3.388 \(\int \frac{x^3 \tan ^{-1}(a x)^3}{c+a^2 c x^2} \, dx\)

Optimal. Leaf size=260 \[ -\frac{3 i \text{PolyLog}\left (2,1-\frac{2}{1+i a x}\right )}{2 a^4 c}-\frac{3 i \text{PolyLog}\left (4,1-\frac{2}{1+i a x}\right )}{4 a^4 c}+\frac{3 i \tan ^{-1}(a x)^2 \text{PolyLog}\left (2,1-\frac{2}{1+i a x}\right )}{2 a^4 c}+\frac{3 \tan ^{-1}(a x) \text{PolyLog}\left (3,1-\frac{2}{1+i a x}\right )}{2 a^4 c}+\frac{x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac{i \tan ^{-1}(a x)^4}{4 a^4 c}+\frac{\tan ^{-1}(a x)^3}{2 a^4 c}-\frac{3 x \tan ^{-1}(a x)^2}{2 a^3 c}-\frac{3 i \tan ^{-1}(a x)^2}{2 a^4 c}+\frac{\log \left (\frac{2}{1+i a x}\right ) \tan ^{-1}(a x)^3}{a^4 c}-\frac{3 \log \left (\frac{2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^4 c} \]

[Out]

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

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Rubi [A]  time = 0.446751, antiderivative size = 260, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4916, 4852, 4846, 4920, 4854, 2402, 2315, 4884, 4994, 4998, 6610} \[ -\frac{3 i \text{PolyLog}\left (2,1-\frac{2}{1+i a x}\right )}{2 a^4 c}-\frac{3 i \text{PolyLog}\left (4,1-\frac{2}{1+i a x}\right )}{4 a^4 c}+\frac{3 i \tan ^{-1}(a x)^2 \text{PolyLog}\left (2,1-\frac{2}{1+i a x}\right )}{2 a^4 c}+\frac{3 \tan ^{-1}(a x) \text{PolyLog}\left (3,1-\frac{2}{1+i a x}\right )}{2 a^4 c}+\frac{x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac{i \tan ^{-1}(a x)^4}{4 a^4 c}+\frac{\tan ^{-1}(a x)^3}{2 a^4 c}-\frac{3 x \tan ^{-1}(a x)^2}{2 a^3 c}-\frac{3 i \tan ^{-1}(a x)^2}{2 a^4 c}+\frac{\log \left (\frac{2}{1+i a x}\right ) \tan ^{-1}(a x)^3}{a^4 c}-\frac{3 \log \left (\frac{2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^4 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTan[a*x]^3)/(c + a^2*c*x^2),x]

[Out]

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

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 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 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 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 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]

Rule 4994

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

Rule 4998

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

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \int \frac{x^3 \tan ^{-1}(a x)^3}{c+a^2 c x^2} \, dx &=-\frac{\int \frac{x \tan ^{-1}(a x)^3}{c+a^2 c x^2} \, dx}{a^2}+\frac{\int x \tan ^{-1}(a x)^3 \, dx}{a^2 c}\\ &=\frac{x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac{i \tan ^{-1}(a x)^4}{4 a^4 c}+\frac{\int \frac{\tan ^{-1}(a x)^3}{i-a x} \, dx}{a^3 c}-\frac{3 \int \frac{x^2 \tan ^{-1}(a x)^2}{1+a^2 x^2} \, dx}{2 a c}\\ &=\frac{x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac{i \tan ^{-1}(a x)^4}{4 a^4 c}+\frac{\tan ^{-1}(a x)^3 \log \left (\frac{2}{1+i a x}\right )}{a^4 c}-\frac{3 \int \tan ^{-1}(a x)^2 \, dx}{2 a^3 c}+\frac{3 \int \frac{\tan ^{-1}(a x)^2}{1+a^2 x^2} \, dx}{2 a^3 c}-\frac{3 \int \frac{\tan ^{-1}(a x)^2 \log \left (\frac{2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^3 c}\\ &=-\frac{3 x \tan ^{-1}(a x)^2}{2 a^3 c}+\frac{\tan ^{-1}(a x)^3}{2 a^4 c}+\frac{x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac{i \tan ^{-1}(a x)^4}{4 a^4 c}+\frac{\tan ^{-1}(a x)^3 \log \left (\frac{2}{1+i a x}\right )}{a^4 c}+\frac{3 i \tan ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{2 a^4 c}-\frac{(3 i) \int \frac{\tan ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^3 c}+\frac{3 \int \frac{x \tan ^{-1}(a x)}{1+a^2 x^2} \, dx}{a^2 c}\\ &=-\frac{3 i \tan ^{-1}(a x)^2}{2 a^4 c}-\frac{3 x \tan ^{-1}(a x)^2}{2 a^3 c}+\frac{\tan ^{-1}(a x)^3}{2 a^4 c}+\frac{x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac{i \tan ^{-1}(a x)^4}{4 a^4 c}+\frac{\tan ^{-1}(a x)^3 \log \left (\frac{2}{1+i a x}\right )}{a^4 c}+\frac{3 i \tan ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{2 a^4 c}+\frac{3 \tan ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1+i a x}\right )}{2 a^4 c}-\frac{3 \int \frac{\text{Li}_3\left (1-\frac{2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{2 a^3 c}-\frac{3 \int \frac{\tan ^{-1}(a x)}{i-a x} \, dx}{a^3 c}\\ &=-\frac{3 i \tan ^{-1}(a x)^2}{2 a^4 c}-\frac{3 x \tan ^{-1}(a x)^2}{2 a^3 c}+\frac{\tan ^{-1}(a x)^3}{2 a^4 c}+\frac{x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac{i \tan ^{-1}(a x)^4}{4 a^4 c}-\frac{3 \tan ^{-1}(a x) \log \left (\frac{2}{1+i a x}\right )}{a^4 c}+\frac{\tan ^{-1}(a x)^3 \log \left (\frac{2}{1+i a x}\right )}{a^4 c}+\frac{3 i \tan ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{2 a^4 c}+\frac{3 \tan ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1+i a x}\right )}{2 a^4 c}-\frac{3 i \text{Li}_4\left (1-\frac{2}{1+i a x}\right )}{4 a^4 c}+\frac{3 \int \frac{\log \left (\frac{2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^3 c}\\ &=-\frac{3 i \tan ^{-1}(a x)^2}{2 a^4 c}-\frac{3 x \tan ^{-1}(a x)^2}{2 a^3 c}+\frac{\tan ^{-1}(a x)^3}{2 a^4 c}+\frac{x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac{i \tan ^{-1}(a x)^4}{4 a^4 c}-\frac{3 \tan ^{-1}(a x) \log \left (\frac{2}{1+i a x}\right )}{a^4 c}+\frac{\tan ^{-1}(a x)^3 \log \left (\frac{2}{1+i a x}\right )}{a^4 c}+\frac{3 i \tan ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{2 a^4 c}+\frac{3 \tan ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1+i a x}\right )}{2 a^4 c}-\frac{3 i \text{Li}_4\left (1-\frac{2}{1+i a x}\right )}{4 a^4 c}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i a x}\right )}{a^4 c}\\ &=-\frac{3 i \tan ^{-1}(a x)^2}{2 a^4 c}-\frac{3 x \tan ^{-1}(a x)^2}{2 a^3 c}+\frac{\tan ^{-1}(a x)^3}{2 a^4 c}+\frac{x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac{i \tan ^{-1}(a x)^4}{4 a^4 c}-\frac{3 \tan ^{-1}(a x) \log \left (\frac{2}{1+i a x}\right )}{a^4 c}+\frac{\tan ^{-1}(a x)^3 \log \left (\frac{2}{1+i a x}\right )}{a^4 c}-\frac{3 i \text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{2 a^4 c}+\frac{3 i \tan ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{2 a^4 c}+\frac{3 \tan ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1+i a x}\right )}{2 a^4 c}-\frac{3 i \text{Li}_4\left (1-\frac{2}{1+i a x}\right )}{4 a^4 c}\\ \end{align*}

Mathematica [A]  time = 0.289916, size = 162, normalized size = 0.62 \[ \frac{6 \tan ^{-1}(a x) \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(a x)}\right )-6 i \left (\tan ^{-1}(a x)^2-1\right ) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(a x)}\right )+3 i \text{PolyLog}\left (4,-e^{2 i \tan ^{-1}(a x)}\right )+2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^3-i \tan ^{-1}(a x)^4-6 a x \tan ^{-1}(a x)^2+6 i \tan ^{-1}(a x)^2+4 \tan ^{-1}(a x)^3 \log \left (1+e^{2 i \tan ^{-1}(a x)}\right )-12 \tan ^{-1}(a x) \log \left (1+e^{2 i \tan ^{-1}(a x)}\right )}{4 a^4 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*ArcTan[a*x]^3)/(c + a^2*c*x^2),x]

[Out]

((6*I)*ArcTan[a*x]^2 - 6*a*x*ArcTan[a*x]^2 + 2*(1 + a^2*x^2)*ArcTan[a*x]^3 - I*ArcTan[a*x]^4 - 12*ArcTan[a*x]*
Log[1 + E^((2*I)*ArcTan[a*x])] + 4*ArcTan[a*x]^3*Log[1 + E^((2*I)*ArcTan[a*x])] - (6*I)*(-1 + ArcTan[a*x]^2)*P
olyLog[2, -E^((2*I)*ArcTan[a*x])] + 6*ArcTan[a*x]*PolyLog[3, -E^((2*I)*ArcTan[a*x])] + (3*I)*PolyLog[4, -E^((2
*I)*ArcTan[a*x])])/(4*a^4*c)

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Maple [A]  time = 5.585, size = 292, normalized size = 1.1 \begin{align*}{\frac{-{\frac{i}{4}} \left ( \arctan \left ( ax \right ) \right ) ^{4}}{{a}^{4}c}}+{\frac{{x}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{3}}{2\,{a}^{2}c}}+{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{3}}{2\,{a}^{4}c}}-{\frac{3\,x \left ( \arctan \left ( ax \right ) \right ) ^{2}}{2\,{a}^{3}c}}+{\frac{{\frac{3\,i}{2}} \left ( \arctan \left ( ax \right ) \right ) ^{2}}{{a}^{4}c}}+{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{3}}{{a}^{4}c}\ln \left ({\frac{ \left ( 1+iax \right ) ^{2}}{{a}^{2}{x}^{2}+1}}+1 \right ) }-{\frac{{\frac{3\,i}{2}} \left ( \arctan \left ( ax \right ) \right ) ^{2}}{{a}^{4}c}{\it polylog} \left ( 2,-{\frac{ \left ( 1+iax \right ) ^{2}}{{a}^{2}{x}^{2}+1}} \right ) }+{\frac{3\,\arctan \left ( ax \right ) }{2\,{a}^{4}c}{\it polylog} \left ( 3,-{\frac{ \left ( 1+iax \right ) ^{2}}{{a}^{2}{x}^{2}+1}} \right ) }+{\frac{{\frac{3\,i}{4}}}{{a}^{4}c}{\it polylog} \left ( 4,-{\frac{ \left ( 1+iax \right ) ^{2}}{{a}^{2}{x}^{2}+1}} \right ) }-3\,{\frac{\arctan \left ( ax \right ) }{{a}^{4}c}\ln \left ({\frac{ \left ( 1+iax \right ) ^{2}}{{a}^{2}{x}^{2}+1}}+1 \right ) }+{\frac{{\frac{3\,i}{2}}}{{a}^{4}c}{\it polylog} \left ( 2,-{\frac{ \left ( 1+iax \right ) ^{2}}{{a}^{2}{x}^{2}+1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctan(a*x)^3/(a^2*c*x^2+c),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \arctan \left (a x\right )^{3}}{a^{2} c x^{2} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(a*x)^3/(a^2*c*x^2+c),x, algorithm="maxima")

[Out]

integrate(x^3*arctan(a*x)^3/(a^2*c*x^2 + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(a*x)^3/(a^2*c*x^2+c),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x^{3} \operatorname{atan}^{3}{\left (a x \right )}}{a^{2} x^{2} + 1}\, dx}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atan(a*x)**3/(a**2*c*x**2+c),x)

[Out]

Integral(x**3*atan(a*x)**3/(a**2*x**2 + 1), x)/c

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \arctan \left (a x\right )^{3}}{a^{2} c x^{2} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(a*x)^3/(a^2*c*x^2+c),x, algorithm="giac")

[Out]

integrate(x^3*arctan(a*x)^3/(a^2*c*x^2 + c), x)