Integrand size = 40, antiderivative size = 69 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {(B-C) \text {arctanh}(\sin (c+d x))}{a d}+\frac {(2 B-C) \tan (c+d x)}{a d}-\frac {(B-C) \tan (c+d x)}{d (a+a \cos (c+d x))} \] Output:
-(B-C)*arctanh(sin(d*x+c))/a/d+(2*B-C)*tan(d*x+c)/a/d-(B-C)*tan(d*x+c)/d/( a+a*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(201\) vs. \(2(69)=138\).
Time = 1.80 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.91 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left ((B-C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \left ((B-C) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\frac {B \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )\right )}{a d (1+\cos (c+d x))} \] Input:
Integrate[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[ c + d*x]),x]
Output:
(2*Cos[(c + d*x)/2]*((B - C)*Sec[c/2]*Sin[(d*x)/2] + Cos[(c + d*x)/2]*((B - C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Si n[(c + d*x)/2]]) + (B*Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2 ])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x) /2])))))/(a*d*(1 + Cos[c + d*x]))
Time = 0.66 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3508, 3042, 3457, 3042, 3227, 3042, 4254, 24, 4257}
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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a \cos (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \frac {\sec ^2(c+d x) (B+C \cos (c+d x))}{a \cos (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int (a (2 B-C)-a (B-C) \cos (c+d x)) \sec ^2(c+d x)dx}{a^2}-\frac {(B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (2 B-C)-a (B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {(B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {a (2 B-C) \int \sec ^2(c+d x)dx-a (B-C) \int \sec (c+d x)dx}{a^2}-\frac {(B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (2 B-C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-a (B-C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {-\frac {a (2 B-C) \int 1d(-\tan (c+d x))}{d}-a (B-C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {a (2 B-C) \tan (c+d x)}{d}-a (B-C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {a (2 B-C) \tan (c+d x)}{d}-\frac {a (B-C) \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {(B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}\) |
Input:
Int[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d* x]),x]
Output:
-(((B - C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x]))) + (-((a*(B - C)*ArcTanh [Sin[c + d*x]])/d) + (a*(2*B - C)*Tan[c + d*x])/d)/a^2
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.36 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.35
| method | result | size |
| parallelrisch | \(\frac {\left (B -C \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (B -C \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (B -\frac {C}{2}\right ) \cos \left (d x +c \right )+\frac {B}{2}\right )}{a d \cos \left (d x +c \right )}\) | \(93\) |
| derivativedivides | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (B -C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-B +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\) | \(100\) |
| default | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (B -C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-B +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\) | \(100\) |
| risch | \(\frac {2 i \left (B \,{\mathrm e}^{2 i \left (d x +c \right )}-C \,{\mathrm e}^{2 i \left (d x +c \right )}+B \,{\mathrm e}^{i \left (d x +c \right )}+2 B -C \right )}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a d}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a d}\) | \(164\) |
| norman | \(\frac {\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}+\frac {\left (3 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}-\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {2 \left (2 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {\left (B -C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}-\frac {\left (B -C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}\) | \(195\) |
Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c)),x,method=_ RETURNVERBOSE)
Output:
((B-C)*cos(d*x+c)*ln(tan(1/2*d*x+1/2*c)-1)-(B-C)*cos(d*x+c)*ln(tan(1/2*d*x +1/2*c)+1)+2*tan(1/2*d*x+1/2*c)*((B-1/2*C)*cos(d*x+c)+1/2*B))/a/d/cos(d*x+ c)
Time = 0.11 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.84 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {{\left ({\left (B - C\right )} \cos \left (d x + c\right )^{2} + {\left (B - C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (B - C\right )} \cos \left (d x + c\right )^{2} + {\left (B - C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (2 \, B - C\right )} \cos \left (d x + c\right ) + B\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c)),x, a lgorithm="fricas")
Output:
-1/2*(((B - C)*cos(d*x + c)^2 + (B - C)*cos(d*x + c))*log(sin(d*x + c) + 1 ) - ((B - C)*cos(d*x + c)^2 + (B - C)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*((2*B - C)*cos(d*x + c) + B)*sin(d*x + c))/(a*d*cos(d*x + c)^2 + a*d* cos(d*x + c))
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\int \frac {B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+a*cos(d*x+c)),x)
Output:
(Integral(B*cos(c + d*x)*sec(c + d*x)**3/(cos(c + d*x) + 1), x) + Integral (C*cos(c + d*x)**2*sec(c + d*x)**3/(cos(c + d*x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (69) = 138\).
Time = 0.04 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.84 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {B {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a - \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - C {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c)),x, a lgorithm="maxima")
Output:
-(B*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d* x + c) + 1) - 1)/a - 2*sin(d*x + c)/((a - a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1))) - C*(log (sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d
Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.59 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\frac {{\left (B - C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {{\left (B - C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} + \frac {2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a}}{d} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c)),x, a lgorithm="giac")
Output:
-((B - C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - (B - C)*log(abs(tan(1/2*d *x + 1/2*c) - 1))/a - (B*tan(1/2*d*x + 1/2*c) - C*tan(1/2*d*x + 1/2*c))/a + 2*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a))/d
Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.13 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B-C\right )}{a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B-C\right )}{a\,d} \] Input:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x ))),x)
Output:
(2*B*tan(c/2 + (d*x)/2))/(d*(a - a*tan(c/2 + (d*x)/2)^2)) - (2*atanh(tan(c /2 + (d*x)/2))*(B - C))/(a*d) + (tan(c/2 + (d*x)/2)*(B - C))/(a*d)
Time = 0.16 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.51 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b -\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) c -\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) c +\cos \left (d x +c \right ) b -\cos \left (d x +c \right ) c +2 \sin \left (d x +c \right )^{2} b -\sin \left (d x +c \right )^{2} c -b +c}{\cos \left (d x +c \right ) \sin \left (d x +c \right ) a d} \] Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c)),x)
Output:
(cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*b - cos(c + d*x)*log( tan((c + d*x)/2) - 1)*sin(c + d*x)*c - cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*b + cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*c + cos(c + d*x)*b - cos(c + d*x)*c + 2*sin(c + d*x)**2*b - sin(c + d*x)**2 *c - b + c)/(cos(c + d*x)*sin(c + d*x)*a*d)