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

Optimal. Leaf size=495 \[ -\frac {x \tan ^{-1}(a x)^3}{a^2 c \sqrt {a^2 c x^2+c}}+\frac {6 x \tan ^{-1}(a x)}{a^2 c \sqrt {a^2 c x^2+c}}+\frac {3 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {3 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {6 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}+\frac {6 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {6 i \sqrt {a^2 x^2+1} \text {Li}_4\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}+\frac {6 i \sqrt {a^2 x^2+1} \text {Li}_4\left (i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}+\frac {6}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {3 \tan ^{-1}(a x)^2}{a^3 c \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 495, 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, 4890, 4888, 4181, 2531, 6609, 2282, 6589, 4898, 4894} \[ \frac {3 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {3 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {6 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}+\frac {6 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {6 i \sqrt {a^2 x^2+1} \text {PolyLog}\left (4,-i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}+\frac {6 i \sqrt {a^2 x^2+1} \text {PolyLog}\left (4,i e^{i \tan ^{-1}(a x)}\right )}{a^3 c \sqrt {a^2 c x^2+c}}+\frac {6}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {x \tan ^{-1}(a x)^3}{a^2 c \sqrt {a^2 c x^2+c}}-\frac {3 \tan ^{-1}(a x)^2}{a^3 c \sqrt {a^2 c x^2+c}}+\frac {6 x \tan ^{-1}(a x)}{a^2 c \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

6/(a^3*c*Sqrt[c + a^2*c*x^2]) + (6*x*ArcTan[a*x])/(a^2*c*Sqrt[c + a^2*c*x^2]) - (3*ArcTan[a*x]^2)/(a^3*c*Sqrt[
c + a^2*c*x^2]) - (x*ArcTan[a*x]^3)/(a^2*c*Sqrt[c + a^2*c*x^2]) - ((2*I)*Sqrt[1 + a^2*x^2]*ArcTan[E^(I*ArcTan[
a*x])]*ArcTan[a*x]^3)/(a^3*c*Sqrt[c + a^2*c*x^2]) + ((3*I)*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*PolyLog[2, (-I)*E^(
I*ArcTan[a*x])])/(a^3*c*Sqrt[c + a^2*c*x^2]) - ((3*I)*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*PolyLog[2, I*E^(I*ArcTan
[a*x])])/(a^3*c*Sqrt[c + a^2*c*x^2]) - (6*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[3, (-I)*E^(I*ArcTan[a*x])])/(a
^3*c*Sqrt[c + a^2*c*x^2]) + (6*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[3, I*E^(I*ArcTan[a*x])])/(a^3*c*Sqrt[c +
a^2*c*x^2]) - ((6*I)*Sqrt[1 + a^2*x^2]*PolyLog[4, (-I)*E^(I*ArcTan[a*x])])/(a^3*c*Sqrt[c + a^2*c*x^2]) + ((6*I
)*Sqrt[1 + a^2*x^2]*PolyLog[4, I*E^(I*ArcTan[a*x])])/(a^3*c*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 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 4894

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

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 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]

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

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Mathematica [A]  time = 1.75, size = 639, normalized size = 1.29 \[ -\frac {\sqrt {a^2 x^2+1} \left (-\frac {384}{\sqrt {a^2 x^2+1}}+\frac {64 a x \tan ^{-1}(a x)^3}{\sqrt {a^2 x^2+1}}+\frac {192 \tan ^{-1}(a x)^2}{\sqrt {a^2 x^2+1}}-\frac {384 a x \tan ^{-1}(a x)}{\sqrt {a^2 x^2+1}}-192 i \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{-i \tan ^{-1}(a x)}\right )-192 i \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )+192 i \pi \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )-384 \tan ^{-1}(a x) \text {Li}_3\left (-i e^{-i \tan ^{-1}(a x)}\right )+384 \tan ^{-1}(a x) \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )-48 i \pi \left (\pi -4 \tan ^{-1}(a x)\right ) \text {Li}_2\left (i e^{-i \tan ^{-1}(a x)}\right )-48 i \pi ^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )+192 \pi \text {Li}_3\left (i e^{-i \tan ^{-1}(a x)}\right )-192 \pi \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )+384 i \text {Li}_4\left (-i e^{-i \tan ^{-1}(a x)}\right )+384 i \text {Li}_4\left (-i e^{i \tan ^{-1}(a x)}\right )-16 i \tan ^{-1}(a x)^4+32 i \pi \tan ^{-1}(a x)^3-24 i \pi ^2 \tan ^{-1}(a x)^2+8 i \pi ^3 \tan ^{-1}(a x)-64 \tan ^{-1}(a x)^3 \log \left (1+i e^{-i \tan ^{-1}(a x)}\right )+64 \tan ^{-1}(a x)^3 \log \left (1+i e^{i \tan ^{-1}(a x)}\right )+96 \pi \tan ^{-1}(a x)^2 \log \left (1-i e^{-i \tan ^{-1}(a x)}\right )-96 \pi \tan ^{-1}(a x)^2 \log \left (1+i e^{i \tan ^{-1}(a x)}\right )-48 \pi ^2 \tan ^{-1}(a x) \log \left (1-i e^{-i \tan ^{-1}(a x)}\right )+48 \pi ^2 \tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right )+8 \pi ^3 \log \left (1+i e^{-i \tan ^{-1}(a x)}\right )-8 \pi ^3 \log \left (1+i e^{i \tan ^{-1}(a x)}\right )-8 \pi ^3 \log \left (\tan \left (\frac {1}{4} \left (2 \tan ^{-1}(a x)+\pi \right )\right )\right )+7 i \pi ^4\right )}{64 a^3 c \sqrt {c \left (a^2 x^2+1\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/64*(Sqrt[1 + a^2*x^2]*((7*I)*Pi^4 - 384/Sqrt[1 + a^2*x^2] + (8*I)*Pi^3*ArcTan[a*x] - (384*a*x*ArcTan[a*x])/
Sqrt[1 + a^2*x^2] - (24*I)*Pi^2*ArcTan[a*x]^2 + (192*ArcTan[a*x]^2)/Sqrt[1 + a^2*x^2] + (32*I)*Pi*ArcTan[a*x]^
3 + (64*a*x*ArcTan[a*x]^3)/Sqrt[1 + a^2*x^2] - (16*I)*ArcTan[a*x]^4 - 48*Pi^2*ArcTan[a*x]*Log[1 - I/E^(I*ArcTa
n[a*x])] + 96*Pi*ArcTan[a*x]^2*Log[1 - I/E^(I*ArcTan[a*x])] + 8*Pi^3*Log[1 + I/E^(I*ArcTan[a*x])] - 64*ArcTan[
a*x]^3*Log[1 + I/E^(I*ArcTan[a*x])] - 8*Pi^3*Log[1 + I*E^(I*ArcTan[a*x])] + 48*Pi^2*ArcTan[a*x]*Log[1 + I*E^(I
*ArcTan[a*x])] - 96*Pi*ArcTan[a*x]^2*Log[1 + I*E^(I*ArcTan[a*x])] + 64*ArcTan[a*x]^3*Log[1 + I*E^(I*ArcTan[a*x
])] - 8*Pi^3*Log[Tan[(Pi + 2*ArcTan[a*x])/4]] - (192*I)*ArcTan[a*x]^2*PolyLog[2, (-I)/E^(I*ArcTan[a*x])] - (48
*I)*Pi*(Pi - 4*ArcTan[a*x])*PolyLog[2, I/E^(I*ArcTan[a*x])] - (48*I)*Pi^2*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] +
(192*I)*Pi*ArcTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] - (192*I)*ArcTan[a*x]^2*PolyLog[2, (-I)*E^(I*ArcTan
[a*x])] - 384*ArcTan[a*x]*PolyLog[3, (-I)/E^(I*ArcTan[a*x])] + 192*Pi*PolyLog[3, I/E^(I*ArcTan[a*x])] - 192*Pi
*PolyLog[3, (-I)*E^(I*ArcTan[a*x])] + 384*ArcTan[a*x]*PolyLog[3, (-I)*E^(I*ArcTan[a*x])] + (384*I)*PolyLog[4,
(-I)/E^(I*ArcTan[a*x])] + (384*I)*PolyLog[4, (-I)*E^(I*ArcTan[a*x])]))/(a^3*c*Sqrt[c*(1 + a^2*x^2)])

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} x^{2} \arctan \left (a x\right )^{3}}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*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^2*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]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [F]  time = 3.40, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \arctan \left (a x \right )^{3}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*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 \[ \int \frac {x^{2} \arctan \left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*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 \[ \int \frac {x^2\,{\mathrm {atan}\left (a\,x\right )}^3}{{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^2*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 \[ \int \frac {x^{2} \operatorname {atan}^{3}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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