\(\int (a+a \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\) [36]

Optimal result

Integrand size = 33, antiderivative size = 207 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=a^4 C x+\frac {a^4 (7 A+12 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^4 (7 A+10 C) \tan (c+d x)}{2 d}+\frac {(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d} \] Output:

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

Mathematica [A] (verified)

Time = 5.14 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.01 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=a^4 C x+\frac {4 a^4 C \coth ^{-1}(\sin (c+d x))}{d}+\frac {7 a^4 A \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a^4 C \text {arctanh}(\sin (c+d x))}{d}+\frac {8 a^4 A \tan (c+d x)}{d}+\frac {7 a^4 C \tan (c+d x)}{d}+\frac {7 a^4 A \sec (c+d x) \tan (c+d x)}{2 d}+\frac {2 a^4 C \sec (c+d x) \tan (c+d x)}{d}+\frac {a^4 A \sec ^3(c+d x) \tan (c+d x)}{d}+\frac {8 a^4 A \tan ^3(c+d x)}{3 d}+\frac {a^4 C \tan ^3(c+d x)}{3 d}+\frac {a^4 A \tan ^5(c+d x)}{5 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]
 

Output:

a^4*C*x + (4*a^4*C*ArcCoth[Sin[c + d*x]])/d + (7*a^4*A*ArcTanh[Sin[c + d*x 
]])/(2*d) + (2*a^4*C*ArcTanh[Sin[c + d*x]])/d + (8*a^4*A*Tan[c + d*x])/d + 
 (7*a^4*C*Tan[c + d*x])/d + (7*a^4*A*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (2 
*a^4*C*Sec[c + d*x]*Tan[c + d*x])/d + (a^4*A*Sec[c + d*x]^3*Tan[c + d*x])/ 
d + (8*a^4*A*Tan[c + d*x]^3)/(3*d) + (a^4*C*Tan[c + d*x]^3)/(3*d) + (a^4*A 
*Tan[c + d*x]^5)/(5*d)
 

Rubi [A] (verified)

Time = 1.66 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.05, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 3523, 3042, 3454, 27, 3042, 3454, 27, 3042, 3454, 27, 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.

\(\displaystyle \int \sec ^6(c+d x) (a \cos (c+d x)+a)^4 \left (A+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\)

\(\Big \downarrow \) 3523

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

Int[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]
 

Output:

(A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((a^3*(7*A 
+ 5*C)*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (a^2*A* 
(a + a*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/d + (5*(((7*A + 8*C)*( 
a^5 + a^5*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (3*(2*a^5*C*x + 
 (a^5*(7*A + 12*C)*ArcTanh[Sin[c + d*x]])/d + (a^5*(7*A + 10*C)*Tan[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 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 = 12.77 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.13

method	result	size
parts	\(-\frac {a^{4} A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (a^{4} A +6 a^{4} C \right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 a^{4} A +4 a^{4} C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (6 a^{4} A +a^{4} C \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{4} C \left (d x +c \right )}{d}+\frac {4 a^{4} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {4 C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{4}}{d}\)	\(233\)
parallelrisch	\(\frac {70 a^{4} \left (-\frac {3 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {12 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {3 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {12 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {3 d x C \cos \left (3 d x +3 c \right )}{14}+\frac {3 d x C \cos \left (5 d x +5 c \right )}{70}+\left (\frac {33 A}{35}+\frac {12 C}{35}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {32 C}{35}+\frac {11 A}{10}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {3 A}{10}+\frac {6 C}{35}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {2 C}{7}+\frac {83 A}{350}\right ) \sin \left (5 d x +5 c \right )+\frac {3 d x C \cos \left (d x +c \right )}{7}+\sin \left (d x +c \right ) \left (A +\frac {22 C}{35}\right )\right )}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(256\)
derivativedivides	\(\frac {a^{4} A \tan \left (d x +c \right )+a^{4} C \left (d x +c \right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-6 a^{4} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 a^{4} C \tan \left (d x +c \right )+4 a^{4} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{4} A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-a^{4} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(265\)
default	\(\frac {a^{4} A \tan \left (d x +c \right )+a^{4} C \left (d x +c \right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-6 a^{4} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 a^{4} C \tan \left (d x +c \right )+4 a^{4} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{4} A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-a^{4} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(265\)
risch	\(a^{4} C x -\frac {i a^{4} \left (105 A \,{\mathrm e}^{9 i \left (d x +c \right )}+60 C \,{\mathrm e}^{9 i \left (d x +c \right )}-30 A \,{\mathrm e}^{8 i \left (d x +c \right )}-180 C \,{\mathrm e}^{8 i \left (d x +c \right )}+330 A \,{\mathrm e}^{7 i \left (d x +c \right )}+120 C \,{\mathrm e}^{7 i \left (d x +c \right )}-480 A \,{\mathrm e}^{6 i \left (d x +c \right )}-780 C \,{\mathrm e}^{6 i \left (d x +c \right )}-1180 A \,{\mathrm e}^{4 i \left (d x +c \right )}-1220 C \,{\mathrm e}^{4 i \left (d x +c \right )}-330 A \,{\mathrm e}^{3 i \left (d x +c \right )}-120 C \,{\mathrm e}^{3 i \left (d x +c \right )}-800 A \,{\mathrm e}^{2 i \left (d x +c \right )}-820 C \,{\mathrm e}^{2 i \left (d x +c \right )}-105 A \,{\mathrm e}^{i \left (d x +c \right )}-60 C \,{\mathrm e}^{i \left (d x +c \right )}-166 A -200 C \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {7 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {7 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\)	\(317\)

Input:

int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x,method=_RETURNVER 
BOSE)
 

Output:

-a^4*A/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+(A*a^4+6*C* 
a^4)/d*tan(d*x+c)+(4*A*a^4+4*C*a^4)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(se 
c(d*x+c)+tan(d*x+c)))-(6*A*a^4+C*a^4)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c) 
+a^4*C/d*(d*x+c)+4*a^4*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)))+4/d*C*ln(sec(d*x+c)+tan(d*x+c))*a^4
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.86 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {60 \, C a^{4} d x \cos \left (d x + c\right )^{5} + 15 \, {\left (7 \, A + 12 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (7 \, A + 12 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (83 \, A + 100 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \, {\left (7 \, A + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (34 \, A + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, A a^{4} \cos \left (d x + c\right ) + 6 \, A a^{4}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \] Input:

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm= 
"fricas")
 

Output:

1/60*(60*C*a^4*d*x*cos(d*x + c)^5 + 15*(7*A + 12*C)*a^4*cos(d*x + c)^5*log 
(sin(d*x + c) + 1) - 15*(7*A + 12*C)*a^4*cos(d*x + c)^5*log(-sin(d*x + c) 
+ 1) + 2*(2*(83*A + 100*C)*a^4*cos(d*x + c)^4 + 15*(7*A + 4*C)*a^4*cos(d*x 
 + c)^3 + 2*(34*A + 5*C)*a^4*cos(d*x + c)^2 + 30*A*a^4*cos(d*x + c) + 6*A* 
a^4)*sin(d*x + c))/(d*cos(d*x + c)^5)
 

Sympy [F(-1)]

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

integrate((a+a*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.52 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {4 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{4} + 120 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 20 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} + 60 \, {\left (d x + c\right )} C a^{4} - 15 \, A a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 60 \, A a^{4} \tan \left (d x + c\right ) + 360 \, C a^{4} \tan \left (d x + c\right )}{60 \, d} \] Input:

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm= 
"maxima")
 

Output:

1/60*(4*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^4 + 1 
20*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 20*(tan(d*x + c)^3 + 3*tan(d* 
x + c))*C*a^4 + 60*(d*x + c)*C*a^4 - 15*A*a^4*(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^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 
 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*C*a^4*(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^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 60*A*a^4*t 
an(d*x + c) + 360*C*a^4*tan(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.24 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {30 \, {\left (d x + c\right )} C a^{4} + 15 \, {\left (7 \, A a^{4} + 12 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (7 \, A a^{4} + 12 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 150 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 490 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 680 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 896 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1180 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 790 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 920 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 375 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 270 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{30 \, d} \] Input:

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm= 
"giac")
 

Output:

1/30*(30*(d*x + c)*C*a^4 + 15*(7*A*a^4 + 12*C*a^4)*log(abs(tan(1/2*d*x + 1 
/2*c) + 1)) - 15*(7*A*a^4 + 12*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 
 2*(105*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 150*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 
490*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 680*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 896* 
A*a^4*tan(1/2*d*x + 1/2*c)^5 + 1180*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 790*A*a 
^4*tan(1/2*d*x + 1/2*c)^3 - 920*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 375*A*a^4*t 
an(1/2*d*x + 1/2*c) + 270*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 = 0.26 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.34 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {7\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {83\,A\,a^4\,\sin \left (c+d\,x\right )}{15\,d\,\cos \left (c+d\,x\right )}+\frac {7\,A\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {34\,A\,a^4\,\sin \left (c+d\,x\right )}{15\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^4}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{5\,d\,{\cos \left (c+d\,x\right )}^5}+\frac {20\,C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,C\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \] Input:

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^4)/cos(c + d*x)^6,x)
 

Output:

(7*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^4*atan(s 
in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*C*a^4*atanh(sin(c/2 + (d*x) 
/2)/cos(c/2 + (d*x)/2)))/d + (83*A*a^4*sin(c + d*x))/(15*d*cos(c + d*x)) + 
 (7*A*a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (34*A*a^4*sin(c + d*x))/(15 
*d*cos(c + d*x)^3) + (A*a^4*sin(c + d*x))/(d*cos(c + d*x)^4) + (A*a^4*sin( 
c + d*x))/(5*d*cos(c + d*x)^5) + (20*C*a^4*sin(c + d*x))/(3*d*cos(c + d*x) 
) + (2*C*a^4*sin(c + d*x))/(d*cos(c + d*x)^2) + (C*a^4*sin(c + d*x))/(3*d* 
cos(c + d*x)^3)
 

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 530, normalized size of antiderivative = 2.56 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:

int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)
 

Output:

(a**4*( - 105*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 1 
80*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*c + 210*cos(c + 
d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 360*cos(c + d*x)*log(ta 
n((c + d*x)/2) - 1)*sin(c + d*x)**2*c - 105*cos(c + d*x)*log(tan((c + d*x) 
/2) - 1)*a - 180*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*c + 105*cos(c + d* 
x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a + 180*cos(c + d*x)*log(tan( 
(c + d*x)/2) + 1)*sin(c + d*x)**4*c - 210*cos(c + d*x)*log(tan((c + d*x)/2 
) + 1)*sin(c + d*x)**2*a - 360*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin( 
c + d*x)**2*c + 105*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a + 180*cos(c + 
 d*x)*log(tan((c + d*x)/2) + 1)*c + 30*cos(c + d*x)*sin(c + d*x)**4*c*d*x 
- 105*cos(c + d*x)*sin(c + d*x)**3*a - 60*cos(c + d*x)*sin(c + d*x)**3*c - 
 60*cos(c + d*x)*sin(c + d*x)**2*c*d*x + 135*cos(c + d*x)*sin(c + d*x)*a + 
 60*cos(c + d*x)*sin(c + d*x)*c + 30*cos(c + d*x)*c*d*x + 166*sin(c + d*x) 
**5*a + 200*sin(c + d*x)**5*c - 400*sin(c + d*x)**3*a - 410*sin(c + d*x)** 
3*c + 240*sin(c + d*x)*a + 210*sin(c + d*x)*c))/(30*cos(c + d*x)*d*(sin(c 
+ d*x)**4 - 2*sin(c + d*x)**2 + 1))