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

Optimal. Leaf size=327 \[ \frac {3 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {3 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {6 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 i \sqrt {a^2 x^2+1} \text {Li}_4\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {6 i \sqrt {a^2 x^2+1} \text {Li}_4\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {4958, 4956, 4183, 2531, 6609, 2282, 6589} \[ \frac {3 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {3 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (3,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {6 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (3,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 i \sqrt {a^2 x^2+1} \text {PolyLog}\left (4,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {6 i \sqrt {a^2 x^2+1} \text {PolyLog}\left (4,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^3*ArcTanh[E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2] + ((3*I)*Sqrt[1 + a^2*x^2]
*ArcTan[a*x]^2*PolyLog[2, -E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2] - ((3*I)*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*Po
lyLog[2, E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2] - (6*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[3, -E^(I*ArcTan[a*
x])])/Sqrt[c + a^2*c*x^2] + (6*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[3, E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2
] - ((6*I)*Sqrt[1 + a^2*x^2]*PolyLog[4, -E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2] + ((6*I)*Sqrt[1 + a^2*x^2]*Po
lyLog[4, E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4956

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

Rule 4958

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

Rule 6589

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

Rule 6609

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

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a x)^3}{x \sqrt {c+a^2 c x^2}} \, dx &=\frac {\sqrt {1+a^2 x^2} \int \frac {\tan ^{-1}(a x)^3}{x \sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int x^3 \csc (x) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \log \left (1-e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \log \left (1+e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (6 i \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (6 i \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \text {Li}_2\left (e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (6 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (6 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (6 i \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (6 i \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 i \sqrt {1+a^2 x^2} \text {Li}_4\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 i \sqrt {1+a^2 x^2} \text {Li}_4\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 208, normalized size = 0.64 \[ -\frac {i \sqrt {a^2 x^2+1} \left (-24 \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{-i \tan ^{-1}(a x)}\right )-24 \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )+48 i \tan ^{-1}(a x) \text {Li}_3\left (e^{-i \tan ^{-1}(a x)}\right )-48 i \tan ^{-1}(a x) \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )+48 \text {Li}_4\left (e^{-i \tan ^{-1}(a x)}\right )+48 \text {Li}_4\left (-e^{i \tan ^{-1}(a x)}\right )-2 \tan ^{-1}(a x)^4+8 i \tan ^{-1}(a x)^3 \log \left (1-e^{-i \tan ^{-1}(a x)}\right )-8 i \tan ^{-1}(a x)^3 \log \left (1+e^{i \tan ^{-1}(a x)}\right )+\pi ^4\right )}{8 \sqrt {c \left (a^2 x^2+1\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/8*I)*Sqrt[1 + a^2*x^2]*(Pi^4 - 2*ArcTan[a*x]^4 + (8*I)*ArcTan[a*x]^3*Log[1 - E^((-I)*ArcTan[a*x])] - (8*I
)*ArcTan[a*x]^3*Log[1 + E^(I*ArcTan[a*x])] - 24*ArcTan[a*x]^2*PolyLog[2, E^((-I)*ArcTan[a*x])] - 24*ArcTan[a*x
]^2*PolyLog[2, -E^(I*ArcTan[a*x])] + (48*I)*ArcTan[a*x]*PolyLog[3, E^((-I)*ArcTan[a*x])] - (48*I)*ArcTan[a*x]*
PolyLog[3, -E^(I*ArcTan[a*x])] + 48*PolyLog[4, E^((-I)*ArcTan[a*x])] + 48*PolyLog[4, -E^(I*ArcTan[a*x])]))/Sqr
t[c*(1 + a^2*x^2)]

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{3}}{a^{2} c x^{3} + c x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^3/x/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^3/x/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.94, size = 261, normalized size = 0.80 \[ \frac {i \left (i \arctan \left (a x \right )^{3} \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-i \arctan \left (a x \right )^{3} \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+3 \arctan \left (a x \right )^{2} \polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+6 i \arctan \left (a x \right ) \polylog \left (3, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-3 \arctan \left (a x \right )^{2} \polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-6 i \arctan \left (a x \right ) \polylog \left (3, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-6 \polylog \left (4, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+6 \polylog \left (4, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)^3/x/(a^2*c*x^2+c)^(1/2),x)

[Out]

I*(I*arctan(a*x)^3*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))-I*arctan(a*x)^3*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2))+3*arcta
n(a*x)^2*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*I*arctan(a*x)*polylog(3,-(1+I*a*x)/(a^2*x^2+1)^(1/2))-3*arc
tan(a*x)^2*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))-6*I*arctan(a*x)*polylog(3,(1+I*a*x)/(a^2*x^2+1)^(1/2))-6*pol
ylog(4,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*polylog(4,(1+I*a*x)/(a^2*x^2+1)^(1/2)))*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*
x^2+1)^(1/2)/c

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x\right )^{3}}{\sqrt {a^{2} c x^{2} + c} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^3/x/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^3}{x\,\sqrt {c\,a^2\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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