Integrand size = 40, antiderivative size = 98 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {(3 B-2 C) x}{2 a}-\frac {2 (B-C) \sin (c+d x)}{a d}+\frac {(3 B-2 C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \] Output:
1/2*(3*B-2*C)*x/a-2*(B-C)*sin(d*x+c)/a/d+1/2*(3*B-2*C)*cos(d*x+c)*sin(d*x+ c)/a/d-(B-C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(197\) vs. \(2(98)=196\).
Time = 0.74 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.01 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (4 (3 B-2 C) d x \cos \left (\frac {d x}{2}\right )+4 (3 B-2 C) d x \cos \left (c+\frac {d x}{2}\right )-20 B \sin \left (\frac {d x}{2}\right )+20 C \sin \left (\frac {d x}{2}\right )-4 B \sin \left (c+\frac {d x}{2}\right )+4 C \sin \left (c+\frac {d x}{2}\right )-3 B \sin \left (c+\frac {3 d x}{2}\right )+4 C \sin \left (c+\frac {3 d x}{2}\right )-3 B \sin \left (2 c+\frac {3 d x}{2}\right )+4 C \sin \left (2 c+\frac {3 d x}{2}\right )+B \sin \left (2 c+\frac {5 d x}{2}\right )+B \sin \left (3 c+\frac {5 d x}{2}\right )\right )}{8 a d (1+\cos (c+d x))} \] Input:
Integrate[(Cos[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[ c + d*x]),x]
Output:
(Cos[(c + d*x)/2]*Sec[c/2]*(4*(3*B - 2*C)*d*x*Cos[(d*x)/2] + 4*(3*B - 2*C) *d*x*Cos[c + (d*x)/2] - 20*B*Sin[(d*x)/2] + 20*C*Sin[(d*x)/2] - 4*B*Sin[c + (d*x)/2] + 4*C*Sin[c + (d*x)/2] - 3*B*Sin[c + (3*d*x)/2] + 4*C*Sin[c + ( 3*d*x)/2] - 3*B*Sin[2*c + (3*d*x)/2] + 4*C*Sin[2*c + (3*d*x)/2] + B*Sin[2* c + (5*d*x)/2] + B*Sin[3*c + (5*d*x)/2]))/(8*a*d*(1 + Cos[c + d*x]))
Time = 0.66 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4560, 3042, 4508, 3042, 4274, 3042, 3115, 24, 3117}
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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\cos ^2(c+d x) (B+C \sec (c+d x))}{a \sec (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\int \cos ^2(c+d x) (a (3 B-2 C)-2 a (B-C) \sec (c+d x))dx}{a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (3 B-2 C)-2 a (B-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {a (3 B-2 C) \int \cos ^2(c+d x)dx-2 a (B-C) \int \cos (c+d x)dx}{a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (3 B-2 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-2 a (B-C) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {a (3 B-2 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-2 a (B-C) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {a (3 B-2 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-2 a (B-C) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle \frac {a (3 B-2 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {2 a (B-C) \sin (c+d x)}{d}}{a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
Input:
Int[(Cos[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d* x]),x]
Output:
-(((B - C)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))) + ((-2*a*( B - C)*Sin[c + d*x])/d + a*(3*B - 2*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/ (2*d)))/a^2
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[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
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_)*(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]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.33 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.62
| method | result | size |
| parallelrisch | \(\frac {\left (B \cos \left (2 d x +2 c \right )+\left (-2 B +4 C \right ) \cos \left (d x +c \right )-7 B +8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+6 d \left (B -\frac {2 C}{3}\right ) x}{4 d a}\) | \(61\) |
| derivativedivides | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {2 \left (-\frac {3 B}{2}+C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \left (-\frac {B}{2}+C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (3 B -2 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(100\) |
| default | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {2 \left (-\frac {3 B}{2}+C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \left (-\frac {B}{2}+C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (3 B -2 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(100\) |
| risch | \(\frac {3 B x}{2 a}-\frac {x C}{a}+\frac {i B \,{\mathrm e}^{i \left (d x +c \right )}}{2 a d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a d}-\frac {i B \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a d}-\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {B \sin \left (2 d x +2 c \right )}{4 a d}\) | \(156\) |
| norman | \(\frac {\frac {\left (2 B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {\left (3 B -2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}+\frac {\left (5 B -4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}-\frac {\left (3 B -2 C \right ) x}{2 a}-\frac {\left (B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}-\frac {\left (3 B -2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {\left (3 B -2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}-\frac {\left (5 B -4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) | \(239\) |
Input:
int(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x,method=_ RETURNVERBOSE)
Output:
1/4*((B*cos(2*d*x+2*c)+(-2*B+4*C)*cos(d*x+c)-7*B+8*C)*tan(1/2*d*x+1/2*c)+6 *d*(B-2/3*C)*x)/d/a
Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {{\left (3 \, B - 2 \, C\right )} d x \cos \left (d x + c\right ) + {\left (3 \, B - 2 \, C\right )} d x + {\left (B \cos \left (d x + c\right )^{2} - {\left (B - 2 \, C\right )} \cos \left (d x + c\right ) - 4 \, B + 4 \, C\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, a lgorithm="fricas")
Output:
1/2*((3*B - 2*C)*d*x*cos(d*x + c) + (3*B - 2*C)*d*x + (B*cos(d*x + c)^2 - (B - 2*C)*cos(d*x + c) - 4*B + 4*C)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)
Output:
(Integral(B*cos(c + d*x)**3*sec(c + d*x)/(sec(c + d*x) + 1), x) + Integral (C*cos(c + d*x)**3*sec(c + d*x)**2/(sec(c + d*x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (94) = 188\).
Time = 0.12 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.30 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {B {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + C {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \] Input:
integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, a lgorithm="maxima")
Output:
-(B*((sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^3/(cos(d*x + c) + 1 )^3)/(a + 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos( d*x + c) + 1)^4) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + sin(d*x + c)/(a*(cos(d*x + c) + 1))) + C*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 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))))/d
Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} {\left (3 \, B - 2 \, C\right )}}{a} - \frac {2 \, {\left (B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a}}{2 \, d} \] Input:
integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, a lgorithm="giac")
Output:
1/2*((d*x + c)*(3*B - 2*C)/a - 2*(B*tan(1/2*d*x + 1/2*c) - C*tan(1/2*d*x + 1/2*c))/a - 2*(3*B*tan(1/2*d*x + 1/2*c)^3 - 2*C*tan(1/2*d*x + 1/2*c)^3 + B*tan(1/2*d*x + 1/2*c) - 2*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^ 2 + 1)^2*a))/d
Time = 12.41 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {x\,\left (3\,B-2\,C\right )}{2\,a}-\frac {\left (3\,B-2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (B-2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B-C\right )}{a\,d} \] Input:
int((cos(c + d*x)^3*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d* x)),x)
Output:
(x*(3*B - 2*C))/(2*a) - (tan(c/2 + (d*x)/2)^3*(3*B - 2*C) + tan(c/2 + (d*x )/2)*(B - 2*C))/(d*(a + 2*a*tan(c/2 + (d*x)/2)^2 + a*tan(c/2 + (d*x)/2)^4) ) - (tan(c/2 + (d*x)/2)*(B - C))/(a*d)
Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b +2 \cos \left (d x +c \right ) b -2 \cos \left (d x +c \right ) c -2 \sin \left (d x +c \right )^{2} b +2 \sin \left (d x +c \right )^{2} c +3 \sin \left (d x +c \right ) b d x -2 \sin \left (d x +c \right ) c d x -2 b +2 c}{2 \sin \left (d x +c \right ) a d} \] Input:
int(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x)
Output:
(cos(c + d*x)*sin(c + d*x)**2*b + 2*cos(c + d*x)*b - 2*cos(c + d*x)*c - 2* sin(c + d*x)**2*b + 2*sin(c + d*x)**2*c + 3*sin(c + d*x)*b*d*x - 2*sin(c + d*x)*c*d*x - 2*b + 2*c)/(2*sin(c + d*x)*a*d)