Integrand size = 21, antiderivative size = 124 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {5 x}{a^2}+\frac {12 \sin (c+d x)}{a^2 d}-\frac {5 \cos (c+d x) \sin (c+d x)}{a^2 d}-\frac {10 \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {4 \sin ^3(c+d x)}{a^2 d} \]
-5*x/a^2+12*sin(d*x+c)/a^2/d-5*cos(d*x+c)*sin(d*x+c)/a^2/d-10/3*cos(d*x+c) ^2*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec( d*x+c))^2-4*sin(d*x+c)^3/a^2/d
Time = 1.65 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.60 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-360 d x \cos \left (\frac {d x}{2}\right )-360 d x \cos \left (c+\frac {d x}{2}\right )-120 d x \cos \left (c+\frac {3 d x}{2}\right )-120 d x \cos \left (2 c+\frac {3 d x}{2}\right )+516 \sin \left (\frac {d x}{2}\right )-156 \sin \left (c+\frac {d x}{2}\right )+342 \sin \left (c+\frac {3 d x}{2}\right )+118 \sin \left (2 c+\frac {3 d x}{2}\right )+30 \sin \left (2 c+\frac {5 d x}{2}\right )+30 \sin \left (3 c+\frac {5 d x}{2}\right )-3 \sin \left (3 c+\frac {7 d x}{2}\right )-3 \sin \left (4 c+\frac {7 d x}{2}\right )+\sin \left (4 c+\frac {9 d x}{2}\right )+\sin \left (5 c+\frac {9 d x}{2}\right )\right )}{192 a^2 d} \]
(Sec[c/2]*Sec[(c + d*x)/2]^3*(-360*d*x*Cos[(d*x)/2] - 360*d*x*Cos[c + (d*x )/2] - 120*d*x*Cos[c + (3*d*x)/2] - 120*d*x*Cos[2*c + (3*d*x)/2] + 516*Sin [(d*x)/2] - 156*Sin[c + (d*x)/2] + 342*Sin[c + (3*d*x)/2] + 118*Sin[2*c + (3*d*x)/2] + 30*Sin[2*c + (5*d*x)/2] + 30*Sin[3*c + (5*d*x)/2] - 3*Sin[3*c + (7*d*x)/2] - 3*Sin[4*c + (7*d*x)/2] + Sin[4*c + (9*d*x)/2] + Sin[5*c + (9*d*x)/2]))/(192*a^2*d)
Time = 0.72 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4304, 27, 3042, 4508, 27, 3042, 4274, 3042, 3113, 2009, 3115, 24}
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.
| \(\displaystyle \int \frac {\cos ^3(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4304 |
| \(\displaystyle -\frac {\int -\frac {2 \cos ^3(c+d x) (3 a-2 a \sec (c+d x))}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \int \frac {\cos ^3(c+d x) (3 a-2 a \sec (c+d x))}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int \frac {3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {2 \left (\frac {\int 3 \cos ^3(c+d x) \left (6 a^2-5 a^2 \sec (c+d x)\right )dx}{a^2}-\frac {5 \sin (c+d x) \cos ^2(c+d x)}{d (\sec (c+d x)+1)}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (\frac {3 \int \cos ^3(c+d x) \left (6 a^2-5 a^2 \sec (c+d x)\right )dx}{a^2}-\frac {5 \sin (c+d x) \cos ^2(c+d x)}{d (\sec (c+d x)+1)}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {3 \int \frac {6 a^2-5 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx}{a^2}-\frac {5 \sin (c+d x) \cos ^2(c+d x)}{d (\sec (c+d x)+1)}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (6 a^2 \int \cos ^3(c+d x)dx-5 a^2 \int \cos ^2(c+d x)dx\right )}{a^2}-\frac {5 \sin (c+d x) \cos ^2(c+d x)}{d (\sec (c+d x)+1)}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (6 a^2 \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx-5 a^2 \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx\right )}{a^2}-\frac {5 \sin (c+d x) \cos ^2(c+d x)}{d (\sec (c+d x)+1)}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (-\frac {6 a^2 \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}-5 a^2 \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx\right )}{a^2}-\frac {5 \sin (c+d x) \cos ^2(c+d x)}{d (\sec (c+d x)+1)}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (-5 a^2 \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {6 a^2 \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{a^2}-\frac {5 \sin (c+d x) \cos ^2(c+d x)}{d (\sec (c+d x)+1)}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (-5 a^2 \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {6 a^2 \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{a^2}-\frac {5 \sin (c+d x) \cos ^2(c+d x)}{d (\sec (c+d x)+1)}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (-\frac {6 a^2 \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}-5 a^2 \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )}{a^2}-\frac {5 \sin (c+d x) \cos ^2(c+d x)}{d (\sec (c+d x)+1)}\right )}{3 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
-1/3*(Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + (2*((-5*Co s[c + d*x]^2*Sin[c + d*x])/(d*(1 + Sec[c + d*x])) + (3*(-5*a^2*(x/2 + (Cos [c + d*x]*Sin[c + d*x])/(2*d)) - (6*a^2*(-Sin[c + d*x] + Sin[c + d*x]^3/3) )/d))/a^2))/(3*a^2)
3.1.60.3.1 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[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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 = 0.51 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.62
| method | result | size |
| parallelrisch | \(\frac {43 \left (\cos \left (d x +c \right )+\frac {14 \cos \left (2 d x +2 c \right )}{129}-\frac {\cos \left (3 d x +3 c \right )}{129}+\frac {\cos \left (4 d x +4 c \right )}{258}+\frac {73}{86}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-40 d x}{8 a^{2} d}\) | \(77\) |
| derivativedivides | \(\frac {-\frac {8 \left (-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2}-\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-20 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}\) | \(101\) |
| default | \(\frac {-\frac {8 \left (-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2}-\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-20 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}\) | \(101\) |
| risch | \(-\frac {5 x}{a^{2}}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{4 a^{2} d}-\frac {15 i {\mathrm e}^{i \left (d x +c \right )}}{8 a^{2} d}+\frac {15 i {\mathrm e}^{-i \left (d x +c \right )}}{8 a^{2} d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}+\frac {2 i \left (15 \,{\mathrm e}^{2 i \left (d x +c \right )}+27 \,{\mathrm e}^{i \left (d x +c \right )}+14\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {\sin \left (3 d x +3 c \right )}{12 a^{2} d}\) | \(143\) |
| norman | \(\frac {-\frac {5 x}{a}+\frac {21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {80 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}-\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {5 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a}\) | \(171\) |
1/8*(43*(cos(d*x+c)+14/129*cos(2*d*x+2*c)-1/129*cos(3*d*x+3*c)+1/258*cos(4 *d*x+4*c)+73/86)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^2-40*d*x)/a^2/d
Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {15 \, d x \cos \left (d x + c\right )^{2} + 30 \, d x \cos \left (d x + c\right ) + 15 \, d x - {\left (\cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 33 \, \cos \left (d x + c\right ) + 24\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
-1/3*(15*d*x*cos(d*x + c)^2 + 30*d*x*cos(d*x + c) + 15*d*x - (cos(d*x + c) ^4 - cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 33*cos(d*x + c) + 24)*sin(d*x + c ))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
Timed out. \[ \int \frac {\cos ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
Time = 0.36 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.67 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {60 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6 \, d} \]
1/6*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c ) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^2 + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^ 2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 60*arctan(sin(d*x + c)/(co s(d*x + c) + 1))/a^2)/d
Time = 0.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {30 \, {\left (d x + c\right )}}{a^{2}} - \frac {4 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
-1/6*(30*(d*x + c)/a^2 - 4*(15*tan(1/2*d*x + 1/2*c)^5 + 20*tan(1/2*d*x + 1 /2*c)^3 + 9*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^2) + ( a^4*tan(1/2*d*x + 1/2*c)^3 - 27*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d
Time = 13.86 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-28\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-60\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+30\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (c+d\,x\right )}{6\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \]