Integrand size = 12, antiderivative size = 115 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=-\frac {43 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {43 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {5 \sin (c+d x)}{32 d (3+5 \cos (c+d x))^2}-\frac {45 \sin (c+d x)}{512 d (3+5 \cos (c+d x))} \]
-43/2048*ln(2*cos(1/2*d*x+1/2*c)-sin(1/2*d*x+1/2*c))/d+43/2048*ln(2*cos(1/ 2*d*x+1/2*c)+sin(1/2*d*x+1/2*c))/d+5/32*sin(d*x+c)/d/(3+5*cos(d*x+c))^2-45 /512*sin(d*x+c)/d/(3+5*cos(d*x+c))
Time = 0.26 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.89 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=-\frac {43 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {43 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {5}{512 d \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {45 \sin \left (\frac {1}{2} (c+d x)\right )}{2048 d \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {5}{512 d \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {45 \sin \left (\frac {1}{2} (c+d x)\right )}{2048 d \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]
(-43*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(2048*d) + (43*Log[2*Cos[ (c + d*x)/2] + Sin[(c + d*x)/2]])/(2048*d) + 5/(512*d*(2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - (45*Sin[(c + d*x)/2])/(2048*d*(2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - 5/(512*d*(2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 ) - (45*Sin[(c + d*x)/2])/(2048*d*(2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
Time = 0.38 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3143, 25, 3042, 3233, 27, 3042, 3138, 219}
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.
| \(\displaystyle \int \frac {1}{(5 \cos (c+d x)+3)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (5 \sin \left (c+d x+\frac {\pi }{2}\right )+3\right )^3}dx\) |
| \(\Big \downarrow \) 3143 |
| \(\displaystyle \frac {1}{32} \int -\frac {6-5 \cos (c+d x)}{(5 \cos (c+d x)+3)^2}dx+\frac {5 \sin (c+d x)}{32 d (5 \cos (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {5 \sin (c+d x)}{32 d (5 \cos (c+d x)+3)^2}-\frac {1}{32} \int \frac {6-5 \cos (c+d x)}{(5 \cos (c+d x)+3)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \sin (c+d x)}{32 d (5 \cos (c+d x)+3)^2}-\frac {1}{32} \int \frac {6-5 \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (5 \sin \left (c+d x+\frac {\pi }{2}\right )+3\right )^2}dx\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {1}{32} \left (-\frac {1}{16} \int -\frac {43}{5 \cos (c+d x)+3}dx-\frac {45 \sin (c+d x)}{16 d (5 \cos (c+d x)+3)}\right )+\frac {5 \sin (c+d x)}{32 d (5 \cos (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{32} \left (\frac {43}{16} \int \frac {1}{5 \cos (c+d x)+3}dx-\frac {45 \sin (c+d x)}{16 d (5 \cos (c+d x)+3)}\right )+\frac {5 \sin (c+d x)}{32 d (5 \cos (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{32} \left (\frac {43}{16} \int \frac {1}{5 \sin \left (c+d x+\frac {\pi }{2}\right )+3}dx-\frac {45 \sin (c+d x)}{16 d (5 \cos (c+d x)+3)}\right )+\frac {5 \sin (c+d x)}{32 d (5 \cos (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {1}{32} \left (\frac {43 \int \frac {1}{8-2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}d\tan \left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {45 \sin (c+d x)}{16 d (5 \cos (c+d x)+3)}\right )+\frac {5 \sin (c+d x)}{32 d (5 \cos (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{32} \left (\frac {43 \text {arctanh}\left (\frac {1}{2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{32 d}-\frac {45 \sin (c+d x)}{16 d (5 \cos (c+d x)+3)}\right )+\frac {5 \sin (c+d x)}{32 d (5 \cos (c+d x)+3)^2}\) |
(5*Sin[c + d*x])/(32*d*(3 + 5*Cos[c + d*x])^2) + ((43*ArcTanh[Tan[(c + d*x )/2]/2])/(32*d) - (45*Sin[c + d*x])/(16*d*(3 + 5*Cos[c + d*x])))/32
3.1.36.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp [1/((n + 1)*(a^2 - b^2)) Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Time = 0.66 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72
| method | result | size |
| norman | \(\frac {-\frac {35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{128 d}+\frac {85 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{512 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-4\right )^{2}}-\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{2048 d}+\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{2048 d}\) | \(83\) |
| derivativedivides | \(\frac {\frac {25}{512 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )^{2}}+\frac {85}{1024 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}-\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{2048}-\frac {25}{512 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )^{2}}+\frac {85}{1024 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}+\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{2048}}{d}\) | \(94\) |
| default | \(\frac {\frac {25}{512 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )^{2}}+\frac {85}{1024 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}-\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{2048}-\frac {25}{512 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )^{2}}+\frac {85}{1024 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}+\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{2048}}{d}\) | \(94\) |
| risch | \(-\frac {i \left (215 \,{\mathrm e}^{3 i \left (d x +c \right )}+387 \,{\mathrm e}^{2 i \left (d x +c \right )}+325 \,{\mathrm e}^{i \left (d x +c \right )}+225\right )}{256 d \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )^{2}}-\frac {43 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {3}{5}-\frac {4 i}{5}\right )}{2048 d}+\frac {43 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {3}{5}+\frac {4 i}{5}\right )}{2048 d}\) | \(107\) |
| parallelrisch | \(\frac {\left (-2580 \cos \left (d x +c \right )-1075 \cos \left (2 d x +2 c \right )-1849\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )+\left (2580 \cos \left (d x +c \right )+1075 \cos \left (2 d x +2 c \right )+1849\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )-440 \sin \left (d x +c \right )-900 \sin \left (2 d x +2 c \right )}{2048 d \left (43+25 \cos \left (2 d x +2 c \right )+60 \cos \left (d x +c \right )\right )}\) | \(117\) |
(-35/128/d*tan(1/2*d*x+1/2*c)+85/512/d*tan(1/2*d*x+1/2*c)^3)/(tan(1/2*d*x+ 1/2*c)^2-4)^2-43/2048/d*ln(tan(1/2*d*x+1/2*c)-2)+43/2048/d*ln(tan(1/2*d*x+ 1/2*c)+2)
Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=\frac {43 \, {\left (25 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) - 43 \, {\left (25 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) - 40 \, {\left (45 \, \cos \left (d x + c\right ) + 11\right )} \sin \left (d x + c\right )}{4096 \, {\left (25 \, d \cos \left (d x + c\right )^{2} + 30 \, d \cos \left (d x + c\right ) + 9 \, d\right )}} \]
1/4096*(43*(25*cos(d*x + c)^2 + 30*cos(d*x + c) + 9)*log(3/2*cos(d*x + c) + 2*sin(d*x + c) + 5/2) - 43*(25*cos(d*x + c)^2 + 30*cos(d*x + c) + 9)*log (3/2*cos(d*x + c) - 2*sin(d*x + c) + 5/2) - 40*(45*cos(d*x + c) + 11)*sin( d*x + c))/(25*d*cos(d*x + c)^2 + 30*d*cos(d*x + c) + 9*d)
Leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (102) = 204\).
Time = 1.18 (sec) , antiderivative size = 474, normalized size of antiderivative = 4.12 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=\begin {cases} \frac {x}{\left (5 \cos {\left (2 \operatorname {atan}{\left (2 \right )} \right )} + 3\right )^{3}} & \text {for}\: c = - d x - 2 \operatorname {atan}{\left (2 \right )} \vee c = - d x + 2 \operatorname {atan}{\left (2 \right )} \\\frac {x}{\left (5 \cos {\left (c \right )} + 3\right )^{3}} & \text {for}\: d = 0 \\- \frac {43 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2048 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32768 d} + \frac {344 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2048 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32768 d} - \frac {688 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )}}{2048 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32768 d} + \frac {43 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2048 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32768 d} - \frac {344 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2048 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32768 d} + \frac {688 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )}}{2048 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32768 d} + \frac {340 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2048 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32768 d} - \frac {560 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2048 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32768 d} & \text {otherwise} \end {cases} \]
Piecewise((x/(5*cos(2*atan(2)) + 3)**3, Eq(c, -d*x - 2*atan(2)) | Eq(c, -d *x + 2*atan(2))), (x/(5*cos(c) + 3)**3, Eq(d, 0)), (-43*log(tan(c/2 + d*x/ 2) - 2)*tan(c/2 + d*x/2)**4/(2048*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 32768*d) + 344*log(tan(c/2 + d*x/2) - 2)*tan(c/2 + d*x/2)**2 /(2048*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 32768*d) - 68 8*log(tan(c/2 + d*x/2) - 2)/(2048*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 32768*d) + 43*log(tan(c/2 + d*x/2) + 2)*tan(c/2 + d*x/2)**4/ (2048*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 32768*d) - 344 *log(tan(c/2 + d*x/2) + 2)*tan(c/2 + d*x/2)**2/(2048*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 32768*d) + 688*log(tan(c/2 + d*x/2) + 2)/ (2048*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 32768*d) + 340 *tan(c/2 + d*x/2)**3/(2048*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/2 + d*x/2 )**2 + 32768*d) - 560*tan(c/2 + d*x/2)/(2048*d*tan(c/2 + d*x/2)**4 - 16384 *d*tan(c/2 + d*x/2)**2 + 32768*d), True))
Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=\frac {\frac {20 \, {\left (\frac {28 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {17 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{\frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 16} + 43 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 2\right ) - 43 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 2\right )}{2048 \, d} \]
1/2048*(20*(28*sin(d*x + c)/(cos(d*x + c) + 1) - 17*sin(d*x + c)^3/(cos(d* x + c) + 1)^3)/(8*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - sin(d*x + c)^4/(co s(d*x + c) + 1)^4 - 16) + 43*log(sin(d*x + c)/(cos(d*x + c) + 1) + 2) - 43 *log(sin(d*x + c)/(cos(d*x + c) + 1) - 2))/d
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=\frac {\frac {20 \, {\left (17 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 28 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4\right )}^{2}} + 43 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right ) - 43 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )}{2048 \, d} \]
1/2048*(20*(17*tan(1/2*d*x + 1/2*c)^3 - 28*tan(1/2*d*x + 1/2*c))/(tan(1/2* d*x + 1/2*c)^2 - 4)^2 + 43*log(abs(tan(1/2*d*x + 1/2*c) + 2)) - 43*log(abs (tan(1/2*d*x + 1/2*c) - 2)))/d
Time = 14.75 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=\frac {43\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{1024\,d}-\frac {\frac {35\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128}-\frac {85\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{512}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+16\right )} \]