3.240 \(\int \frac {\csc (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=98 \[ \frac {8 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}+\frac {3 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}+\frac {2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac {A \tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

[Out]

-A*arctanh(cos(d*x+c))/a^3/d+2/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^3+3/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^2+8
/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))

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Rubi [A]  time = 0.16, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2966, 3770, 2650, 2648} \[ \frac {8 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}+\frac {3 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}+\frac {2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac {A \tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

-((A*ArcTanh[Cos[c + d*x]])/(a^3*d)) + (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) + (3*A*Cos[c + d*x])/
(5*a^3*d*(1 + Sin[c + d*x])^2) + (8*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x]))

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\csc (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx &=\int \left (\frac {A \csc (c+d x)}{a^3}-\frac {2 A}{a^3 (1+\sin (c+d x))^3}-\frac {A}{a^3 (1+\sin (c+d x))^2}-\frac {A}{a^3 (1+\sin (c+d x))}\right ) \, dx\\ &=\frac {A \int \csc (c+d x) \, dx}{a^3}-\frac {A \int \frac {1}{(1+\sin (c+d x))^2} \, dx}{a^3}-\frac {A \int \frac {1}{1+\sin (c+d x)} \, dx}{a^3}-\frac {(2 A) \int \frac {1}{(1+\sin (c+d x))^3} \, dx}{a^3}\\ &=-\frac {A \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}+\frac {A \cos (c+d x)}{a^3 d (1+\sin (c+d x))}-\frac {A \int \frac {1}{1+\sin (c+d x)} \, dx}{3 a^3}-\frac {(4 A) \int \frac {1}{(1+\sin (c+d x))^2} \, dx}{5 a^3}\\ &=-\frac {A \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {3 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}+\frac {4 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))}-\frac {(4 A) \int \frac {1}{1+\sin (c+d x)} \, dx}{15 a^3}\\ &=-\frac {A \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {3 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}+\frac {8 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 1.01, size = 313, normalized size = 3.19 \[ \frac {(A-A \sin (c+d x)) \left (2 \sin \left (\frac {d x}{2}\right ) (-19 \sin (c+d x)+4 \cos (2 (c+d x))-17)+\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (3 \cos \left (\frac {c}{2}\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2-3 \sin \left (\frac {c}{2}\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2-5 \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4+5 \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4-2 \sin \left (\frac {c}{2}\right )+2 \cos \left (\frac {c}{2}\right )\right )\right )}{5 a^3 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(2*Cos[c/2] - 2*Sin[c/2] + 3*Cos[c/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2])^2 - 3*Sin[c/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 5*Log[Cos[(c + d*x)/2]]*(Cos[c/2] + Sin[c/2])*(
Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + 5*Log[Sin[(c + d*x)/2]]*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin
[(c + d*x)/2])^4) + 2*Sin[(d*x)/2]*(-17 + 4*Cos[2*(c + d*x)] - 19*Sin[c + d*x]))*(A - A*Sin[c + d*x]))/(5*a^3*
d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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fricas [B]  time = 0.44, size = 310, normalized size = 3.16 \[ \frac {16 \, A \cos \left (d x + c\right )^{3} - 22 \, A \cos \left (d x + c\right )^{2} - 42 \, A \cos \left (d x + c\right ) - 5 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) - 4 \, A\right )} \sin \left (d x + c\right ) - 4 \, A\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 5 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) - 4 \, A\right )} \sin \left (d x + c\right ) - 4 \, A\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (8 \, A \cos \left (d x + c\right )^{2} + 19 \, A \cos \left (d x + c\right ) - 2 \, A\right )} \sin \left (d x + c\right ) - 4 \, A}{10 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/10*(16*A*cos(d*x + c)^3 - 22*A*cos(d*x + c)^2 - 42*A*cos(d*x + c) - 5*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2
- 2*A*cos(d*x + c) + (A*cos(d*x + c)^2 - 2*A*cos(d*x + c) - 4*A)*sin(d*x + c) - 4*A)*log(1/2*cos(d*x + c) + 1
/2) + 5*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + (A*cos(d*x + c)^2 - 2*A*cos(d*x + c) - 4*A
)*sin(d*x + c) - 4*A)*log(-1/2*cos(d*x + c) + 1/2) - 2*(8*A*cos(d*x + c)^2 + 19*A*cos(d*x + c) - 2*A)*sin(d*x
+ c) - 4*A)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*cos(d*x +
c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))

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giac [A]  time = 0.19, size = 99, normalized size = 1.01 \[ \frac {\frac {5 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {2 \, {\left (20 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 55 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 75 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 13 \, A\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/5*(5*A*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 2*(20*A*tan(1/2*d*x + 1/2*c)^4 + 55*A*tan(1/2*d*x + 1/2*c)^3 + 7
5*A*tan(1/2*d*x + 1/2*c)^2 + 45*A*tan(1/2*d*x + 1/2*c) + 13*A)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^5))/d

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maple [A]  time = 0.73, size = 130, normalized size = 1.33 \[ \frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {16 A}{5 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {8 A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {12 A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {10 A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8 A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x)

[Out]

1/d*A/a^3*ln(tan(1/2*d*x+1/2*c))+16/5/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^5-8/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^4+12/d
*A/a^3/(tan(1/2*d*x+1/2*c)+1)^3-10/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^2+8/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.59, size = 433, normalized size = 4.42 \[ \frac {A {\left (\frac {2 \, {\left (\frac {115 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {185 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {135 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {45 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 32\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {15 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + \frac {2 \, A {\left (\frac {20 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 7\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/15*(A*(2*(115*sin(d*x + c)/(cos(d*x + c) + 1) + 185*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 135*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 + 45*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 32)/(a^3 + 5*a^3*sin(d*x + c)/(cos(d*x + c) +
1) + 10*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*a^3*sin(d*x
+ c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 15*log(sin(d*x + c)/(cos(d*x + c) + 1
))/a^3) + 2*A*(20*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 30*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 7)/(a^3 + 5*a^3*sin(d*x + c)/(cos(d*x + c) +
1) + 10*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*a^3*sin(d*x +
c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5))/d

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mupad [B]  time = 14.77, size = 199, normalized size = 2.03 \[ \frac {A\,\left (5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )+90\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+150\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+110\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+25\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+50\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+50\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+25\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+26\right )}{5\,a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A - A*sin(c + d*x))/(sin(c + d*x)*(a + a*sin(c + d*x))^3),x)

[Out]

(A*(5*log(tan(c/2 + (d*x)/2)) + 90*tan(c/2 + (d*x)/2) + 150*tan(c/2 + (d*x)/2)^2 + 110*tan(c/2 + (d*x)/2)^3 +
40*tan(c/2 + (d*x)/2)^4 + 25*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2) + 50*log(tan(c/2 + (d*x)/2))*tan(c/2 +
(d*x)/2)^2 + 50*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^3 + 25*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^
4 + 5*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^5 + 26))/(5*a^3*d*(tan(c/2 + (d*x)/2) + 1)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {A \left (\int \left (- \frac {\csc {\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\right )\, dx + \int \frac {\sin {\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)

[Out]

-A*(Integral(-csc(c + d*x)/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x) + Integral(sin(c + d
*x)*csc(c + d*x)/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x))/a**3

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