\(\int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [554]

Optimal result

Integrand size = 29, antiderivative size = 102 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4 \log (1+\sin (c+d x))}{a^3 d}+\frac {4 \sin (c+d x)}{a^3 d}-\frac {2 \sin ^2(c+d x)}{a^3 d}+\frac {4 \sin ^3(c+d x)}{3 a^3 d}-\frac {3 \sin ^4(c+d x)}{4 a^3 d}+\frac {\sin ^5(c+d x)}{5 a^3 d} \]

[Out]

-4*ln(1+sin(d*x+c))/a^3/d+4*sin(d*x+c)/a^3/d-2*sin(d*x+c)^2/a^3/d+4/3*sin(d*x+c)^3/a^3/d-3/4*sin(d*x+c)^4/a^3/
d+1/5*sin(d*x+c)^5/a^3/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 90} \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^5(c+d x)}{5 a^3 d}-\frac {3 \sin ^4(c+d x)}{4 a^3 d}+\frac {4 \sin ^3(c+d x)}{3 a^3 d}-\frac {2 \sin ^2(c+d x)}{a^3 d}+\frac {4 \sin (c+d x)}{a^3 d}-\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]

[Out]

(-4*Log[1 + Sin[c + d*x]])/(a^3*d) + (4*Sin[c + d*x])/(a^3*d) - (2*Sin[c + d*x]^2)/(a^3*d) + (4*Sin[c + d*x]^3
)/(3*a^3*d) - (3*Sin[c + d*x]^4)/(4*a^3*d) + Sin[c + d*x]^5/(5*a^3*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^2 x^3}{a^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {(a-x)^2 x^3}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^8 d} \\ & = \frac {\text {Subst}\left (\int \left (4 a^4-4 a^3 x+4 a^2 x^2-3 a x^3+x^4-\frac {4 a^5}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^8 d} \\ & = -\frac {4 \log (1+\sin (c+d x))}{a^3 d}+\frac {4 \sin (c+d x)}{a^3 d}-\frac {2 \sin ^2(c+d x)}{a^3 d}+\frac {4 \sin ^3(c+d x)}{3 a^3 d}-\frac {3 \sin ^4(c+d x)}{4 a^3 d}+\frac {\sin ^5(c+d x)}{5 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {45-3840 \log (1+\sin (c+d x))+3840 \sin (c+d x)-1920 \sin ^2(c+d x)+1280 \sin ^3(c+d x)-720 \sin ^4(c+d x)+192 \sin ^5(c+d x)}{960 a^3 d} \]

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]

[Out]

(45 - 3840*Log[1 + Sin[c + d*x]] + 3840*Sin[c + d*x] - 1920*Sin[c + d*x]^2 + 1280*Sin[c + d*x]^3 - 720*Sin[c +
d*x]^4 + 192*Sin[c + d*x]^5)/(960*a^3*d)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{3}-2 \left (\sin ^{2}\left (d x +c \right )\right )+4 \sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(68\)
default	\(\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{3}-2 \left (\sin ^{2}\left (d x +c \right )\right )+4 \sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(68\)
parallelrisch	\(\frac {-3840 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+1920 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-615+2460 \sin \left (d x +c \right )-190 \sin \left (3 d x +3 c \right )+6 \sin \left (5 d x +5 c \right )+660 \cos \left (2 d x +2 c \right )-45 \cos \left (4 d x +4 c \right )}{480 d \,a^{3}}\)	\(91\)
risch	\(\frac {4 i x}{a^{3}}-\frac {41 i {\mathrm e}^{i \left (d x +c \right )}}{16 d \,a^{3}}+\frac {41 i {\mathrm e}^{-i \left (d x +c \right )}}{16 d \,a^{3}}+\frac {8 i c}{d \,a^{3}}-\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}+\frac {\sin \left (5 d x +5 c \right )}{80 d \,a^{3}}-\frac {3 \cos \left (4 d x +4 c \right )}{32 d \,a^{3}}-\frac {19 \sin \left (3 d x +3 c \right )}{48 d \,a^{3}}+\frac {11 \cos \left (2 d x +2 c \right )}{8 d \,a^{3}}\)	\(144\)

[In]

int(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d/a^3*(1/5*sin(d*x+c)^5-3/4*sin(d*x+c)^4+4/3*sin(d*x+c)^3-2*sin(d*x+c)^2+4*sin(d*x+c)-4*ln(1+sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {45 \, \cos \left (d x + c\right )^{4} - 210 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 26 \, \cos \left (d x + c\right )^{2} + 83\right )} \sin \left (d x + c\right ) + 240 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{60 \, a^{3} d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(45*cos(d*x + c)^4 - 210*cos(d*x + c)^2 - 4*(3*cos(d*x + c)^4 - 26*cos(d*x + c)^2 + 83)*sin(d*x + c) + 2
40*log(sin(d*x + c) + 1))/(a^3*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2558 vs. \(2 (94) = 188\).

Time = 84.90 (sec) , antiderivative size = 2558, normalized size of antiderivative = 25.08 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)**3/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((-120*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**10/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan
(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2
)**2 + 15*a**3*d) - 600*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**8/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**
3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2
+ d*x/2)**2 + 15*a**3*d) - 1200*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**6/(15*a**3*d*tan(c/2 + d*x/2)**10
+ 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*
tan(c/2 + d*x/2)**2 + 15*a**3*d) - 1200*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(15*a**3*d*tan(c/2 + d*x
/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75
*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) - 600*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(15*a**3*d*tan(c/
2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)*
*4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) - 120*log(tan(c/2 + d*x/2) + 1)/(15*a**3*d*tan(c/2 + d*x/2)**1
0 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*
d*tan(c/2 + d*x/2)**2 + 15*a**3*d) + 60*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**10/(15*a**3*d*tan(c/2 +
d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4
+ 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) + 300*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**8/(15*a**3*d
*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 +
d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) + 600*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**6/
(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*
tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) + 600*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 +
d*x/2)**4/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 1
50*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) + 300*log(tan(c/2 + d*x/2)**2 + 1)*
tan(c/2 + d*x/2)**2/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x
/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) + 60*log(tan(c/2 + d*x/2)
**2 + 1)/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 15
0*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) + 120*tan(c/2 + d*x/2)**9/(15*a**3*d
*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 +
d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) - 120*tan(c/2 + d*x/2)**8/(15*a**3*d*tan(c/2 + d*x/2)**
10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3
*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) + 640*tan(c/2 + d*x/2)**7/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(
c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)
**2 + 15*a**3*d) - 540*tan(c/2 + d*x/2)**6/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 1
50*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) +
1136*tan(c/2 + d*x/2)**5/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2
+ d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) - 540*tan(c/2 + d*x/
2)**4/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a
**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) + 640*tan(c/2 + d*x/2)**3/(15*a**3*d*ta
n(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x
/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) - 120*tan(c/2 + d*x/2)**2/(15*a**3*d*tan(c/2 + d*x/2)**10
+ 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*
tan(c/2 + d*x/2)**2 + 15*a**3*d) + 120*tan(c/2 + d*x/2)/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 +
d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 +
15*a**3*d), Ne(d, 0)), (x*sin(c)**3*cos(c)**5/(a*sin(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {12 \, \sin \left (d x + c\right )^{5} - 45 \, \sin \left (d x + c\right )^{4} + 80 \, \sin \left (d x + c\right )^{3} - 120 \, \sin \left (d x + c\right )^{2} + 240 \, \sin \left (d x + c\right )}{a^{3}} - \frac {240 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*((12*sin(d*x + c)^5 - 45*sin(d*x + c)^4 + 80*sin(d*x + c)^3 - 120*sin(d*x + c)^2 + 240*sin(d*x + c))/a^3
- 240*log(sin(d*x + c) + 1)/a^3)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (96) = 192\).

Time = 0.83 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.89 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {60 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}{a^{3}} - \frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {137 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 805 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 640 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1910 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 1136 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1910 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 640 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 805 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 137}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} a^{3}}}{15 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/15*(60*log(tan(1/2*d*x + 1/2*c)^2 + 1)/a^3 - 120*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (137*tan(1/2*d*x +
1/2*c)^10 - 120*tan(1/2*d*x + 1/2*c)^9 + 805*tan(1/2*d*x + 1/2*c)^8 - 640*tan(1/2*d*x + 1/2*c)^7 + 1910*tan(1
/2*d*x + 1/2*c)^6 - 1136*tan(1/2*d*x + 1/2*c)^5 + 1910*tan(1/2*d*x + 1/2*c)^4 - 640*tan(1/2*d*x + 1/2*c)^3 + 8
05*tan(1/2*d*x + 1/2*c)^2 - 120*tan(1/2*d*x + 1/2*c) + 137)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*a^3))/d

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {4\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3}-\frac {4\,\sin \left (c+d\,x\right )}{a^3}+\frac {2\,{\sin \left (c+d\,x\right )}^2}{a^3}-\frac {4\,{\sin \left (c+d\,x\right )}^3}{3\,a^3}+\frac {3\,{\sin \left (c+d\,x\right )}^4}{4\,a^3}-\frac {{\sin \left (c+d\,x\right )}^5}{5\,a^3}}{d} \]

[In]

int((cos(c + d*x)^5*sin(c + d*x)^3)/(a + a*sin(c + d*x))^3,x)

[Out]

-((4*log(sin(c + d*x) + 1))/a^3 - (4*sin(c + d*x))/a^3 + (2*sin(c + d*x)^2)/a^3 - (4*sin(c + d*x)^3)/(3*a^3) +
(3*sin(c + d*x)^4)/(4*a^3) - sin(c + d*x)^5/(5*a^3))/d