3.320 \(\int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=106 \[ \frac {3 \cot (c+d x)}{a^3 d}+\frac {17 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)}+\frac {2 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)^2}-\frac {11 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^3 d} \]

[Out]

-11/2*arctanh(cos(d*x+c))/a^3/d+3*cot(d*x+c)/a^3/d-1/2*cot(d*x+c)*csc(d*x+c)/a^3/d+2/3*cos(d*x+c)/a^3/d/(1+sin
(d*x+c))^2+17/3*cos(d*x+c)/a^3/d/(1+sin(d*x+c))

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Rubi [A]  time = 0.27, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2874, 2966, 3770, 3767, 8, 3768, 2650, 2648} \[ \frac {3 \cot (c+d x)}{a^3 d}+\frac {17 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)}+\frac {2 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)^2}-\frac {11 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^2*Csc[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(-11*ArcTanh[Cos[c + d*x]])/(2*a^3*d) + (3*Cot[c + d*x])/(a^3*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a^3*d) + (2*
Cos[c + d*x])/(3*a^3*d*(1 + Sin[c + d*x])^2) + (17*Cos[c + d*x])/(3*a^3*d*(1 + Sin[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2874

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

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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*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 {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx}{a^2}\\ &=\frac {\int \left (\frac {5 \csc (c+d x)}{a}-\frac {3 \csc ^2(c+d x)}{a}+\frac {\csc ^3(c+d x)}{a}-\frac {2}{a (1+\sin (c+d x))^2}-\frac {5}{a (1+\sin (c+d x))}\right ) \, dx}{a^2}\\ &=\frac {\int \csc ^3(c+d x) \, dx}{a^3}-\frac {2 \int \frac {1}{(1+\sin (c+d x))^2} \, dx}{a^3}-\frac {3 \int \csc ^2(c+d x) \, dx}{a^3}+\frac {5 \int \csc (c+d x) \, dx}{a^3}-\frac {5 \int \frac {1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=-\frac {5 \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac {2 \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}+\frac {5 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac {\int \csc (c+d x) \, dx}{2 a^3}-\frac {2 \int \frac {1}{1+\sin (c+d x)} \, dx}{3 a^3}+\frac {3 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=-\frac {11 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac {3 \cot (c+d x)}{a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac {2 \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}+\frac {17 \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 5.95, size = 308, normalized size = 2.91 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (-32 \sin \left (\frac {1}{2} (c+d x)\right )-272 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2+16 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+3 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\tan \left (\frac {1}{2} (c+d x)\right )+1\right )^3-3 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cot \left (\frac {1}{2} (c+d x)\right )+1\right )^3-132 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3+132 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3-36 \tan \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3+36 \cot \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3\right )}{24 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^2*Csc[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(-32*Sin[(c + d*x)/2] - 3*(1 + Cot[(c + d*x)/2])^3*Sin[(c + d*x)/2] +
16*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) - 272*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 36*
Cot[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - 132*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(
c + d*x)/2])^3 + 132*Log[Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - 36*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^3*Tan[(c + d*x)/2] + 3*Cos[(c + d*x)/2]*(1 + Tan[(c + d*x)/2])^3))/(24*a^3*d*(1 + Sin[c + d*x]
)^3)

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fricas [B]  time = 0.52, size = 365, normalized size = 3.44 \[ -\frac {104 \, \cos \left (d x + c\right )^{4} + 142 \, \cos \left (d x + c\right )^{3} - 90 \, \cos \left (d x + c\right )^{2} + 33 \, {\left (\cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) + 2\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 33 \, {\left (\cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) + 2\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (52 \, \cos \left (d x + c\right )^{3} - 19 \, \cos \left (d x + c\right )^{2} - 64 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right ) - 136 \, \cos \left (d x + c\right ) - 8}{12 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - a^{3} d \cos \left (d x + c\right )^{3} - 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right ) + 2 \, a^{3} d - {\left (a^{3} d \cos \left (d x + c\right )^{3} + 2 \, a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - 2 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(104*cos(d*x + c)^4 + 142*cos(d*x + c)^3 - 90*cos(d*x + c)^2 + 33*(cos(d*x + c)^4 - cos(d*x + c)^3 - 3*c
os(d*x + c)^2 - (cos(d*x + c)^3 + 2*cos(d*x + c)^2 - cos(d*x + c) - 2)*sin(d*x + c) + cos(d*x + c) + 2)*log(1/
2*cos(d*x + c) + 1/2) - 33*(cos(d*x + c)^4 - cos(d*x + c)^3 - 3*cos(d*x + c)^2 - (cos(d*x + c)^3 + 2*cos(d*x +
c)^2 - cos(d*x + c) - 2)*sin(d*x + c) + cos(d*x + c) + 2)*log(-1/2*cos(d*x + c) + 1/2) + 2*(52*cos(d*x + c)^3
- 19*cos(d*x + c)^2 - 64*cos(d*x + c) + 4)*sin(d*x + c) - 136*cos(d*x + c) - 8)/(a^3*d*cos(d*x + c)^4 - a^3*d
*cos(d*x + c)^3 - 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c) + 2*a^3*d - (a^3*d*cos(d*x + c)^3 + 2*a^3*d*cos(
d*x + c)^2 - a^3*d*cos(d*x + c) - 2*a^3*d)*sin(d*x + c))

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giac [A]  time = 0.22, size = 143, normalized size = 1.35 \[ \frac {\frac {132 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {3 \, {\left (66 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {3 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{6}} + \frac {16 \, {\left (21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 19\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(132*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 3*(66*tan(1/2*d*x + 1/2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 1)/(a^
3*tan(1/2*d*x + 1/2*c)^2) + 3*(a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a^3*tan(1/2*d*x + 1/2*c))/a^6 + 16*(21*tan(1/2*
d*x + 1/2*c)^2 + 36*tan(1/2*d*x + 1/2*c) + 19)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^3))/d

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maple [A]  time = 0.69, size = 157, normalized size = 1.48 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{3}}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{3}}-\frac {1}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {3}{2 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {11 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}}+\frac {8}{3 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {4}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d/a^3*tan(1/2*d*x+1/2*c)^2-3/2/d/a^3*tan(1/2*d*x+1/2*c)-1/8/d/a^3/tan(1/2*d*x+1/2*c)^2+3/2/d/a^3/tan(1/2*d
*x+1/2*c)+11/2/d/a^3*ln(tan(1/2*d*x+1/2*c))+8/3/d/a^3/(tan(1/2*d*x+1/2*c)+1)^3-4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^
2+14/d/a^3/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.34, size = 247, normalized size = 2.33 \[ \frac {\frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {403 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {681 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {372 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 3}{\frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} - \frac {3 \, {\left (\frac {12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{a^{3}} + \frac {132 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((27*sin(d*x + c)/(cos(d*x + c) + 1) + 403*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 681*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 + 372*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 3)/(a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^
3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*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) - 3*(12*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^3 + 132*log(si
n(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 8.66, size = 178, normalized size = 1.68 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}+\frac {11\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^3\,d}+\frac {62\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {227\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {403\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{6}+\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {1}{2}}{d\,\left (4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^3*d) + (11*log(tan(c/2 + (d*x)/2)))/(2*a^3*d) + ((9*tan(c/2 + (d*x)/2))/2 + (403*tan
(c/2 + (d*x)/2)^2)/6 + (227*tan(c/2 + (d*x)/2)^3)/2 + 62*tan(c/2 + (d*x)/2)^4 - 1/2)/(d*(4*a^3*tan(c/2 + (d*x)
/2)^2 + 12*a^3*tan(c/2 + (d*x)/2)^3 + 12*a^3*tan(c/2 + (d*x)/2)^4 + 4*a^3*tan(c/2 + (d*x)/2)^5)) - (3*tan(c/2
+ (d*x)/2))/(2*a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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