Integrand size = 31, antiderivative size = 96 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {(c-d)^2 (a+a \sin (e+f x))^{1+m}}{a f (1+m)}+\frac {2 (c-d) d (a+a \sin (e+f x))^{2+m}}{a^2 f (2+m)}+\frac {d^2 (a+a \sin (e+f x))^{3+m}}{a^3 f (3+m)} \] Output:
(c-d)^2*(a+a*sin(f*x+e))^(1+m)/a/f/(1+m)+2*(c-d)*d*(a+a*sin(f*x+e))^(2+m)/ a^2/f/(2+m)+d^2*(a+a*sin(f*x+e))^(3+m)/a^3/f/(3+m)
Time = 0.43 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {(a (1+\sin (e+f x)))^{1+m} \left (\frac {a^2 (c-d)^2}{1+m}+\frac {2 a^2 (c-d) d (1+\sin (e+f x))}{2+m}+\frac {d^2 (a+a \sin (e+f x))^2}{3+m}\right )}{a^3 f} \] Input:
Integrate[Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^2,x]
Output:
((a*(1 + Sin[e + f*x]))^(1 + m)*((a^2*(c - d)^2)/(1 + m) + (2*a^2*(c - d)* d*(1 + Sin[e + f*x]))/(2 + m) + (d^2*(a + a*Sin[e + f*x])^2)/(3 + m)))/(a^ 3*f)
Time = 0.35 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3312, 27, 53, 2009}
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.
| \(\displaystyle \int \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2dx\) |
| \(\Big \downarrow \) 3312 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^m (a c+a d \sin (e+f x))^2}{a^2}d(a \sin (e+f x))}{a f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^m (a c+a d \sin (e+f x))^2d(a \sin (e+f x))}{a^3 f}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {\int \left (a^2 (c-d)^2 (\sin (e+f x) a+a)^m+2 a (c-d) d (\sin (e+f x) a+a)^{m+1}+d^2 (\sin (e+f x) a+a)^{m+2}\right )d(a \sin (e+f x))}{a^3 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^2 (c-d)^2 (a \sin (e+f x)+a)^{m+1}}{m+1}+\frac {2 a d (c-d) (a \sin (e+f x)+a)^{m+2}}{m+2}+\frac {d^2 (a \sin (e+f x)+a)^{m+3}}{m+3}}{a^3 f}\) |
Input:
Int[Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^2,x]
Output:
((a^2*(c - d)^2*(a + a*Sin[e + f*x])^(1 + m))/(1 + m) + (2*a*(c - d)*d*(a + a*Sin[e + f*x])^(2 + m))/(2 + m) + (d^2*(a + a*Sin[e + f*x])^(3 + m))/(3 + m))/(a^3*f)
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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 2.32 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.60
| method | result | size |
| parallelrisch | \(-\frac {\left (4 \left (1+m \right ) d \left (\frac {d m}{2}+c \left (3+m \right )\right ) \cos \left (2 f x +2 e \right )+\sin \left (3 f x +3 e \right ) d^{2} \left (2+m \right ) \left (1+m \right )+\left (\left (-3 m^{2}-m -6\right ) d^{2}-8 c m \left (3+m \right ) d -4 c^{2} \left (3+m \right ) \left (2+m \right )\right ) \sin \left (f x +e \right )+\left (-2 m^{2}-2 m -8\right ) d^{2}-4 c \left (3+m \right ) \left (-1+m \right ) d -4 c^{2} \left (3+m \right ) \left (2+m \right )\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{m}}{4 \left (m^{3}+6 m^{2}+11 m +6\right ) f}\) | \(154\) |
| derivativedivides | \(\frac {\left (c^{2} m^{2}+5 c^{2} m -2 c d m +6 c^{2}-6 c d +2 d^{2}\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{3}+6 m^{2}+11 m +6\right )}+\frac {d^{2} \sin \left (f x +e \right )^{3} {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (3+m \right )}+\frac {\left (c^{2} m^{2}+2 c d \,m^{2}+5 c^{2} m +6 c d m -2 d^{2} m +6 c^{2}\right ) \sin \left (f x +e \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{3}+6 m^{2}+11 m +6\right )}+\frac {\left (2 c m +d m +6 c \right ) d \sin \left (f x +e \right )^{2} {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{2}+5 m +6\right )}\) | \(226\) |
| default | \(\frac {\left (c^{2} m^{2}+5 c^{2} m -2 c d m +6 c^{2}-6 c d +2 d^{2}\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{3}+6 m^{2}+11 m +6\right )}+\frac {d^{2} \sin \left (f x +e \right )^{3} {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (3+m \right )}+\frac {\left (c^{2} m^{2}+2 c d \,m^{2}+5 c^{2} m +6 c d m -2 d^{2} m +6 c^{2}\right ) \sin \left (f x +e \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{3}+6 m^{2}+11 m +6\right )}+\frac {\left (2 c m +d m +6 c \right ) d \sin \left (f x +e \right )^{2} {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{2}+5 m +6\right )}\) | \(226\) |
Input:
int(cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^2,x,method=_RETURNVERBO SE)
Output:
-1/4*(4*(1+m)*d*(1/2*d*m+c*(3+m))*cos(2*f*x+2*e)+sin(3*f*x+3*e)*d^2*(2+m)* (1+m)+((-3*m^2-m-6)*d^2-8*c*m*(3+m)*d-4*c^2*(3+m)*(2+m))*sin(f*x+e)+(-2*m^ 2-2*m-8)*d^2-4*c*(3+m)*(-1+m)*d-4*c^2*(3+m)*(2+m))*(a*(1+sin(f*x+e)))^m/(m ^3+6*m^2+11*m+6)/f
Time = 0.09 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.97 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {{\left ({\left (c^{2} + 2 \, c d + d^{2}\right )} m^{2} - {\left ({\left (2 \, c d + d^{2}\right )} m^{2} + 6 \, c d + {\left (8 \, c d + d^{2}\right )} m\right )} \cos \left (f x + e\right )^{2} + 6 \, c^{2} + 2 \, d^{2} + {\left (5 \, c^{2} + 6 \, c d + d^{2}\right )} m + {\left ({\left (c^{2} + 2 \, c d + d^{2}\right )} m^{2} - {\left (d^{2} m^{2} + 3 \, d^{2} m + 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, c^{2} + 2 \, d^{2} + {\left (5 \, c^{2} + 6 \, c d + d^{2}\right )} m\right )} \sin \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{f m^{3} + 6 \, f m^{2} + 11 \, f m + 6 \, f} \] Input:
integrate(cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^2,x, algorithm="f ricas")
Output:
((c^2 + 2*c*d + d^2)*m^2 - ((2*c*d + d^2)*m^2 + 6*c*d + (8*c*d + d^2)*m)*c os(f*x + e)^2 + 6*c^2 + 2*d^2 + (5*c^2 + 6*c*d + d^2)*m + ((c^2 + 2*c*d + d^2)*m^2 - (d^2*m^2 + 3*d^2*m + 2*d^2)*cos(f*x + e)^2 + 6*c^2 + 2*d^2 + (5 *c^2 + 6*c*d + d^2)*m)*sin(f*x + e))*(a*sin(f*x + e) + a)^m/(f*m^3 + 6*f*m ^2 + 11*f*m + 6*f)
Leaf count of result is larger than twice the leaf count of optimal. 1622 vs. \(2 (80) = 160\).
Time = 2.72 (sec) , antiderivative size = 1622, normalized size of antiderivative = 16.90 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\text {Too large to display} \] Input:
integrate(cos(f*x+e)*(a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**2,x)
Output:
Piecewise((x*(c + d*sin(e))**2*(a*sin(e) + a)**m*cos(e), Eq(f, 0)), (-c**2 /(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) - 4*c*d*sin (e + f*x)/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) - 2*c*d/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) + 2*d* *2*log(sin(e + f*x) + 1)*sin(e + f*x)**2/(2*a**3*f*sin(e + f*x)**2 + 4*a** 3*f*sin(e + f*x) + 2*a**3*f) + 4*d**2*log(sin(e + f*x) + 1)*sin(e + f*x)/( 2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) + 2*d**2*log( sin(e + f*x) + 1)/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a* *3*f) + 4*d**2*sin(e + f*x)/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f *x) + 2*a**3*f) + 3*d**2/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f), Eq(m, -3)), (-c**2/(a**2*f*sin(e + f*x) + a**2*f) + 2*c*d*lo g(sin(e + f*x) + 1)*sin(e + f*x)/(a**2*f*sin(e + f*x) + a**2*f) + 2*c*d*lo g(sin(e + f*x) + 1)/(a**2*f*sin(e + f*x) + a**2*f) + 2*c*d/(a**2*f*sin(e + f*x) + a**2*f) - 2*d**2*log(sin(e + f*x) + 1)*sin(e + f*x)/(a**2*f*sin(e + f*x) + a**2*f) - 2*d**2*log(sin(e + f*x) + 1)/(a**2*f*sin(e + f*x) + a** 2*f) + d**2*sin(e + f*x)**2/(a**2*f*sin(e + f*x) + a**2*f) - 2*d**2/(a**2* f*sin(e + f*x) + a**2*f), Eq(m, -2)), (c**2*log(sin(e + f*x) + 1)/(a*f) - 2*c*d*log(sin(e + f*x) + 1)/(a*f) + 2*c*d*sin(e + f*x)/(a*f) + d**2*log(si n(e + f*x) + 1)/(a*f) + d**2*sin(e + f*x)**2/(2*a*f) - d**2*sin(e + f*x)/( a*f), Eq(m, -1)), (c**2*m**2*(a*sin(e + f*x) + a)**m*sin(e + f*x)/(f*m*...
Time = 0.04 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.78 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {\frac {2 \, {\left (a^{m} {\left (m + 1\right )} \sin \left (f x + e\right )^{2} + a^{m} m \sin \left (f x + e\right ) - a^{m}\right )} c d {\left (\sin \left (f x + e\right ) + 1\right )}^{m}}{m^{2} + 3 \, m + 2} + \frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} a^{m} \sin \left (f x + e\right )^{3} + {\left (m^{2} + m\right )} a^{m} \sin \left (f x + e\right )^{2} - 2 \, a^{m} m \sin \left (f x + e\right ) + 2 \, a^{m}\right )} d^{2} {\left (\sin \left (f x + e\right ) + 1\right )}^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} + \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m + 1} c^{2}}{a {\left (m + 1\right )}}}{f} \] Input:
integrate(cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^2,x, algorithm="m axima")
Output:
(2*(a^m*(m + 1)*sin(f*x + e)^2 + a^m*m*sin(f*x + e) - a^m)*c*d*(sin(f*x + e) + 1)^m/(m^2 + 3*m + 2) + ((m^2 + 3*m + 2)*a^m*sin(f*x + e)^3 + (m^2 + m )*a^m*sin(f*x + e)^2 - 2*a^m*m*sin(f*x + e) + 2*a^m)*d^2*(sin(f*x + e) + 1 )^m/(m^3 + 6*m^2 + 11*m + 6) + (a*sin(f*x + e) + a)^(m + 1)*c^2/(a*(m + 1) ))/f
Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (96) = 192\).
Time = 0.15 (sec) , antiderivative size = 293, normalized size of antiderivative = 3.05 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {\frac {2 \, {\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m} m \sin \left (f x + e\right )^{2} + {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m \sin \left (f x + e\right ) + {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sin \left (f x + e\right )^{2} - {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\right )} c d}{m^{2} + 3 \, m + 2} + \frac {{\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m} m^{2} \sin \left (f x + e\right )^{3} + {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m^{2} \sin \left (f x + e\right )^{2} + 3 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m \sin \left (f x + e\right )^{3} + {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m \sin \left (f x + e\right )^{2} + 2 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sin \left (f x + e\right )^{3} - 2 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m \sin \left (f x + e\right ) + 2 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\right )} d^{2}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} + \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m + 1} c^{2}}{a {\left (m + 1\right )}}}{f} \] Input:
integrate(cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^2,x, algorithm="g iac")
Output:
(2*((a*sin(f*x + e) + a)^m*m*sin(f*x + e)^2 + (a*sin(f*x + e) + a)^m*m*sin (f*x + e) + (a*sin(f*x + e) + a)^m*sin(f*x + e)^2 - (a*sin(f*x + e) + a)^m )*c*d/(m^2 + 3*m + 2) + ((a*sin(f*x + e) + a)^m*m^2*sin(f*x + e)^3 + (a*si n(f*x + e) + a)^m*m^2*sin(f*x + e)^2 + 3*(a*sin(f*x + e) + a)^m*m*sin(f*x + e)^3 + (a*sin(f*x + e) + a)^m*m*sin(f*x + e)^2 + 2*(a*sin(f*x + e) + a)^ m*sin(f*x + e)^3 - 2*(a*sin(f*x + e) + a)^m*m*sin(f*x + e) + 2*(a*sin(f*x + e) + a)^m)*d^2/(m^3 + 6*m^2 + 11*m + 6) + (a*sin(f*x + e) + a)^(m + 1)*c ^2/(a*(m + 1)))/f
Time = 34.21 (sec) , antiderivative size = 305, normalized size of antiderivative = 3.18 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (20\,c^2\,m-12\,c\,d+2\,d^2\,m+24\,c^2\,\sin \left (e+f\,x\right )+6\,d^2\,\sin \left (e+f\,x\right )+24\,c^2+8\,d^2+4\,c^2\,m^2+2\,d^2\,m^2-2\,d^2\,\sin \left (3\,e+3\,f\,x\right )+20\,c^2\,m\,\sin \left (e+f\,x\right )+d^2\,m\,\sin \left (e+f\,x\right )-2\,d^2\,m\,\cos \left (2\,e+2\,f\,x\right )+4\,c^2\,m^2\,\sin \left (e+f\,x\right )-3\,d^2\,m\,\sin \left (3\,e+3\,f\,x\right )+3\,d^2\,m^2\,\sin \left (e+f\,x\right )+8\,c\,d\,m-2\,d^2\,m^2\,\cos \left (2\,e+2\,f\,x\right )-d^2\,m^2\,\sin \left (3\,e+3\,f\,x\right )-12\,c\,d\,\cos \left (2\,e+2\,f\,x\right )+4\,c\,d\,m^2+24\,c\,d\,m\,\sin \left (e+f\,x\right )-16\,c\,d\,m\,\cos \left (2\,e+2\,f\,x\right )+8\,c\,d\,m^2\,\sin \left (e+f\,x\right )-4\,c\,d\,m^2\,\cos \left (2\,e+2\,f\,x\right )\right )}{4\,f\,\left (m^3+6\,m^2+11\,m+6\right )} \] Input:
int(cos(e + f*x)*(a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^2,x)
Output:
((a*(sin(e + f*x) + 1))^m*(20*c^2*m - 12*c*d + 2*d^2*m + 24*c^2*sin(e + f* x) + 6*d^2*sin(e + f*x) + 24*c^2 + 8*d^2 + 4*c^2*m^2 + 2*d^2*m^2 - 2*d^2*s in(3*e + 3*f*x) + 20*c^2*m*sin(e + f*x) + d^2*m*sin(e + f*x) - 2*d^2*m*cos (2*e + 2*f*x) + 4*c^2*m^2*sin(e + f*x) - 3*d^2*m*sin(3*e + 3*f*x) + 3*d^2* m^2*sin(e + f*x) + 8*c*d*m - 2*d^2*m^2*cos(2*e + 2*f*x) - d^2*m^2*sin(3*e + 3*f*x) - 12*c*d*cos(2*e + 2*f*x) + 4*c*d*m^2 + 24*c*d*m*sin(e + f*x) - 1 6*c*d*m*cos(2*e + 2*f*x) + 8*c*d*m^2*sin(e + f*x) - 4*c*d*m^2*cos(2*e + 2* f*x)))/(4*f*(11*m + 6*m^2 + m^3 + 6))
Time = 0.17 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.56 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (\sin \left (f x +e \right )^{3} d^{2} m^{2}+3 \sin \left (f x +e \right )^{3} d^{2} m +2 \sin \left (f x +e \right )^{3} d^{2}+2 \sin \left (f x +e \right )^{2} c d \,m^{2}+8 \sin \left (f x +e \right )^{2} c d m +6 \sin \left (f x +e \right )^{2} c d +\sin \left (f x +e \right )^{2} d^{2} m^{2}+\sin \left (f x +e \right )^{2} d^{2} m +\sin \left (f x +e \right ) c^{2} m^{2}+5 \sin \left (f x +e \right ) c^{2} m +6 \sin \left (f x +e \right ) c^{2}+2 \sin \left (f x +e \right ) c d \,m^{2}+6 \sin \left (f x +e \right ) c d m -2 \sin \left (f x +e \right ) d^{2} m +c^{2} m^{2}+5 c^{2} m +6 c^{2}-2 c d m -6 c d +2 d^{2}\right )}{f \left (m^{3}+6 m^{2}+11 m +6\right )} \] Input:
int(cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^2,x)
Output:
((sin(e + f*x)*a + a)**m*(sin(e + f*x)**3*d**2*m**2 + 3*sin(e + f*x)**3*d* *2*m + 2*sin(e + f*x)**3*d**2 + 2*sin(e + f*x)**2*c*d*m**2 + 8*sin(e + f*x )**2*c*d*m + 6*sin(e + f*x)**2*c*d + sin(e + f*x)**2*d**2*m**2 + sin(e + f *x)**2*d**2*m + sin(e + f*x)*c**2*m**2 + 5*sin(e + f*x)*c**2*m + 6*sin(e + f*x)*c**2 + 2*sin(e + f*x)*c*d*m**2 + 6*sin(e + f*x)*c*d*m - 2*sin(e + f* x)*d**2*m + c**2*m**2 + 5*c**2*m + 6*c**2 - 2*c*d*m - 6*c*d + 2*d**2))/(f* (m**3 + 6*m**2 + 11*m + 6))