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

Optimal. Leaf size=403 \[ \frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3}{a^4 c^2}-\frac {6 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {a^2 c x^2+c}}+\frac {6 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {a^2 c x^2+c}}+\frac {6 \sqrt {a^2 x^2+1} \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {a^2 c x^2+c}}-\frac {6 \sqrt {a^2 x^2+1} \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {a^2 c x^2+c}}+\frac {\tan ^{-1}(a x)^3}{a^4 c \sqrt {a^2 c x^2+c}}+\frac {6 i \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^4 c \sqrt {a^2 c x^2+c}}-\frac {6 \tan ^{-1}(a x)}{a^4 c \sqrt {a^2 c x^2+c}}+\frac {6 x}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {3 x \tan ^{-1}(a x)^2}{a^3 c \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.52, antiderivative size = 403, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {4964, 4930, 4890, 4888, 4181, 2531, 2282, 6589, 4898, 191} \[ -\frac {6 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {a^2 c x^2+c}}+\frac {6 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {a^2 c x^2+c}}+\frac {6 \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {a^2 c x^2+c}}-\frac {6 \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3}{a^4 c^2}+\frac {6 x}{a^3 c \sqrt {a^2 c x^2+c}}+\frac {\tan ^{-1}(a x)^3}{a^4 c \sqrt {a^2 c x^2+c}}+\frac {6 i \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^4 c \sqrt {a^2 c x^2+c}}-\frac {3 x \tan ^{-1}(a x)^2}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {6 \tan ^{-1}(a x)}{a^4 c \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*x)/(a^3*c*Sqrt[c + a^2*c*x^2]) - (6*ArcTan[a*x])/(a^4*c*Sqrt[c + a^2*c*x^2]) - (3*x*ArcTan[a*x]^2)/(a^3*c*S
qrt[c + a^2*c*x^2]) + ((6*I)*Sqrt[1 + a^2*x^2]*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^2)/(a^4*c*Sqrt[c + a^2*c*
x^2]) + ArcTan[a*x]^3/(a^4*c*Sqrt[c + a^2*c*x^2]) + (Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^3)/(a^4*c^2) - ((6*I)*Sqr
t[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/(a^4*c*Sqrt[c + a^2*c*x^2]) + ((6*I)*Sqrt[1 + a
^2*x^2]*ArcTan[a*x]*PolyLog[2, I*E^(I*ArcTan[a*x])])/(a^4*c*Sqrt[c + a^2*c*x^2]) + (6*Sqrt[1 + a^2*x^2]*PolyLo
g[3, (-I)*E^(I*ArcTan[a*x])])/(a^4*c*Sqrt[c + a^2*c*x^2]) - (6*Sqrt[1 + a^2*x^2]*PolyLog[3, I*E^(I*ArcTan[a*x]
)])/(a^4*c*Sqrt[c + a^2*c*x^2])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

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 4181

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

Rule 4888

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c*Sqrt[d]), Subst
[Int[(a + b*x)^p*Sec[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 4890

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/Sq
rt[d + e*x^2], Int[(a + b*ArcTan[c*x])^p/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 4898

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(b*p*(a + b*ArcTan[
c*x])^(p - 1))/(c*d*Sqrt[d + e*x^2]), x] + (-Dist[b^2*p*(p - 1), Int[(a + b*ArcTan[c*x])^(p - 2)/(d + e*x^2)^(
3/2), x], x] + Simp[(x*(a + b*ArcTan[c*x])^p)/(d*Sqrt[d + e*x^2]), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e,
c^2*d] && GtQ[p, 1]

Rule 4930

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

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int[
x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*Arc
Tan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && IGtQ[m
, 1] && NeQ[p, -1]

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]

Rubi steps

\begin {align*} \int \frac {x^3 \tan ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^{3/2}} \, dx &=-\frac {\int \frac {x \tan ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^2}+\frac {\int \frac {x \tan ^{-1}(a x)^3}{\sqrt {c+a^2 c x^2}} \, dx}{a^2 c}\\ &=\frac {\tan ^{-1}(a x)^3}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^4 c^2}-\frac {3 \int \frac {\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^3}-\frac {3 \int \frac {\tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx}{a^3 c}\\ &=-\frac {6 \tan ^{-1}(a x)}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {3 x \tan ^{-1}(a x)^2}{a^3 c \sqrt {c+a^2 c x^2}}+\frac {\tan ^{-1}(a x)^3}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^4 c^2}+\frac {6 \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^3}-\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{a^3 c \sqrt {c+a^2 c x^2}}\\ &=\frac {6 x}{a^3 c \sqrt {c+a^2 c x^2}}-\frac {6 \tan ^{-1}(a x)}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {3 x \tan ^{-1}(a x)^2}{a^3 c \sqrt {c+a^2 c x^2}}+\frac {\tan ^{-1}(a x)^3}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^4 c^2}-\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \sec (x) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c \sqrt {c+a^2 c x^2}}\\ &=\frac {6 x}{a^3 c \sqrt {c+a^2 c x^2}}-\frac {6 \tan ^{-1}(a x)}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {3 x \tan ^{-1}(a x)^2}{a^3 c \sqrt {c+a^2 c x^2}}+\frac {6 i \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\tan ^{-1}(a x)^3}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^4 c^2}+\frac {\left (6 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {\left (6 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c \sqrt {c+a^2 c x^2}}\\ &=\frac {6 x}{a^3 c \sqrt {c+a^2 c x^2}}-\frac {6 \tan ^{-1}(a x)}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {3 x \tan ^{-1}(a x)^2}{a^3 c \sqrt {c+a^2 c x^2}}+\frac {6 i \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\tan ^{-1}(a x)^3}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^4 c^2}-\frac {6 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {6 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\left (6 i \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {\left (6 i \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c \sqrt {c+a^2 c x^2}}\\ &=\frac {6 x}{a^3 c \sqrt {c+a^2 c x^2}}-\frac {6 \tan ^{-1}(a x)}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {3 x \tan ^{-1}(a x)^2}{a^3 c \sqrt {c+a^2 c x^2}}+\frac {6 i \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\tan ^{-1}(a x)^3}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^4 c^2}-\frac {6 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {6 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\left (6 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {\left (6 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {c+a^2 c x^2}}\\ &=\frac {6 x}{a^3 c \sqrt {c+a^2 c x^2}}-\frac {6 \tan ^{-1}(a x)}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {3 x \tan ^{-1}(a x)^2}{a^3 c \sqrt {c+a^2 c x^2}}+\frac {6 i \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\tan ^{-1}(a x)^3}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^4 c^2}-\frac {6 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {6 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {6 \sqrt {1+a^2 x^2} \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {6 \sqrt {1+a^2 x^2} \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a^4 c \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.89, size = 308, normalized size = 0.76 \[ \frac {\sqrt {a^2 x^2+1} \left (\frac {6 a x}{\sqrt {a^2 x^2+1}}+\frac {3}{2} \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^3-\frac {3 a x \tan ^{-1}(a x)^2}{\sqrt {a^2 x^2+1}}-3 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)+\frac {1}{2} \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^3 \cos \left (2 \tan ^{-1}(a x)\right )-3 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \cos \left (2 \tan ^{-1}(a x)\right )-6 i \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )+6 i \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )+6 \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )-6 \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )-3 \tan ^{-1}(a x)^2 \log \left (1-i e^{i \tan ^{-1}(a x)}\right )+3 \tan ^{-1}(a x)^2 \log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right )}{a^4 c \sqrt {c \left (a^2 x^2+1\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 + a^2*x^2]*((6*a*x)/Sqrt[1 + a^2*x^2] - 3*Sqrt[1 + a^2*x^2]*ArcTan[a*x] - (3*a*x*ArcTan[a*x]^2)/Sqrt[1
+ a^2*x^2] + (3*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^3)/2 - 3*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*Cos[2*ArcTan[a*x]] + (Sq
rt[1 + a^2*x^2]*ArcTan[a*x]^3*Cos[2*ArcTan[a*x]])/2 - 3*ArcTan[a*x]^2*Log[1 - I*E^(I*ArcTan[a*x])] + 3*ArcTan[
a*x]^2*Log[1 + I*E^(I*ArcTan[a*x])] - (6*I)*ArcTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] + (6*I)*ArcTan[a*x]
*PolyLog[2, I*E^(I*ArcTan[a*x])] + 6*PolyLog[3, (-I)*E^(I*ArcTan[a*x])] - 6*PolyLog[3, I*E^(I*ArcTan[a*x])]))/
(a^4*c*Sqrt[c*(1 + a^2*x^2)])

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} x^{3} \arctan \left (a x\right )^{3}}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a^2*c*x^2 + c)*x^3*arctan(a*x)^3/(a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 3.35, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \arctan \left (a x \right )^{3}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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