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

Optimal. Leaf size=262 \[ -\frac{3 i a^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c}-\frac{3 i a^2 \text{PolyLog}\left (4,-1+\frac{2}{1-i a x}\right )}{4 c}+\frac{3 i a^2 \tan ^{-1}(a x)^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c}-\frac{3 a^2 \tan ^{-1}(a x) \text{PolyLog}\left (3,-1+\frac{2}{1-i a x}\right )}{2 c}+\frac{i a^2 \tan ^{-1}(a x)^4}{4 c}-\frac{a^2 \tan ^{-1}(a x)^3}{2 c}-\frac{3 i a^2 \tan ^{-1}(a x)^2}{2 c}-\frac{a^2 \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)^3}{c}+\frac{3 a^2 \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)}{c}-\frac{\tan ^{-1}(a x)^3}{2 c x^2}-\frac{3 a \tan ^{-1}(a x)^2}{2 c x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.510552, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {4918, 4852, 4924, 4868, 2447, 4884, 4992, 4996, 6610} \[ -\frac{3 i a^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c}-\frac{3 i a^2 \text{PolyLog}\left (4,-1+\frac{2}{1-i a x}\right )}{4 c}+\frac{3 i a^2 \tan ^{-1}(a x)^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c}-\frac{3 a^2 \tan ^{-1}(a x) \text{PolyLog}\left (3,-1+\frac{2}{1-i a x}\right )}{2 c}+\frac{i a^2 \tan ^{-1}(a x)^4}{4 c}-\frac{a^2 \tan ^{-1}(a x)^3}{2 c}-\frac{3 i a^2 \tan ^{-1}(a x)^2}{2 c}-\frac{a^2 \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)^3}{c}+\frac{3 a^2 \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)}{c}-\frac{\tan ^{-1}(a x)^3}{2 c x^2}-\frac{3 a \tan ^{-1}(a x)^2}{2 c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), 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] && LtQ[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 4924

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

Rule 4868

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTan[c*x]
)^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2 - 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 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 4992

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a + b*ArcT
an[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 4996

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

Mathematica [A]  time = 0.397151, size = 189, normalized size = 0.72 \[ \frac{i a^2 \left (-96 \tan ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(a x)}\right )+96 i \tan ^{-1}(a x) \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(a x)}\right )-96 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(a x)}\right )+48 \text{PolyLog}\left (4,e^{-2 i \tan ^{-1}(a x)}\right )+\frac{32 i \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^3}{a^2 x^2}-16 \tan ^{-1}(a x)^4+\frac{96 i \tan ^{-1}(a x)^2}{a x}-96 \tan ^{-1}(a x)^2+64 i \tan ^{-1}(a x)^3 \log \left (1-e^{-2 i \tan ^{-1}(a x)}\right )-192 i \tan ^{-1}(a x) \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )+\pi ^4\right )}{64 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((I/64)*a^2*(Pi^4 - 96*ArcTan[a*x]^2 + ((96*I)*ArcTan[a*x]^2)/(a*x) + ((32*I)*(1 + a^2*x^2)*ArcTan[a*x]^3)/(a^
2*x^2) - 16*ArcTan[a*x]^4 + (64*I)*ArcTan[a*x]^3*Log[1 - E^((-2*I)*ArcTan[a*x])] - (192*I)*ArcTan[a*x]*Log[1 -
E^((2*I)*ArcTan[a*x])] - 96*ArcTan[a*x]^2*PolyLog[2, E^((-2*I)*ArcTan[a*x])] - 96*PolyLog[2, E^((2*I)*ArcTan[
a*x])] + (96*I)*ArcTan[a*x]*PolyLog[3, E^((-2*I)*ArcTan[a*x])] + 48*PolyLog[4, E^((-2*I)*ArcTan[a*x])]))/c

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Maple [B]  time = 7.375, size = 479, normalized size = 1.8 \begin{align*}{\frac{-{\frac{3\,i}{2}}{a}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{2}}{c}}-{\frac{{a}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{3}}{2\,c}}-{\frac{6\,i{a}^{2}}{c}{\it polylog} \left ( 4,{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{3\,a \left ( \arctan \left ( ax \right ) \right ) ^{2}}{2\,cx}}-{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{3}}{2\,c{x}^{2}}}-{\frac{3\,i{a}^{2}}{c}{\it polylog} \left ( 2,{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{a}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{3}}{c}\ln \left ( 1-{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{i}{4}}{a}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{4}}{c}}-6\,{\frac{{a}^{2}\arctan \left ( ax \right ) }{c}{\it polylog} \left ( 3,{\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) }+{\frac{3\,i{a}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{2}}{c}{\it polylog} \left ( 2,-{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{a}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{3}}{c}\ln \left ( 1+{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{3\,i{a}^{2}}{c}{\it polylog} \left ( 2,-{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) }-6\,{\frac{{a}^{2}\arctan \left ( ax \right ) }{c}{\it polylog} \left ( 3,-{\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) }-{\frac{6\,i{a}^{2}}{c}{\it polylog} \left ( 4,-{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{3\,i{a}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{2}}{c}{\it polylog} \left ( 2,{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) }+3\,{\frac{{a}^{2}\arctan \left ( ax \right ) }{c}\ln \left ( 1-{\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) }+3\,{\frac{{a}^{2}\arctan \left ( ax \right ) }{c}\ln \left ( 1+{\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*I*a^2*arctan(a*x)^2/c-1/2*a^2*arctan(a*x)^3/c-6*I*a^2/c*polylog(4,(1+I*a*x)/(a^2*x^2+1)^(1/2))-3/2*a*arct
an(a*x)^2/c/x-1/2*arctan(a*x)^3/c/x^2-3*I*a^2/c*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))-a^2/c*arctan(a*x)^3*ln(
1-(1+I*a*x)/(a^2*x^2+1)^(1/2))+1/4*I*a^2*arctan(a*x)^4/c-6*a^2/c*arctan(a*x)*polylog(3,(1+I*a*x)/(a^2*x^2+1)^(
1/2))+3*I*a^2/c*arctan(a*x)^2*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))-a^2/c*arctan(a*x)^3*ln(1+(1+I*a*x)/(a^2*
x^2+1)^(1/2))-3*I*a^2/c*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))-6*a^2/c*arctan(a*x)*polylog(3,-(1+I*a*x)/(a^2*
x^2+1)^(1/2))-6*I*a^2/c*polylog(4,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+3*I*a^2/c*arctan(a*x)^2*polylog(2,(1+I*a*x)/(a
^2*x^2+1)^(1/2))+3*a^2/c*arctan(a*x)*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2))+3*a^2/c*arctan(a*x)*ln(1+(1+I*a*x)/(a^2
*x^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arctan \left (a x\right )^{3}}{a^{2} c x^{5} + c x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(a*x)^3/(a^2*c*x^5 + c*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{atan}^{3}{\left (a x \right )}}{a^{2} x^{5} + x^{3}}\, dx}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(a*x)**3/(a**2*x**5 + x**3), x)/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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