Integrand size = 27, antiderivative size = 86 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=2 a^2 x-\frac {4 a^2 \cos (c+d x)}{3 d}-\frac {2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2} \] Output:
2*a^2*x-4/3*a^2*cos(d*x+c)/d-2*a^2*cos(d*x+c)/d/(1-sin(d*x+c))+1/3*a^4*cos (d*x+c)*sin(d*x+c)^2/d/(a-a*sin(d*x+c))^2
Time = 2.16 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.52 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {a^2 (1+\sin (c+d x))^2 \left (6 c+6 d x-3 \cos (c+d x)+\frac {1}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right ) (-7+8 \sin (c+d x))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \] Input:
Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^3,x]
Output:
(a^2*(1 + Sin[c + d*x])^2*(6*c + 6*d*x - 3*Cos[c + d*x] + (Cos[(c + d*x)/2 ] - Sin[(c + d*x)/2])^(-2) + (2*Sin[(c + d*x)/2]*(-7 + 8*Sin[c + d*x]))/(C os[(c + d*x)/2] - Sin[(c + d*x)/2])^3))/(3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)
Time = 0.77 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, Rules used = {3042, 3348, 3042, 3244, 27, 3042, 3447, 3042, 3502, 27, 3042, 3214, 3042, 3127}
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.
| \(\displaystyle \int \tan ^3(c+d x) \sec (c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 (a \sin (c+d x)+a)^2}{\cos (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3348 |
| \(\displaystyle a^4 \int \frac {\sin ^3(c+d x)}{(a-a \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \int \frac {\sin (c+d x)^3}{(a-a \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3244 |
| \(\displaystyle a^4 \left (\frac {\int -\frac {2 \sin (c+d x) (2 \sin (c+d x) a+a)}{a-a \sin (c+d x)}dx}{3 a^2}+\frac {\sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle a^4 \left (\frac {\sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 \int \frac {\sin (c+d x) (2 \sin (c+d x) a+a)}{a-a \sin (c+d x)}dx}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \left (\frac {\sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 \int \frac {\sin (c+d x) (2 \sin (c+d x) a+a)}{a-a \sin (c+d x)}dx}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle a^4 \left (\frac {\sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 \int \frac {2 a \sin ^2(c+d x)+a \sin (c+d x)}{a-a \sin (c+d x)}dx}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \left (\frac {\sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 \int \frac {2 a \sin (c+d x)^2+a \sin (c+d x)}{a-a \sin (c+d x)}dx}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle a^4 \left (\frac {\sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 \left (\frac {2 \cos (c+d x)}{d}-\frac {\int -\frac {3 a^2 \sin (c+d x)}{a-a \sin (c+d x)}dx}{a}\right )}{3 a^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle a^4 \left (\frac {\sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 \left (3 a \int \frac {\sin (c+d x)}{a-a \sin (c+d x)}dx+\frac {2 \cos (c+d x)}{d}\right )}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \left (\frac {\sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 \left (3 a \int \frac {\sin (c+d x)}{a-a \sin (c+d x)}dx+\frac {2 \cos (c+d x)}{d}\right )}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle a^4 \left (\frac {\sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 \left (3 a \left (\int \frac {1}{a-a \sin (c+d x)}dx-\frac {x}{a}\right )+\frac {2 \cos (c+d x)}{d}\right )}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \left (\frac {\sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 \left (3 a \left (\int \frac {1}{a-a \sin (c+d x)}dx-\frac {x}{a}\right )+\frac {2 \cos (c+d x)}{d}\right )}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle a^4 \left (\frac {\sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 \left (3 a \left (\frac {\cos (c+d x)}{d (a-a \sin (c+d x))}-\frac {x}{a}\right )+\frac {2 \cos (c+d x)}{d}\right )}{3 a^2}\right )\) |
Input:
Int[Sec[c + d*x]*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^3,x]
Output:
a^4*((Cos[c + d*x]*Sin[c + d*x]^2)/(3*d*(a - a*Sin[c + d*x])^2) - (2*((2*C os[c + d*x])/d + 3*a*(-(x/a) + Cos[c + d*x]/(d*(a - a*Sin[c + d*x])))))/(3 *a^2))
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_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* (2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a^(2*m) Int[(d* Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]
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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Result contains complex when optimal does not.
Time = 1.98 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03
| method | result | size |
| risch | \(2 a^{2} x -\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {2 a^{2} \left (-15 i {\mathrm e}^{i \left (d x +c \right )}+9 \,{\mathrm e}^{2 i \left (d x +c \right )}-8\right )}{3 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3}}\) | \(89\) |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+2 a^{2} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )+a^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )}{d}\) | \(162\) |
| default | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+2 a^{2} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )+a^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )}{d}\) | \(162\) |
Input:
int(sec(d*x+c)*(a+a*sin(d*x+c))^2*tan(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
2*a^2*x-1/2*a^2/d*exp(I*(d*x+c))-1/2*a^2/d*exp(-I*(d*x+c))-2/3*a^2*(-15*I* exp(I*(d*x+c))+9*exp(2*I*(d*x+c))-8)/d/(exp(I*(d*x+c))-I)^3
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (81) = 162\).
Time = 0.08 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.95 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {3 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} d x - {\left (6 \, a^{2} d x + 11 \, a^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + {\left (6 \, a^{2} d x - 13 \, a^{2}\right )} \cos \left (d x + c\right ) - {\left (12 \, a^{2} d x - 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, {\left (3 \, a^{2} d x - 7 \, a^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))^2*tan(d*x+c)^3,x, algorithm="fricas" )
Output:
-1/3*(3*a^2*cos(d*x + c)^3 + 12*a^2*d*x - (6*a^2*d*x + 11*a^2)*cos(d*x + c )^2 + a^2 + (6*a^2*d*x - 13*a^2)*cos(d*x + c) - (12*a^2*d*x - 3*a^2*cos(d* x + c)^2 - a^2 + 2*(3*a^2*d*x - 7*a^2)*cos(d*x + c))*sin(d*x + c))/(d*cos( d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*sin(d*x + c) - 2*d)
\[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=a^{2} \left (\int \tan ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))**2*tan(d*x+c)**3,x)
Output:
a**2*(Integral(tan(c + d*x)**3*sec(c + d*x), x) + Integral(2*sin(c + d*x)* tan(c + d*x)**3*sec(c + d*x), x) + Integral(sin(c + d*x)**2*tan(c + d*x)** 3*sec(c + d*x), x))
Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.10 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} - a^{2} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))^2*tan(d*x+c)^3,x, algorithm="maxima" )
Output:
1/3*(2*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^2 - a^2*((6*cos(d *x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c)) - (3*cos(d*x + c)^2 - 1)*a ^2/cos(d*x + c)^3)/d
Time = 0.43 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {2 \, {\left (3 \, {\left (d x + c\right )} a^{2} - \frac {3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}\right )}}{3 \, d} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))^2*tan(d*x+c)^3,x, algorithm="giac")
Output:
2/3*(3*(d*x + c)*a^2 - 3*a^2/(tan(1/2*d*x + 1/2*c)^2 + 1) + (6*a^2*tan(1/2 *d*x + 1/2*c)^2 - 15*a^2*tan(1/2*d*x + 1/2*c) + 7*a^2)/(tan(1/2*d*x + 1/2* c) - 1)^3)/d
Time = 36.16 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.47 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=2\,a^2\,x+\frac {2\,a^2\,\left (c+d\,x\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2\,\left (c+d\,x\right )-\frac {2\,a^2\,\left (9\,c+9\,d\,x-24\right )}{3}\right )-\frac {2\,a^2\,\left (3\,c+3\,d\,x-10\right )}{3}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (6\,a^2\,\left (c+d\,x\right )-\frac {2\,a^2\,\left (9\,c+9\,d\,x-6\right )}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (8\,a^2\,\left (c+d\,x\right )-\frac {2\,a^2\,\left (12\,c+12\,d\,x-18\right )}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (8\,a^2\,\left (c+d\,x\right )-\frac {2\,a^2\,\left (12\,c+12\,d\,x-22\right )}{3}\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int((tan(c + d*x)^3*(a + a*sin(c + d*x))^2)/cos(c + d*x),x)
Output:
2*a^2*x + (2*a^2*(c + d*x) - tan(c/2 + (d*x)/2)*(6*a^2*(c + d*x) - (2*a^2* (9*c + 9*d*x - 24))/3) - (2*a^2*(3*c + 3*d*x - 10))/3 + tan(c/2 + (d*x)/2) ^4*(6*a^2*(c + d*x) - (2*a^2*(9*c + 9*d*x - 6))/3) - tan(c/2 + (d*x)/2)^3* (8*a^2*(c + d*x) - (2*a^2*(12*c + 12*d*x - 18))/3) + tan(c/2 + (d*x)/2)^2* (8*a^2*(c + d*x) - (2*a^2*(12*c + 12*d*x - 22))/3))/(d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2)^2 + 1))
Time = 0.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.97 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {a^{2} \left (-3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x +8 \cos \left (d x +c \right ) \sin \left (d x +c \right )-6 \cos \left (d x +c \right ) d x -4 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )^{2} d x -17 \sin \left (d x +c \right )^{2}-12 \sin \left (d x +c \right ) d x +8 \sin \left (d x +c \right )+6 d x +4\right )}{3 d \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )-\cos \left (d x +c \right )+\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)*(a+a*sin(d*x+c))^2*tan(d*x+c)^3,x)
Output:
(a**2*( - 3*cos(c + d*x)*sin(c + d*x)**2 + 6*cos(c + d*x)*sin(c + d*x)*d*x + 8*cos(c + d*x)*sin(c + d*x) - 6*cos(c + d*x)*d*x - 4*cos(c + d*x) + 3*s in(c + d*x)**3 + 6*sin(c + d*x)**2*d*x - 17*sin(c + d*x)**2 - 12*sin(c + d *x)*d*x + 8*sin(c + d*x) + 6*d*x + 4))/(3*d*(cos(c + d*x)*sin(c + d*x) - c os(c + d*x) + sin(c + d*x)**2 - 2*sin(c + d*x) + 1))