 48*sin(c + d*x)**3*c + 32*sin(c + d*x)* 
a + 24*sin(c + d*x)*c))/(8*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x 
)**2 + 1))