\(\int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx\) [1046]

Optimal result

Integrand size = 27, antiderivative size = 105 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {b x}{16}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}+\frac {b \cos (c+d x) \sin (c+d x)}{16 d}-\frac {b \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {b \cos ^3(c+d x) \sin ^3(c+d x)}{6 d} \] Output:

1/16*b*x-1/3*a*cos(d*x+c)^3/d+1/5*a*cos(d*x+c)^5/d+1/16*b*cos(d*x+c)*sin(d 
*x+c)/d-1/8*b*cos(d*x+c)^3*sin(d*x+c)/d-1/6*b*cos(d*x+c)^3*sin(d*x+c)^3/d
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {60 b d x-120 a \cos (c+d x)-20 a \cos (3 (c+d x))+12 a \cos (5 (c+d x))-15 b \sin (2 (c+d x))-15 b \sin (4 (c+d x))+5 b \sin (6 (c+d x))}{960 d} \] Input:

Integrate[Cos[c + d*x]^2*Sin[c + d*x]^3*(a + b*Sin[c + d*x]),x]
 

Output:

(60*b*d*x - 120*a*Cos[c + d*x] - 20*a*Cos[3*(c + d*x)] + 12*a*Cos[5*(c + d 
*x)] - 15*b*Sin[2*(c + d*x)] - 15*b*Sin[4*(c + d*x)] + 5*b*Sin[6*(c + d*x) 
])/(960*d)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3317, 3042, 3045, 244, 2009, 3048, 3042, 3048, 3042, 3115, 24}

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.

\(\displaystyle \int \sin ^3(c+d x) \cos ^2(c+d x) (a+b \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^3 \cos (c+d x)^2 (a+b \sin (c+d x))dx\)

\(\Big \downarrow \) 3317

\(\displaystyle a \int \cos ^2(c+d x) \sin ^3(c+d x)dx+b \int \cos ^2(c+d x) \sin ^4(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \cos (c+d x)^2 \sin (c+d x)^3dx+b \int \cos (c+d x)^2 \sin (c+d x)^4dx\)

\(\Big \downarrow \) 3045

\(\displaystyle b \int \cos (c+d x)^2 \sin (c+d x)^4dx-\frac {a \int \cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 244

\(\displaystyle b \int \cos (c+d x)^2 \sin (c+d x)^4dx-\frac {a \int \left (\cos ^2(c+d x)-\cos ^4(c+d x)\right )d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle b \int \cos (c+d x)^2 \sin (c+d x)^4dx-\frac {a \left (\frac {1}{3} \cos ^3(c+d x)-\frac {1}{5} \cos ^5(c+d x)\right )}{d}\)

\(\Big \downarrow \) 3048

\(\displaystyle b \left (\frac {1}{2} \int \cos ^2(c+d x) \sin ^2(c+d x)dx-\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{3} \cos ^3(c+d x)-\frac {1}{5} \cos ^5(c+d x)\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle b \left (\frac {1}{2} \int \cos (c+d x)^2 \sin (c+d x)^2dx-\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{3} \cos ^3(c+d x)-\frac {1}{5} \cos ^5(c+d x)\right )}{d}\)

\(\Big \downarrow \) 3048

\(\displaystyle b \left (\frac {1}{2} \left (\frac {1}{4} \int \cos ^2(c+d x)dx-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{3} \cos ^3(c+d x)-\frac {1}{5} \cos ^5(c+d x)\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle b \left (\frac {1}{2} \left (\frac {1}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{3} \cos ^3(c+d x)-\frac {1}{5} \cos ^5(c+d x)\right )}{d}\)

\(\Big \downarrow \) 3115

\(\displaystyle b \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{3} \cos ^3(c+d x)-\frac {1}{5} \cos ^5(c+d x)\right )}{d}\)

\(\Big \downarrow \) 24

\(\displaystyle b \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{3} \cos ^3(c+d x)-\frac {1}{5} \cos ^5(c+d x)\right )}{d}\)

Input:

Int[Cos[c + d*x]^2*Sin[c + d*x]^3*(a + b*Sin[c + d*x]),x]
 

Output:

-((a*(Cos[c + d*x]^3/3 - Cos[c + d*x]^5/5))/d) + b*(-1/6*(Cos[c + d*x]^3*S 
in[c + d*x]^3)/d + (-1/4*(Cos[c + d*x]^3*Sin[c + d*x])/d + (x/2 + (Cos[c + 
 d*x]*Sin[c + d*x])/(2*d))/4)/2)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3045

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> Simp[-(a*f)^(-1)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], 
x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && 
 !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
 

rule 3048

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 
1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n))   Int[(b*Cos[e + f*x])^n 
*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] 
 && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
 

rule 3115

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]
 

rule 3317

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a   Int[(g*Co 
s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d   Int[(g*Cos[e + f*x])^ 
p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
 

Maple [A] (verified)

Time = 6.80 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.80

method	result	size
parallelrisch	\(\frac {60 b x d +5 b \sin \left (6 d x +6 c \right )+12 a \cos \left (5 d x +5 c \right )-15 b \sin \left (4 d x +4 c \right )-20 a \cos \left (3 d x +3 c \right )-15 b \sin \left (2 d x +2 c \right )-120 a \cos \left (d x +c \right )-128 a}{960 d}\)	\(84\)
risch	\(\frac {b x}{16}-\frac {a \cos \left (d x +c \right )}{8 d}+\frac {b \sin \left (6 d x +6 c \right )}{192 d}+\frac {a \cos \left (5 d x +5 c \right )}{80 d}-\frac {b \sin \left (4 d x +4 c \right )}{64 d}-\frac {a \cos \left (3 d x +3 c \right )}{48 d}-\frac {b \sin \left (2 d x +2 c \right )}{64 d}\)	\(93\)
derivativedivides	\(\frac {a \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+b \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{3}}{6}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{8}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )}{d}\)	\(95\)
default	\(\frac {a \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+b \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{3}}{6}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{8}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )}{d}\)	\(95\)
norman	\(\frac {\frac {b x}{16}-\frac {4 a}{15 d}-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {17 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}+\frac {19 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}-\frac {19 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {17 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}+\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}+\frac {3 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16}+\frac {5 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16}+\frac {3 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{16}-\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}-\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5 d}-\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}\)	\(269\)
orering	\(\text {Expression too large to display}\)	\(1810\)

Input:

int(cos(d*x+c)^2*sin(d*x+c)^3*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/960*(60*b*x*d+5*b*sin(6*d*x+6*c)+12*a*cos(5*d*x+5*c)-15*b*sin(4*d*x+4*c) 
-20*a*cos(3*d*x+3*c)-15*b*sin(2*d*x+2*c)-120*a*cos(d*x+c)-128*a)/d
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.70 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {48 \, a \cos \left (d x + c\right )^{5} - 80 \, a \cos \left (d x + c\right )^{3} + 15 \, b d x + 5 \, {\left (8 \, b \cos \left (d x + c\right )^{5} - 14 \, b \cos \left (d x + c\right )^{3} + 3 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="fricas" 
)
 

Output:

1/240*(48*a*cos(d*x + c)^5 - 80*a*cos(d*x + c)^3 + 15*b*d*x + 5*(8*b*cos(d 
*x + c)^5 - 14*b*cos(d*x + c)^3 + 3*b*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (92) = 184\).

Time = 0.34 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.83 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx=\begin {cases} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 a \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac {b x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {b x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \sin ^{3}{\left (c \right )} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**2*sin(d*x+c)**3*(a+b*sin(d*x+c)),x)
 

Output:

Piecewise((-a*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 2*a*cos(c + d*x)**5/ 
(15*d) + b*x*sin(c + d*x)**6/16 + 3*b*x*sin(c + d*x)**4*cos(c + d*x)**2/16 
 + 3*b*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + b*x*cos(c + d*x)**6/16 + b*s 
in(c + d*x)**5*cos(c + d*x)/(16*d) - b*sin(c + d*x)**3*cos(c + d*x)**3/(6* 
d) - b*sin(c + d*x)*cos(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*sin(c))*s 
in(c)**3*cos(c)**2, True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.62 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {64 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b}{960 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="maxima" 
)
 

Output:

1/960*(64*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a - 5*(4*sin(2*d*x + 2*c)^ 
3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*c))*b)/d
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {1}{16} \, b x + \frac {a \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {a \cos \left (d x + c\right )}{8 \, d} + \frac {b \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {b \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac {b \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="giac")
 

Output:

1/16*b*x + 1/80*a*cos(5*d*x + 5*c)/d - 1/48*a*cos(3*d*x + 3*c)/d - 1/8*a*c 
os(d*x + c)/d + 1/192*b*sin(6*d*x + 6*c)/d - 1/64*b*sin(4*d*x + 4*c)/d - 1 
/64*b*sin(2*d*x + 2*c)/d
 

Mupad [B] (verification not implemented)

Time = 38.22 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.46 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {b\,x}{16}-\frac {-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}-\frac {17\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {19\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}-\frac {19\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {17\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {4\,a}{15}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \] Input:

int(cos(c + d*x)^2*sin(c + d*x)^3*(a + b*sin(c + d*x)),x)
 

Output:

(b*x)/16 - ((4*a)/15 + (b*tan(c/2 + (d*x)/2))/8 + (8*a*tan(c/2 + (d*x)/2)^ 
2)/5 + (8*a*tan(c/2 + (d*x)/2)^6)/3 + 4*a*tan(c/2 + (d*x)/2)^8 + (17*b*tan 
(c/2 + (d*x)/2)^3)/24 - (19*b*tan(c/2 + (d*x)/2)^5)/4 + (19*b*tan(c/2 + (d 
*x)/2)^7)/4 - (17*b*tan(c/2 + (d*x)/2)^9)/24 - (b*tan(c/2 + (d*x)/2)^11)/8 
)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^6)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.01 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b +48 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a -10 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b -16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a -15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -32 \cos \left (d x +c \right ) a +32 a +15 b d x}{240 d} \] Input:

int(cos(d*x+c)^2*sin(d*x+c)^3*(a+b*sin(d*x+c)),x)
 

Output:

(40*cos(c + d*x)*sin(c + d*x)**5*b + 48*cos(c + d*x)*sin(c + d*x)**4*a - 1 
0*cos(c + d*x)*sin(c + d*x)**3*b - 16*cos(c + d*x)*sin(c + d*x)**2*a - 15* 
cos(c + d*x)*sin(c + d*x)*b - 32*cos(c + d*x)*a + 32*a + 15*b*d*x)/(240*d)