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

Optimal. Leaf size=373 \[ -\frac{i c \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{i c \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{c \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}-\frac{c \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{\left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}+\frac{i c \sqrt{a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^2 \sqrt{a^2 c x^2+c}}-\frac{x \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 a}+\frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{a^2}-\frac{\sqrt{c} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{a^2} \]

[Out]

(Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/a^2 - (x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(2*a) + (I*c*Sqrt[1 + a^2*x^2]*A
rcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^2)/(a^2*Sqrt[c + a^2*c*x^2]) + ((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^3)/(3*a
^2*c) - (Sqrt[c]*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/a^2 - (I*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[
2, (-I)*E^(I*ArcTan[a*x])])/(a^2*Sqrt[c + a^2*c*x^2]) + (I*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, I*E^(I*A
rcTan[a*x])])/(a^2*Sqrt[c + a^2*c*x^2]) + (c*Sqrt[1 + a^2*x^2]*PolyLog[3, (-I)*E^(I*ArcTan[a*x])])/(a^2*Sqrt[c
+ a^2*c*x^2]) - (c*Sqrt[1 + a^2*x^2]*PolyLog[3, I*E^(I*ArcTan[a*x])])/(a^2*Sqrt[c + a^2*c*x^2])

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Rubi [A]  time = 0.355473, antiderivative size = 373, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {4930, 4880, 4890, 4888, 4181, 2531, 2282, 6589, 217, 206} \[ -\frac{i c \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{i c \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{c \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}-\frac{c \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{\left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}+\frac{i c \sqrt{a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^2 \sqrt{a^2 c x^2+c}}-\frac{x \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 a}+\frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{a^2}-\frac{\sqrt{c} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/a^2 - (x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(2*a) + (I*c*Sqrt[1 + a^2*x^2]*A
rcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^2)/(a^2*Sqrt[c + a^2*c*x^2]) + ((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^3)/(3*a
^2*c) - (Sqrt[c]*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/a^2 - (I*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[
2, (-I)*E^(I*ArcTan[a*x])])/(a^2*Sqrt[c + a^2*c*x^2]) + (I*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, I*E^(I*A
rcTan[a*x])])/(a^2*Sqrt[c + a^2*c*x^2]) + (c*Sqrt[1 + a^2*x^2]*PolyLog[3, (-I)*E^(I*ArcTan[a*x])])/(a^2*Sqrt[c
+ a^2*c*x^2]) - (c*Sqrt[1 + a^2*x^2]*PolyLog[3, I*E^(I*ArcTan[a*x])])/(a^2*Sqrt[c + a^2*c*x^2])

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 4880

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

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 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 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 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 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 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.581, size = 206, normalized size = 0.55 \[ \frac{\sqrt{a^2 c x^2+c} \left (\left (a^2 x^2+1\right ) \tan ^{-1}(a x) \left (4 \tan ^{-1}(a x)^2-3 \tan ^{-1}(a x) \sin \left (2 \tan ^{-1}(a x)\right )+6 \cos \left (2 \tan ^{-1}(a x)\right )+6\right )+\frac{12 \left (-i \tan ^{-1}(a x) \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )+i \tan ^{-1}(a x) \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )+\text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )-\text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )-\tanh ^{-1}\left (\frac{a x}{\sqrt{a^2 x^2+1}}\right )+i \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2\right )}{\sqrt{a^2 x^2+1}}\right )}{12 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c + a^2*c*x^2]*((12*(I*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^2 - ArcTanh[(a*x)/Sqrt[1 + a^2*x^2]] - I*Ar
cTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] + I*ArcTan[a*x]*PolyLog[2, I*E^(I*ArcTan[a*x])] + PolyLog[3, (-I)
*E^(I*ArcTan[a*x])] - PolyLog[3, I*E^(I*ArcTan[a*x])]))/Sqrt[1 + a^2*x^2] + (1 + a^2*x^2)*ArcTan[a*x]*(6 + 4*A
rcTan[a*x]^2 + 6*Cos[2*ArcTan[a*x]] - 3*ArcTan[a*x]*Sin[2*ArcTan[a*x]])))/(12*a^2)

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Maple [A]  time = 1.884, size = 370, normalized size = 1. \begin{align*}{\frac{\arctan \left ( ax \right ) \left ( 2\, \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{2}{a}^{2}-3\,\arctan \left ( ax \right ) xa+2\, \left ( \arctan \left ( ax \right ) \right ) ^{2}+6 \right ) }{6\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{1}{6\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) } \left ( i \left ( \arctan \left ( ax \right ) \right ) ^{3}-3\, \left ( \arctan \left ( ax \right ) \right ) ^{2}\ln \left ( 1+{\frac{i \left ( 1+iax \right ) }{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) +6\,i\arctan \left ( ax \right ){\it polylog} \left ( 2,{-i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -6\,{\it polylog} \left ( 3,{\frac{-i \left ( 1+iax \right ) }{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{1}{6\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) } \left ( i \left ( \arctan \left ( ax \right ) \right ) ^{3}+6\,i\arctan \left ( ax \right ){\it polylog} \left ( 2,{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -3\, \left ( \arctan \left ( ax \right ) \right ) ^{2}\ln \left ( 1-{\frac{i \left ( 1+iax \right ) }{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) -6\,{\it polylog} \left ( 3,{\frac{i \left ( 1+iax \right ) }{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{2\,i}{{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }\arctan \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError