Integrand size = 19, antiderivative size = 126 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^4} \, dx=-\frac {4 x}{a^4}+\frac {664 \sin (c+d x)}{105 a^4 d}-\frac {88 \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {4 \sin (c+d x)}{a^4 d (1+\sec (c+d x))}-\frac {\sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {12 \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3} \]
-4*x/a^4+664/105*sin(d*x+c)/a^4/d-88/105*sin(d*x+c)/a^4/d/(1+sec(d*x+c))^2 -4*sin(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*sin(d*x+c)/d/(a+a*sec(d*x+c))^4-12/ 35*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^3
Time = 2.77 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\sin (c+d x) \left (6720 \arcsin (\cos (c+d x)) \cos ^8\left (\frac {1}{2} (c+d x)\right )+\left (664+2236 \cos (c+d x)+2636 \cos ^2(c+d x)+1184 \cos ^3(c+d x)+105 \cos ^4(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{105 a^4 d \sqrt {1-\cos (c+d x)} (1+\cos (c+d x))^{9/2}} \]
(Sin[c + d*x]*(6720*ArcSin[Cos[c + d*x]]*Cos[(c + d*x)/2]^8 + (664 + 2236* Cos[c + d*x] + 2636*Cos[c + d*x]^2 + 1184*Cos[c + d*x]^3 + 105*Cos[c + d*x ]^4)*Sqrt[Sin[c + d*x]^2]))/(105*a^4*d*Sqrt[1 - Cos[c + d*x]]*(1 + Cos[c + d*x])^(9/2))
Time = 1.00 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.19, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.737, Rules used = {3042, 4304, 27, 3042, 4508, 3042, 4508, 3042, 4508, 3042, 4274, 24, 3042, 3117}
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.
| \(\displaystyle \int \frac {\cos (c+d x)}{(a \sec (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 4304 |
| \(\displaystyle -\frac {\int -\frac {4 \cos (c+d x) (2 a-a \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 \int \frac {\cos (c+d x) (2 a-a \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \int \frac {2 a-a \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {4 \left (\frac {\int \frac {\cos (c+d x) \left (13 a^2-9 a^2 \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {3 a \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {\int \frac {13 a^2-9 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {3 a \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\int \frac {\cos (c+d x) \left (61 a^3-44 a^3 \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {22 \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {3 a \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\int \frac {61 a^3-44 a^3 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {22 \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {3 a \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {\int \cos (c+d x) \left (166 a^4-105 a^4 \sec (c+d x)\right )dx}{a^2}-\frac {105 a^3 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {22 \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {3 a \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {\int \frac {166 a^4-105 a^4 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}-\frac {105 a^3 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {22 \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {3 a \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {166 a^4 \int \cos (c+d x)dx-105 a^4 \int 1dx}{a^2}-\frac {105 a^3 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {22 \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {3 a \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {166 a^4 \int \cos (c+d x)dx-105 a^4 x}{a^2}-\frac {105 a^3 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {22 \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {3 a \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {166 a^4 \int \sin \left (c+d x+\frac {\pi }{2}\right )dx-105 a^4 x}{a^2}-\frac {105 a^3 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {22 \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {3 a \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {\frac {166 a^4 \sin (c+d x)}{d}-105 a^4 x}{a^2}-\frac {105 a^3 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {22 \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {3 a \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
-1/7*Sin[c + d*x]/(d*(a + a*Sec[c + d*x])^4) + (4*((-3*a*Sin[c + d*x])/(5* d*(a + a*Sec[c + d*x])^3) + ((-22*Sin[c + d*x])/(3*d*(1 + Sec[c + d*x])^2) + ((-105*a^3*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])) + (-105*a^4*x + (166* a^4*Sin[c + d*x])/d)/a^2)/(3*a^2))/(5*a^2)))/(7*a^2)
3.1.78.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(a + b*Csc[e + f*x])^m*((d*Csc [e + f*x])^n/(f*(2*m + 1))), x] + Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ [m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B , 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
Time = 0.54 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.61
| method | result | size |
| parallelrisch | \(\frac {781 \left (\cos \left (d x +c \right )+\frac {2741 \cos \left (2 d x +2 c \right )}{6248}+\frac {74 \cos \left (3 d x +3 c \right )}{781}+\frac {105 \cos \left (4 d x +4 c \right )}{24992}+\frac {16171}{24992}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-840 d x}{210 a^{4} d}\) | \(77\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-64 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}\) | \(98\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-64 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}\) | \(98\) |
| risch | \(-\frac {4 x}{a^{4}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 a^{4} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 a^{4} d}+\frac {4 i \left (525 \,{\mathrm e}^{6 i \left (d x +c \right )}+2625 \,{\mathrm e}^{5 i \left (d x +c \right )}+5950 \,{\mathrm e}^{4 i \left (d x +c \right )}+7420 \,{\mathrm e}^{3 i \left (d x +c \right )}+5397 \,{\mathrm e}^{2 i \left (d x +c \right )}+2149 \,{\mathrm e}^{i \left (d x +c \right )}+382\right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\) | \(134\) |
| norman | \(\frac {-\frac {4 x}{a}+\frac {65 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}-\frac {47 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{70 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{56 a d}-\frac {4 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a^{3}}\) | \(137\) |
1/210*(781*(cos(d*x+c)+2741/6248*cos(2*d*x+2*c)+74/781*cos(3*d*x+3*c)+105/ 24992*cos(4*d*x+4*c)+16171/24992)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^6- 840*d*x)/a^4/d
Time = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.29 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^4} \, dx=-\frac {420 \, d x \cos \left (d x + c\right )^{4} + 1680 \, d x \cos \left (d x + c\right )^{3} + 2520 \, d x \cos \left (d x + c\right )^{2} + 1680 \, d x \cos \left (d x + c\right ) + 420 \, d x - {\left (105 \, \cos \left (d x + c\right )^{4} + 1184 \, \cos \left (d x + c\right )^{3} + 2636 \, \cos \left (d x + c\right )^{2} + 2236 \, \cos \left (d x + c\right ) + 664\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]
-1/105*(420*d*x*cos(d*x + c)^4 + 1680*d*x*cos(d*x + c)^3 + 2520*d*x*cos(d* x + c)^2 + 1680*d*x*cos(d*x + c) + 420*d*x - (105*cos(d*x + c)^4 + 1184*co s(d*x + c)^3 + 2636*cos(d*x + c)^2 + 2236*cos(d*x + c) + 664)*sin(d*x + c) )/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)
\[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
Integral(cos(c + d*x)/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x )**2 + 4*sec(c + d*x) + 1), x)/a**4
Time = 0.43 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.25 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} + \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {6720 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}}{840 \, d} \]
1/840*(1680*sin(d*x + c)/((a^4 + a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)* (cos(d*x + c) + 1)) + (5145*sin(d*x + c)/(cos(d*x + c) + 1) - 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 15 *sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 6720*arctan(sin(d*x + c)/(cos( d*x + c) + 1))/a^4)/d
Time = 0.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.89 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^4} \, dx=-\frac {\frac {3360 \, {\left (d x + c\right )}}{a^{4}} - \frac {1680 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{4}} + \frac {15 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 147 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5145 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]
-1/840*(3360*(d*x + c)/a^4 - 1680*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2 *c)^2 + 1)*a^4) + (15*a^24*tan(1/2*d*x + 1/2*c)^7 - 147*a^24*tan(1/2*d*x + 1/2*c)^5 + 805*a^24*tan(1/2*d*x + 1/2*c)^3 - 5145*a^24*tan(1/2*d*x + 1/2* c))/a^28)/d
Time = 13.71 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.09 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^4} \, dx=-\frac {15\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-192\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+1144\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-6112\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-1680\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+3360\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (c+d\,x\right )}{840\,a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7} \]