3.26 \(\int x^3 (a+b \tan ^{-1}(c x))^3 \, dx\)

Optimal. Leaf size=194 \[ \frac{i b^3 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^4}+\frac{b^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{4 c^2}+\frac{2 b^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4}+\frac{3 b x \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{4 c^4}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}-\frac{b^3 x}{4 c^3}+\frac{b^3 \tan ^{-1}(c x)}{4 c^4} \]

[Out]

-(b^3*x)/(4*c^3) + (b^3*ArcTan[c*x])/(4*c^4) + (b^2*x^2*(a + b*ArcTan[c*x]))/(4*c^2) + (I*b*(a + b*ArcTan[c*x]
)^2)/c^4 + (3*b*x*(a + b*ArcTan[c*x])^2)/(4*c^3) - (b*x^3*(a + b*ArcTan[c*x])^2)/(4*c) - (a + b*ArcTan[c*x])^3
/(4*c^4) + (x^4*(a + b*ArcTan[c*x])^3)/4 + (2*b^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c^4 + (I*b^3*PolyLog
[2, 1 - 2/(1 + I*c*x)])/c^4

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Rubi [A]  time = 0.545385, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 10, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {4852, 4916, 321, 203, 4920, 4854, 2402, 2315, 4846, 4884} \[ \frac{i b^3 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^4}+\frac{b^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{4 c^2}+\frac{2 b^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4}+\frac{3 b x \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{4 c^4}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}-\frac{b^3 x}{4 c^3}+\frac{b^3 \tan ^{-1}(c x)}{4 c^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTan[c*x])^3,x]

[Out]

-(b^3*x)/(4*c^3) + (b^3*ArcTan[c*x])/(4*c^4) + (b^2*x^2*(a + b*ArcTan[c*x]))/(4*c^2) + (I*b*(a + b*ArcTan[c*x]
)^2)/c^4 + (3*b*x*(a + b*ArcTan[c*x])^2)/(4*c^3) - (b*x^3*(a + b*ArcTan[c*x])^2)/(4*c) - (a + b*ArcTan[c*x])^3
/(4*c^4) + (x^4*(a + b*ArcTan[c*x])^3)/4 + (2*b^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c^4 + (I*b^3*PolyLog
[2, 1 - 2/(1 + I*c*x)])/c^4

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4916

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 4920

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[c*x])^(p + 1))/(b*e*(p + 1)), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b,
c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 + c^2*x^
2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 4884

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rubi steps

\begin{align*} \int x^3 \left (a+b \tan ^{-1}(c x)\right )^3 \, dx &=\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{1}{4} (3 b c) \int \frac{x^4 \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx\\ &=\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{(3 b) \int x^2 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{4 c}+\frac{(3 b) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{4 c}\\ &=-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{1}{2} b^2 \int \frac{x^3 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx+\frac{(3 b) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{4 c^3}-\frac{(3 b) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{4 c^3}\\ &=\frac{3 b x \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^3}-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{b^2 \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c^2}-\frac{b^2 \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{2 c^2}-\frac{\left (3 b^2\right ) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{2 c^2}\\ &=\frac{b^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{4 c^2}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{c^4}+\frac{3 b x \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^3}-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{b^2 \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{2 c^3}+\frac{\left (3 b^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{2 c^3}-\frac{b^3 \int \frac{x^2}{1+c^2 x^2} \, dx}{4 c}\\ &=-\frac{b^3 x}{4 c^3}+\frac{b^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{4 c^2}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{c^4}+\frac{3 b x \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^3}-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{2 b^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^4}+\frac{b^3 \int \frac{1}{1+c^2 x^2} \, dx}{4 c^3}-\frac{b^3 \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{2 c^3}-\frac{\left (3 b^3\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{2 c^3}\\ &=-\frac{b^3 x}{4 c^3}+\frac{b^3 \tan ^{-1}(c x)}{4 c^4}+\frac{b^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{4 c^2}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{c^4}+\frac{3 b x \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^3}-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{2 b^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^4}+\frac{\left (i b^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{2 c^4}+\frac{\left (3 i b^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{2 c^4}\\ &=-\frac{b^3 x}{4 c^3}+\frac{b^3 \tan ^{-1}(c x)}{4 c^4}+\frac{b^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{4 c^2}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{c^4}+\frac{3 b x \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^3}-\frac{b x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{2 b^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^4}+\frac{i b^3 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^4}\\ \end{align*}

Mathematica [A]  time = 0.523544, size = 225, normalized size = 1.16 \[ \frac{-4 i b^3 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+b \tan ^{-1}(c x) \left (3 a^2 \left (c^4 x^4-1\right )-2 a b c x \left (c^2 x^2-3\right )+b^2 \left (c^2 x^2+1\right )+8 b^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )-a^2 b c^3 x^3+3 a^2 b c x+a^3 c^4 x^4+a b^2 c^2 x^2-4 a b^2 \log \left (c^2 x^2+1\right )-b^2 \tan ^{-1}(c x)^2 \left (a \left (3-3 c^4 x^4\right )+b \left (c^3 x^3-3 c x+4 i\right )\right )+a b^2+b^3 \left (c^4 x^4-1\right ) \tan ^{-1}(c x)^3-b^3 c x}{4 c^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*(a + b*ArcTan[c*x])^3,x]

[Out]

(a*b^2 + 3*a^2*b*c*x - b^3*c*x + a*b^2*c^2*x^2 - a^2*b*c^3*x^3 + a^3*c^4*x^4 - b^2*(b*(4*I - 3*c*x + c^3*x^3)
+ a*(3 - 3*c^4*x^4))*ArcTan[c*x]^2 + b^3*(-1 + c^4*x^4)*ArcTan[c*x]^3 + b*ArcTan[c*x]*(-2*a*b*c*x*(-3 + c^2*x^
2) + b^2*(1 + c^2*x^2) + 3*a^2*(-1 + c^4*x^4) + 8*b^2*Log[1 + E^((2*I)*ArcTan[c*x])]) - 4*a*b^2*Log[1 + c^2*x^
2] - (4*I)*b^3*PolyLog[2, -E^((2*I)*ArcTan[c*x])])/(4*c^4)

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Maple [B]  time = 0.019, size = 445, normalized size = 2.3 \begin{align*}{\frac{a{x}^{2}{b}^{2}}{4\,{c}^{2}}}+{\frac{3\,{a}^{2}xb}{4\,{c}^{3}}}-{\frac{3\,a{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{4\,{c}^{4}}}+{\frac{3\,{x}^{4}{a}^{2}b\arctan \left ( cx \right ) }{4}}+{\frac{3\,a{b}^{2}{x}^{4} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{4}}-{\frac{{\frac{i}{2}}{b}^{3}{\it dilog} \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{{c}^{4}}}+{\frac{{\frac{i}{2}}{b}^{3}{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{4}}}+{\frac{{\frac{i}{4}}{b}^{3} \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{{c}^{4}}}-{\frac{{\frac{i}{4}}{b}^{3} \left ( \ln \left ( cx+i \right ) \right ) ^{2}}{{c}^{4}}}+{\frac{3\,{b}^{3} \left ( \arctan \left ( cx \right ) \right ) ^{2}x}{4\,{c}^{3}}}+{\frac{{b}^{3}\arctan \left ( cx \right ){x}^{2}}{4\,{c}^{2}}}-{\frac{a{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{4}}}-{\frac{3\,{a}^{2}b\arctan \left ( cx \right ) }{4\,{c}^{4}}}-{\frac{{b}^{3}\arctan \left ( cx \right ) \ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{4}}}-{\frac{{b}^{3} \left ( \arctan \left ( cx \right ) \right ) ^{2}{x}^{3}}{4\,c}}-{\frac{{a}^{2}b{x}^{3}}{4\,c}}-{\frac{{b}^{3}x}{4\,{c}^{3}}}+{\frac{{b}^{3}\arctan \left ( cx \right ) }{4\,{c}^{4}}}+{\frac{{b}^{3}{x}^{4} \left ( \arctan \left ( cx \right ) \right ) ^{3}}{4}}-{\frac{{b}^{3} \left ( \arctan \left ( cx \right ) \right ) ^{3}}{4\,{c}^{4}}}+{\frac{{\frac{i}{2}}{b}^{3}\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{4}}}-{\frac{{\frac{i}{2}}{b}^{3}\ln \left ( cx+i \right ) \ln \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{{c}^{4}}}+{\frac{3\,a{b}^{2}x\arctan \left ( cx \right ) }{2\,{c}^{3}}}-{\frac{a{b}^{2}{x}^{3}\arctan \left ( cx \right ) }{2\,c}}+{\frac{{\frac{i}{2}}{b}^{3}\ln \left ( cx+i \right ) \ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{4}}}-{\frac{{\frac{i}{2}}{b}^{3}\ln \left ( cx-i \right ) \ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{4}}}+{\frac{{x}^{4}{a}^{3}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctan(c*x))^3,x)

[Out]

1/4/c^2*x^2*a*b^2+3/4/c^3*x*a^2*b-3/4/c^4*a*b^2*arctan(c*x)^2+3/4*x^4*a^2*b*arctan(c*x)+3/4*a*b^2*x^4*arctan(c
*x)^2-1/2*I/c^4*b^3*dilog(1/2*I*(c*x-I))+1/2*I/c^4*b^3*dilog(-1/2*I*(c*x+I))+1/4*I/c^4*b^3*ln(c*x-I)^2-1/4*I/c
^4*b^3*ln(c*x+I)^2+3/4/c^3*b^3*arctan(c*x)^2*x+1/4/c^2*b^3*arctan(c*x)*x^2-1/c^4*a*b^2*ln(c^2*x^2+1)-3/4/c^4*a
^2*b*arctan(c*x)-1/c^4*b^3*arctan(c*x)*ln(c^2*x^2+1)-1/4/c*b^3*arctan(c*x)^2*x^3-1/4/c*a^2*b*x^3-1/4*b^3*x/c^3
+1/4*b^3*arctan(c*x)/c^4+1/4*b^3*x^4*arctan(c*x)^3-1/4/c^4*b^3*arctan(c*x)^3+1/2*I/c^4*b^3*ln(c*x-I)*ln(-1/2*I
*(c*x+I))-1/2*I/c^4*b^3*ln(c*x+I)*ln(1/2*I*(c*x-I))+3/2/c^3*a*b^2*x*arctan(c*x)-1/2/c*a*b^2*x^3*arctan(c*x)+1/
2*I/c^4*b^3*ln(c*x+I)*ln(c^2*x^2+1)-1/2*I/c^4*b^3*ln(c*x-I)*ln(c^2*x^2+1)+1/4*x^4*a^3

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{3} x^{3} \arctan \left (c x\right )^{3} + 3 \, a b^{2} x^{3} \arctan \left (c x\right )^{2} + 3 \, a^{2} b x^{3} \arctan \left (c x\right ) + a^{3} x^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b^3*x^3*arctan(c*x)^3 + 3*a*b^2*x^3*arctan(c*x)^2 + 3*a^2*b*x^3*arctan(c*x) + a^3*x^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \left (a + b \operatorname{atan}{\left (c x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atan(c*x))**3,x)

[Out]

Integral(x**3*(a + b*atan(c*x))**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arctan \left (c x\right ) + a\right )}^{3} x^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctan(c*x) + a)^3*x^3, x)