Integrand size = 21, antiderivative size = 114 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {7 x}{2 a^2}-\frac {16 \sin (c+d x)}{3 a^2 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {8 \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:
7/2*x/a^2-16/3*sin(d*x+c)/a^2/d+7/2*cos(d*x+c)*sin(d*x+c)/a^2/d-8/3*cos(d* x+c)^2*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*cos(d*x+c)^3*sin(d*x+c)/d/(a+a* cos(d*x+c))^2
Time = 0.83 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\sin (c+d x) \left (-84 \arcsin (\cos (c+d x)) \cos ^4\left (\frac {1}{2} (c+d x)\right )+\left (-32-43 \cos (c+d x)-6 \cos ^2(c+d x)+3 \cos ^3(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{6 a^2 d \sqrt {1-\cos (c+d x)} (1+\cos (c+d x))^{5/2}} \] Input:
Integrate[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^2,x]
Output:
(Sin[c + d*x]*(-84*ArcSin[Cos[c + d*x]]*Cos[(c + d*x)/2]^4 + (-32 - 43*Cos [c + d*x] - 6*Cos[c + d*x]^2 + 3*Cos[c + d*x]^3)*Sqrt[Sin[c + d*x]^2]))/(6 *a^2*d*Sqrt[1 - Cos[c + d*x]]*(1 + Cos[c + d*x])^(5/2))
Time = 0.52 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3244, 3042, 3456, 3042, 3213}
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.
| \(\displaystyle \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3244 |
| \(\displaystyle -\frac {\int \frac {\cos ^2(c+d x) (3 a-5 a \cos (c+d x))}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (3 a-5 a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle -\frac {\frac {\int \cos (c+d x) \left (16 a^2-21 a^2 \cos (c+d x)\right )dx}{a^2}+\frac {8 \sin (c+d x) \cos ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (16 a^2-21 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {8 \sin (c+d x) \cos ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3213 |
| \(\displaystyle -\frac {\frac {\frac {16 a^2 \sin (c+d x)}{d}-\frac {21 a^2 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {21 a^2 x}{2}}{a^2}+\frac {8 \sin (c+d x) \cos ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
Input:
Int[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^2,x]
Output:
-1/3*(Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) - ((8*Cos[c + d*x]^2*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + ((-21*a^2*x)/2 + (16*a^2*S in[c + d*x])/d - (21*a^2*Cos[c + d*x]*Sin[c + d*x])/(2*d))/a^2)/(3*a^2)
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 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 - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{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] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 0])
Time = 0.94 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.58
| method | result | size |
| parallelrisch | \(\frac {-163 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos \left (d x +c \right )+\frac {12 \cos \left (2 d x +2 c \right )}{163}-\frac {3 \cos \left (3 d x +3 c \right )}{163}+\frac {140}{163}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+168 d x}{48 a^{2} d}\) | \(66\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+14 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(88\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+14 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(88\) |
| risch | \(\frac {7 x}{2 a^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{a^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{2} d}-\frac {2 i \left (12 \,{\mathrm e}^{2 i \left (d x +c \right )}+21 \,{\mathrm e}^{i \left (d x +c \right )}+11\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\) | \(126\) |
| norman | \(\frac {\frac {7 x}{2 a}-\frac {13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {149 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}-\frac {100 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}-\frac {18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 a d}+\frac {14 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {21 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}+\frac {14 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}+\frac {7 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}}{a \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\) | \(207\) |
Input:
int(cos(d*x+c)^4/(a+a*cos(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/48*(-163*tan(1/2*d*x+1/2*c)*(cos(d*x+c)+12/163*cos(2*d*x+2*c)-3/163*cos( 3*d*x+3*c)+140/163)*sec(1/2*d*x+1/2*c)^2+168*d*x)/a^2/d
Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {21 \, d x \cos \left (d x + c\right )^{2} + 42 \, d x \cos \left (d x + c\right ) + 21 \, d x + {\left (3 \, \cos \left (d x + c\right )^{3} - 6 \, \cos \left (d x + c\right )^{2} - 43 \, \cos \left (d x + c\right ) - 32\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^2,x, algorithm="fricas")
Output:
1/6*(21*d*x*cos(d*x + c)^2 + 42*d*x*cos(d*x + c) + 21*d*x + (3*cos(d*x + c )^3 - 6*cos(d*x + c)^2 - 43*cos(d*x + c) - 32)*sin(d*x + c))/(a^2*d*cos(d* x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 413 vs. \(2 (107) = 214\).
Time = 1.56 (sec) , antiderivative size = 413, normalized size of antiderivative = 3.62 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\begin {cases} \frac {21 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {42 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {21 d x}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {\tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {19 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {71 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {39 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{4}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**4/(a+a*cos(d*x+c))**2,x)
Output:
Piecewise((21*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a **2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 42*d*x*tan(c/2 + d*x/2)**2/(6*a**2 *d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*d* x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d ) + tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 19*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2 )**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 71*tan(c/2 + d*x/2)**3/ (6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 39*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d), Ne(d, 0)), (x*cos(c)**4/(a*cos(c) + a)**2, True))
Time = 0.12 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.44 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^2,x, algorithm="maxima")
Output:
-1/6*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c ) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d* x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.33 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {21 \, {\left (d x + c\right )}}{a^{2}} - \frac {6 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^2,x, algorithm="giac")
Output:
1/6*(21*(d*x + c)/a^2 - 6*(5*tan(1/2*d*x + 1/2*c)^3 + 3*tan(1/2*d*x + 1/2* c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + (a^4*tan(1/2*d*x + 1/2*c)^3 - 2 1*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d
Time = 41.86 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-22\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-30\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+12\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+21\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (c+d\,x\right )}{6\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \] Input:
int(cos(c + d*x)^4/(a + a*cos(c + d*x))^2,x)
Output:
(sin(c/2 + (d*x)/2) - 22*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) - 30*cos( c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2) + 12*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d *x)/2) + 21*cos(c/2 + (d*x)/2)^3*(c + d*x))/(6*a^2*d*cos(c/2 + (d*x)/2)^3)
Time = 0.16 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {-9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+21 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x -2 \cos \left (d x +c \right )-3 \sin \left (d x +c \right )^{4}-31 \sin \left (d x +c \right )^{2}+21 \sin \left (d x +c \right ) d x +2}{6 \sin \left (d x +c \right ) a^{2} d \left (\cos \left (d x +c \right )+1\right )} \] Input:
int(cos(d*x+c)^4/(a+a*cos(d*x+c))^2,x)
Output:
( - 9*cos(c + d*x)*sin(c + d*x)**2 + 21*cos(c + d*x)*sin(c + d*x)*d*x - 2* cos(c + d*x) - 3*sin(c + d*x)**4 - 31*sin(c + d*x)**2 + 21*sin(c + d*x)*d* x + 2)/(6*sin(c + d*x)*a**2*d*(cos(c + d*x) + 1))