\(\int \cos (c+d x) \sin ^3(a+b x) \, dx\) [199]

Optimal result

Integrand size = 15, antiderivative size = 97 \[ \int \cos (c+d x) \sin ^3(a+b x) \, dx=-\frac {3 \cos (a-c+(b-d) x)}{8 (b-d)}+\frac {\cos (3 a-c+(3 b-d) x)}{8 (3 b-d)}-\frac {3 \cos (a+c+(b+d) x)}{8 (b+d)}+\frac {\cos (3 a+c+(3 b+d) x)}{8 (3 b+d)} \] Output:

-3*cos(a-c+(b-d)*x)/(8*b-8*d)+cos(3*a-c+(3*b-d)*x)/(24*b-8*d)-3*cos(a+c+(b 
+d)*x)/(8*b+8*d)+cos(3*a+c+(3*b+d)*x)/(24*b+8*d)
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.93 \[ \int \cos (c+d x) \sin ^3(a+b x) \, dx=\frac {1}{8} \left (-\frac {3 \cos (a-c+b x-d x)}{b-d}+\frac {\cos (3 a-c+3 b x-d x)}{3 b-d}+\frac {\cos (3 a+c+3 b x+d x)}{3 b+d}-\frac {3 \cos (a+c+(b+d) x)}{b+d}\right ) \] Input:

Integrate[Cos[c + d*x]*Sin[a + b*x]^3,x]
 

Output:

((-3*Cos[a - c + b*x - d*x])/(b - d) + Cos[3*a - c + 3*b*x - d*x]/(3*b - d 
) + Cos[3*a + c + 3*b*x + d*x]/(3*b + d) - (3*Cos[a + c + (b + d)*x])/(b + 
 d))/8
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5085, 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.

\(\displaystyle \int \sin ^3(a+b x) \cos (c+d x) \, dx\)

\(\Big \downarrow \) 5085

\(\displaystyle \int \left (\frac {3}{8} \sin (a+x (b-d)-c)-\frac {1}{8} \sin (3 a+x (3 b-d)-c)+\frac {3}{8} \sin (a+x (b+d)+c)-\frac {1}{8} \sin (3 a+x (3 b+d)+c)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 \cos (a+x (b-d)-c)}{8 (b-d)}+\frac {\cos (3 a+x (3 b-d)-c)}{8 (3 b-d)}-\frac {3 \cos (a+x (b+d)+c)}{8 (b+d)}+\frac {\cos (3 a+x (3 b+d)+c)}{8 (3 b+d)}\)

Input:

Int[Cos[c + d*x]*Sin[a + b*x]^3,x]
 

Output:

(-3*Cos[a - c + (b - d)*x])/(8*(b - d)) + Cos[3*a - c + (3*b - d)*x]/(8*(3 
*b - d)) - (3*Cos[a + c + (b + d)*x])/(8*(b + d)) + Cos[3*a + c + (3*b + d 
)*x]/(8*(3*b + d))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5085

Int[Cos[w_]^(q_.)*Sin[v_]^(p_.), x_Symbol] :> Int[ExpandTrigReduce[Sin[v]^p 
*Cos[w]^q, x], x] /; IGtQ[p, 0] && IGtQ[q, 0] && ((PolynomialQ[v, x] && Pol 
ynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/w], x]))
 

Maple [A] (verified)

Time = 4.00 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {3 \cos \left (a -c +\left (b -d \right ) x \right )}{8 \left (b -d \right )}-\frac {3 \cos \left (a +c +\left (b +d \right ) x \right )}{8 \left (b +d \right )}+\frac {\cos \left (3 a -c +\left (3 b -d \right ) x \right )}{24 b -8 d}+\frac {\cos \left (3 a +c +\left (3 b +d \right ) x \right )}{24 b +8 d}\)	\(90\)
parallelrisch	\(\frac {-12 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{6} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{3}-24 b^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{5}+12 \left (\left (-3 b^{3}+b \,d^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-b \,d^{2}\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+16 \left (-4 b^{2} d +d^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}+12 \left (-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b \,d^{2}-3 b^{3}+b \,d^{2}\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-24 b^{2} d \tan \left (\frac {a}{2}+\frac {b x}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-12 b^{3}}{\left (b -d \right ) \left (b +d \right ) \left (3 b -d \right ) \left (3 b +d \right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )^{3}}\)	\(260\)
risch	\(-\frac {27 \cos \left (b x -d x +a -c \right ) b^{3}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (-b +d \right ) \left (-3 b +d \right )}-\frac {27 \cos \left (b x -d x +a -c \right ) b^{2} d}{8 \left (b +d \right ) \left (3 b +d \right ) \left (-b +d \right ) \left (-3 b +d \right )}+\frac {3 \cos \left (b x -d x +a -c \right ) b \,d^{2}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (-b +d \right ) \left (-3 b +d \right )}+\frac {3 \cos \left (b x -d x +a -c \right ) d^{3}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (-b +d \right ) \left (-3 b +d \right )}-\frac {27 \cos \left (b x +d x +a +c \right ) b^{3}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (b -d \right ) \left (3 b -d \right )}+\frac {27 \cos \left (b x +d x +a +c \right ) b^{2} d}{8 \left (b +d \right ) \left (3 b +d \right ) \left (b -d \right ) \left (3 b -d \right )}+\frac {3 \cos \left (b x +d x +a +c \right ) b \,d^{2}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (b -d \right ) \left (3 b -d \right )}-\frac {3 \cos \left (b x +d x +a +c \right ) d^{3}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (b -d \right ) \left (3 b -d \right )}+\frac {3 \cos \left (3 b x -d x +3 a -c \right ) b^{3}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (-b +d \right ) \left (-3 b +d \right )}+\frac {\cos \left (3 b x -d x +3 a -c \right ) b^{2} d}{8 \left (b +d \right ) \left (3 b +d \right ) \left (-b +d \right ) \left (-3 b +d \right )}-\frac {3 \cos \left (3 b x -d x +3 a -c \right ) b \,d^{2}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (-b +d \right ) \left (-3 b +d \right )}-\frac {\cos \left (3 b x -d x +3 a -c \right ) d^{3}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (-b +d \right ) \left (-3 b +d \right )}+\frac {3 \cos \left (3 b x +d x +3 a +c \right ) b^{3}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (b -d \right ) \left (3 b -d \right )}-\frac {\cos \left (3 b x +d x +3 a +c \right ) b^{2} d}{8 \left (b +d \right ) \left (3 b +d \right ) \left (b -d \right ) \left (3 b -d \right )}-\frac {3 \cos \left (3 b x +d x +3 a +c \right ) b \,d^{2}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (b -d \right ) \left (3 b -d \right )}+\frac {\cos \left (3 b x +d x +3 a +c \right ) d^{3}}{8 \left (b +d \right ) \left (3 b +d \right ) \left (b -d \right ) \left (3 b -d \right )}\)	\(730\)
orering	\(\text {Expression too large to display}\)	\(950\)

Input:

int(cos(d*x+c)*sin(b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

-3/8*cos(a-c+(b-d)*x)/(b-d)-3/8/(b+d)*cos(a+c+(b+d)*x)+1/8/(3*b-d)*cos(3*a 
-c+(3*b-d)*x)+1/8/(3*b+d)*cos(3*a+c+(3*b+d)*x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.20 \[ \int \cos (c+d x) \sin ^3(a+b x) \, dx=-\frac {{\left (7 \, b^{2} d - d^{3} - {\left (b^{2} d - d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right ) \sin \left (d x + c\right ) - 3 \, {\left ({\left (b^{3} - b d^{2}\right )} \cos \left (b x + a\right )^{3} - {\left (3 \, b^{3} - b d^{2}\right )} \cos \left (b x + a\right )\right )} \cos \left (d x + c\right )}{9 \, b^{4} - 10 \, b^{2} d^{2} + d^{4}} \] Input:

integrate(cos(d*x+c)*sin(b*x+a)^3,x, algorithm="fricas")
 

Output:

-((7*b^2*d - d^3 - (b^2*d - d^3)*cos(b*x + a)^2)*sin(b*x + a)*sin(d*x + c) 
 - 3*((b^3 - b*d^2)*cos(b*x + a)^3 - (3*b^3 - b*d^2)*cos(b*x + a))*cos(d*x 
 + c))/(9*b^4 - 10*b^2*d^2 + d^4)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 933 vs. \(2 (76) = 152\).

Time = 2.09 (sec) , antiderivative size = 933, normalized size of antiderivative = 9.62 \[ \int \cos (c+d x) \sin ^3(a+b x) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)*sin(b*x+a)**3,x)
 

Output:

Piecewise((x*sin(a)**3*cos(c), Eq(b, 0) & Eq(d, 0)), (3*x*sin(a - d*x)**3* 
cos(c + d*x)/8 + 3*x*sin(a - d*x)**2*sin(c + d*x)*cos(a - d*x)/8 + 3*x*sin 
(a - d*x)*cos(a - d*x)**2*cos(c + d*x)/8 + 3*x*sin(c + d*x)*cos(a - d*x)** 
3/8 - sin(a - d*x)**3*sin(c + d*x)/(8*d) + 3*sin(a - d*x)**2*cos(a - d*x)* 
cos(c + d*x)/(4*d) + 3*cos(a - d*x)**3*cos(c + d*x)/(8*d), Eq(b, -d)), (x* 
sin(a - d*x/3)**3*cos(c + d*x)/8 + 3*x*sin(a - d*x/3)**2*sin(c + d*x)*cos( 
a - d*x/3)/8 - 3*x*sin(a - d*x/3)*cos(a - d*x/3)**2*cos(c + d*x)/8 - x*sin 
(c + d*x)*cos(a - d*x/3)**3/8 + 9*sin(a - d*x/3)**3*sin(c + d*x)/(8*d) - 3 
*sin(a - d*x/3)**2*cos(a - d*x/3)*cos(c + d*x)/(4*d) - cos(a - d*x/3)**3*c 
os(c + d*x)/(8*d), Eq(b, -d/3)), (x*sin(a + d*x/3)**3*cos(c + d*x)/8 - 3*x 
*sin(a + d*x/3)**2*sin(c + d*x)*cos(a + d*x/3)/8 - 3*x*sin(a + d*x/3)*cos( 
a + d*x/3)**2*cos(c + d*x)/8 + x*sin(c + d*x)*cos(a + d*x/3)**3/8 + 9*sin( 
a + d*x/3)**3*sin(c + d*x)/(8*d) + 3*sin(a + d*x/3)**2*cos(a + d*x/3)*cos( 
c + d*x)/(4*d) + cos(a + d*x/3)**3*cos(c + d*x)/(8*d), Eq(b, d/3)), (3*x*s 
in(a + d*x)**3*cos(c + d*x)/8 - 3*x*sin(a + d*x)**2*sin(c + d*x)*cos(a + d 
*x)/8 + 3*x*sin(a + d*x)*cos(a + d*x)**2*cos(c + d*x)/8 - 3*x*sin(c + d*x) 
*cos(a + d*x)**3/8 + 5*sin(a + d*x)**3*sin(c + d*x)/(8*d) + 3*sin(a + d*x) 
*sin(c + d*x)*cos(a + d*x)**2/(4*d) + 3*cos(a + d*x)**3*cos(c + d*x)/(8*d) 
, Eq(b, d)), (-9*b**3*sin(a + b*x)**2*cos(a + b*x)*cos(c + d*x)/(9*b**4 - 
10*b**2*d**2 + d**4) - 6*b**3*cos(a + b*x)**3*cos(c + d*x)/(9*b**4 - 10...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 785 vs. \(2 (89) = 178\).

Time = 0.12 (sec) , antiderivative size = 785, normalized size of antiderivative = 8.09 \[ \int \cos (c+d x) \sin ^3(a+b x) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)*sin(b*x+a)^3,x, algorithm="maxima")
 

Output:

1/16*((3*b^3*cos(c) - b^2*d*cos(c) - 3*b*d^2*cos(c) + d^3*cos(c))*cos((3*b 
 + d)*x + 3*a + 2*c) + (3*b^3*cos(c) - b^2*d*cos(c) - 3*b*d^2*cos(c) + d^3 
*cos(c))*cos((3*b + d)*x + 3*a) + (3*b^3*cos(c) + b^2*d*cos(c) - 3*b*d^2*c 
os(c) - d^3*cos(c))*cos(-(3*b - d)*x - 3*a + 2*c) + (3*b^3*cos(c) + b^2*d* 
cos(c) - 3*b*d^2*cos(c) - d^3*cos(c))*cos(-(3*b - d)*x - 3*a) - 3*(9*b^3*c 
os(c) - 9*b^2*d*cos(c) - b*d^2*cos(c) + d^3*cos(c))*cos((b + d)*x + a + 2* 
c) - 3*(9*b^3*cos(c) - 9*b^2*d*cos(c) - b*d^2*cos(c) + d^3*cos(c))*cos((b 
+ d)*x + a) - 3*(9*b^3*cos(c) + 9*b^2*d*cos(c) - b*d^2*cos(c) - d^3*cos(c) 
)*cos(-(b - d)*x - a + 2*c) - 3*(9*b^3*cos(c) + 9*b^2*d*cos(c) - b*d^2*cos 
(c) - d^3*cos(c))*cos(-(b - d)*x - a) + (3*b^3*sin(c) - b^2*d*sin(c) - 3*b 
*d^2*sin(c) + d^3*sin(c))*sin((3*b + d)*x + 3*a + 2*c) - (3*b^3*sin(c) - b 
^2*d*sin(c) - 3*b*d^2*sin(c) + d^3*sin(c))*sin((3*b + d)*x + 3*a) + (3*b^3 
*sin(c) + b^2*d*sin(c) - 3*b*d^2*sin(c) - d^3*sin(c))*sin(-(3*b - d)*x - 3 
*a + 2*c) - (3*b^3*sin(c) + b^2*d*sin(c) - 3*b*d^2*sin(c) - d^3*sin(c))*si 
n(-(3*b - d)*x - 3*a) - 3*(9*b^3*sin(c) - 9*b^2*d*sin(c) - b*d^2*sin(c) + 
d^3*sin(c))*sin((b + d)*x + a + 2*c) + 3*(9*b^3*sin(c) - 9*b^2*d*sin(c) - 
b*d^2*sin(c) + d^3*sin(c))*sin((b + d)*x + a) - 3*(9*b^3*sin(c) + 9*b^2*d* 
sin(c) - b*d^2*sin(c) - d^3*sin(c))*sin(-(b - d)*x - a + 2*c) + 3*(9*b^3*s 
in(c) + 9*b^2*d*sin(c) - b*d^2*sin(c) - d^3*sin(c))*sin(-(b - d)*x - a))/( 
9*b^4*cos(c)^2 + 9*b^4*sin(c)^2 + (cos(c)^2 + sin(c)^2)*d^4 - 10*(b^2*c...
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.92 \[ \int \cos (c+d x) \sin ^3(a+b x) \, dx=\frac {\cos \left (3 \, b x + d x + 3 \, a + c\right )}{8 \, {\left (3 \, b + d\right )}} + \frac {\cos \left (3 \, b x - d x + 3 \, a - c\right )}{8 \, {\left (3 \, b - d\right )}} - \frac {3 \, \cos \left (b x + d x + a + c\right )}{8 \, {\left (b + d\right )}} - \frac {3 \, \cos \left (b x - d x + a - c\right )}{8 \, {\left (b - d\right )}} \] Input:

integrate(cos(d*x+c)*sin(b*x+a)^3,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

1/8*cos(3*b*x + d*x + 3*a + c)/(3*b + d) + 1/8*cos(3*b*x - d*x + 3*a - c)/ 
(3*b - d) - 3/8*cos(b*x + d*x + a + c)/(b + d) - 3/8*cos(b*x - d*x + a - c 
)/(b - d)
 

Mupad [B] (verification not implemented)

Time = 18.12 (sec) , antiderivative size = 471, normalized size of antiderivative = 4.86 \[ \int \cos (c+d x) \sin ^3(a+b x) \, dx=-{\mathrm {e}}^{a\,3{}\mathrm {i}-c\,1{}\mathrm {i}+b\,x\,3{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\left (\frac {-3\,b^3-b^2\,d+3\,b\,d^2+d^3}{144\,b^4-160\,b^2\,d^2+16\,d^4}+\frac {{\mathrm {e}}^{-a\,6{}\mathrm {i}-b\,x\,6{}\mathrm {i}}\,\left (-3\,b^3+b^2\,d+3\,b\,d^2-d^3\right )}{144\,b^4-160\,b^2\,d^2+16\,d^4}-\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (-27\,b^3-27\,b^2\,d+3\,b\,d^2+3\,d^3\right )}{144\,b^4-160\,b^2\,d^2+16\,d^4}-\frac {{\mathrm {e}}^{-a\,4{}\mathrm {i}-b\,x\,4{}\mathrm {i}}\,\left (-27\,b^3+27\,b^2\,d+3\,b\,d^2-3\,d^3\right )}{144\,b^4-160\,b^2\,d^2+16\,d^4}\right )-{\mathrm {e}}^{a\,3{}\mathrm {i}+c\,1{}\mathrm {i}+b\,x\,3{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {-3\,b^3+b^2\,d+3\,b\,d^2-d^3}{144\,b^4-160\,b^2\,d^2+16\,d^4}+\frac {{\mathrm {e}}^{-a\,6{}\mathrm {i}-b\,x\,6{}\mathrm {i}}\,\left (-3\,b^3-b^2\,d+3\,b\,d^2+d^3\right )}{144\,b^4-160\,b^2\,d^2+16\,d^4}-\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (-27\,b^3+27\,b^2\,d+3\,b\,d^2-3\,d^3\right )}{144\,b^4-160\,b^2\,d^2+16\,d^4}-\frac {{\mathrm {e}}^{-a\,4{}\mathrm {i}-b\,x\,4{}\mathrm {i}}\,\left (-27\,b^3-27\,b^2\,d+3\,b\,d^2+3\,d^3\right )}{144\,b^4-160\,b^2\,d^2+16\,d^4}\right ) \] Input:

int(cos(c + d*x)*sin(a + b*x)^3,x)
 

Output:

- exp(a*3i - c*1i + b*x*3i - d*x*1i)*((3*b*d^2 - b^2*d - 3*b^3 + d^3)/(144 
*b^4 + 16*d^4 - 160*b^2*d^2) + (exp(- a*6i - b*x*6i)*(3*b*d^2 + b^2*d - 3* 
b^3 - d^3))/(144*b^4 + 16*d^4 - 160*b^2*d^2) - (exp(- a*2i - b*x*2i)*(3*b* 
d^2 - 27*b^2*d - 27*b^3 + 3*d^3))/(144*b^4 + 16*d^4 - 160*b^2*d^2) - (exp( 
- a*4i - b*x*4i)*(3*b*d^2 + 27*b^2*d - 27*b^3 - 3*d^3))/(144*b^4 + 16*d^4 
- 160*b^2*d^2)) - exp(a*3i + c*1i + b*x*3i + d*x*1i)*((3*b*d^2 + b^2*d - 3 
*b^3 - d^3)/(144*b^4 + 16*d^4 - 160*b^2*d^2) + (exp(- a*6i - b*x*6i)*(3*b* 
d^2 - b^2*d - 3*b^3 + d^3))/(144*b^4 + 16*d^4 - 160*b^2*d^2) - (exp(- a*2i 
 - b*x*2i)*(3*b*d^2 + 27*b^2*d - 27*b^3 - 3*d^3))/(144*b^4 + 16*d^4 - 160* 
b^2*d^2) - (exp(- a*4i - b*x*4i)*(3*b*d^2 - 27*b^2*d - 27*b^3 + 3*d^3))/(1 
44*b^4 + 16*d^4 - 160*b^2*d^2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.61 \[ \int \cos (c+d x) \sin ^3(a+b x) \, dx=\frac {-3 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right )^{2} b^{3}+3 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right )^{2} b \,d^{2}-6 \cos \left (b x +a \right ) \cos \left (d x +c \right ) b^{3}-\sin \left (b x +a \right )^{3} \sin \left (d x +c \right ) b^{2} d +\sin \left (b x +a \right )^{3} \sin \left (d x +c \right ) d^{3}-6 \sin \left (b x +a \right ) \sin \left (d x +c \right ) b^{2} d -6 b^{3}+4 b \,d^{2}}{9 b^{4}-10 b^{2} d^{2}+d^{4}} \] Input:

int(cos(d*x+c)*sin(b*x+a)^3,x)
 

Output:

( - 3*cos(a + b*x)*cos(c + d*x)*sin(a + b*x)**2*b**3 + 3*cos(a + b*x)*cos( 
c + d*x)*sin(a + b*x)**2*b*d**2 - 6*cos(a + b*x)*cos(c + d*x)*b**3 - sin(a 
 + b*x)**3*sin(c + d*x)*b**2*d + sin(a + b*x)**3*sin(c + d*x)*d**3 - 6*sin 
(a + b*x)*sin(c + d*x)*b**2*d - 6*b**3 + 4*b*d**2)/(9*b**4 - 10*b**2*d**2 
+ d**4)