3.348 \(\int \frac{x^4 \tan ^{-1}(a x)^2}{(c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=444 \[ \frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{a^2 c x^2+c}}+\frac{2 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{a^2 c x^2+c}}+\frac{22 x}{9 a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{x \tan ^{-1}(a x)^2}{a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^5 c^2 \sqrt{a^2 c x^2+c}}-\frac{22 \tan ^{-1}(a x)}{9 a^5 c^2 \sqrt{a^2 c x^2+c}}+\frac{2 x^3}{27 a^2 c \left (a^2 c x^2+c\right )^{3/2}}-\frac{x^3 \tan ^{-1}(a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}-\frac{2 x^2 \tan ^{-1}(a x)}{9 a^3 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

(2*x^3)/(27*a^2*c*(c + a^2*c*x^2)^(3/2)) + (22*x)/(9*a^4*c^2*Sqrt[c + a^2*c*x^2]) - (2*x^2*ArcTan[a*x])/(9*a^3
*c*(c + a^2*c*x^2)^(3/2)) - (22*ArcTan[a*x])/(9*a^5*c^2*Sqrt[c + a^2*c*x^2]) - (x^3*ArcTan[a*x]^2)/(3*a^2*c*(c
+ a^2*c*x^2)^(3/2)) - (x*ArcTan[a*x]^2)/(a^4*c^2*Sqrt[c + a^2*c*x^2]) - ((2*I)*Sqrt[1 + a^2*x^2]*ArcTan[E^(I*
ArcTan[a*x])]*ArcTan[a*x]^2)/(a^5*c^2*Sqrt[c + a^2*c*x^2]) + ((2*I)*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, (
-I)*E^(I*ArcTan[a*x])])/(a^5*c^2*Sqrt[c + a^2*c*x^2]) - ((2*I)*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, I*E^(I
*ArcTan[a*x])])/(a^5*c^2*Sqrt[c + a^2*c*x^2]) - (2*Sqrt[1 + a^2*x^2]*PolyLog[3, (-I)*E^(I*ArcTan[a*x])])/(a^5*
c^2*Sqrt[c + a^2*c*x^2]) + (2*Sqrt[1 + a^2*x^2]*PolyLog[3, I*E^(I*ArcTan[a*x])])/(a^5*c^2*Sqrt[c + a^2*c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.769034, antiderivative size = 444, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 12, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4964, 4890, 4888, 4181, 2531, 2282, 6589, 4898, 191, 4944, 4938, 4930} \[ \frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{a^2 c x^2+c}}+\frac{2 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{a^2 c x^2+c}}+\frac{22 x}{9 a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{x \tan ^{-1}(a x)^2}{a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^5 c^2 \sqrt{a^2 c x^2+c}}-\frac{22 \tan ^{-1}(a x)}{9 a^5 c^2 \sqrt{a^2 c x^2+c}}+\frac{2 x^3}{27 a^2 c \left (a^2 c x^2+c\right )^{3/2}}-\frac{x^3 \tan ^{-1}(a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}-\frac{2 x^2 \tan ^{-1}(a x)}{9 a^3 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*ArcTan[a*x]^2)/(c + a^2*c*x^2)^(5/2),x]

[Out]

(2*x^3)/(27*a^2*c*(c + a^2*c*x^2)^(3/2)) + (22*x)/(9*a^4*c^2*Sqrt[c + a^2*c*x^2]) - (2*x^2*ArcTan[a*x])/(9*a^3
*c*(c + a^2*c*x^2)^(3/2)) - (22*ArcTan[a*x])/(9*a^5*c^2*Sqrt[c + a^2*c*x^2]) - (x^3*ArcTan[a*x]^2)/(3*a^2*c*(c
+ a^2*c*x^2)^(3/2)) - (x*ArcTan[a*x]^2)/(a^4*c^2*Sqrt[c + a^2*c*x^2]) - ((2*I)*Sqrt[1 + a^2*x^2]*ArcTan[E^(I*
ArcTan[a*x])]*ArcTan[a*x]^2)/(a^5*c^2*Sqrt[c + a^2*c*x^2]) + ((2*I)*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, (
-I)*E^(I*ArcTan[a*x])])/(a^5*c^2*Sqrt[c + a^2*c*x^2]) - ((2*I)*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, I*E^(I
*ArcTan[a*x])])/(a^5*c^2*Sqrt[c + a^2*c*x^2]) - (2*Sqrt[1 + a^2*x^2]*PolyLog[3, (-I)*E^(I*ArcTan[a*x])])/(a^5*
c^2*Sqrt[c + a^2*c*x^2]) + (2*Sqrt[1 + a^2*x^2]*PolyLog[3, I*E^(I*ArcTan[a*x])])/(a^5*c^2*Sqrt[c + a^2*c*x^2])

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 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 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 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 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 1)), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 4938

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

Rubi steps

\begin{align*} \int \frac{x^4 \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=-\frac{\int \frac{x^2 \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx}{a^2}+\frac{\int \frac{x^2 \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^2 c}\\ &=-\frac{x^3 \tan ^{-1}(a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{2 \int \frac{x^3 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx}{3 a}+\frac{\int \frac{\tan ^{-1}(a x)^2}{\sqrt{c+a^2 c x^2}} \, dx}{a^4 c^2}-\frac{\int \frac{\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^4 c}\\ &=\frac{2 x^3}{27 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{2 x^2 \tan ^{-1}(a x)}{9 a^3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{2 \tan ^{-1}(a x)}{a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{x^3 \tan ^{-1}(a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{x \tan ^{-1}(a x)^2}{a^4 c^2 \sqrt{c+a^2 c x^2}}+\frac{2 \int \frac{1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^4 c}+\frac{4 \int \frac{x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{9 a^3 c}+\frac{\sqrt{1+a^2 x^2} \int \frac{\tan ^{-1}(a x)^2}{\sqrt{1+a^2 x^2}} \, dx}{a^4 c^2 \sqrt{c+a^2 c x^2}}\\ &=\frac{2 x^3}{27 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{2 x}{a^4 c^2 \sqrt{c+a^2 c x^2}}-\frac{2 x^2 \tan ^{-1}(a x)}{9 a^3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{22 \tan ^{-1}(a x)}{9 a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{x^3 \tan ^{-1}(a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{x \tan ^{-1}(a x)^2}{a^4 c^2 \sqrt{c+a^2 c x^2}}+\frac{4 \int \frac{1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{9 a^4 c}+\frac{\sqrt{1+a^2 x^2} \operatorname{Subst}\left (\int x^2 \sec (x) \, dx,x,\tan ^{-1}(a x)\right )}{a^5 c^2 \sqrt{c+a^2 c x^2}}\\ &=\frac{2 x^3}{27 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{22 x}{9 a^4 c^2 \sqrt{c+a^2 c x^2}}-\frac{2 x^2 \tan ^{-1}(a x)}{9 a^3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{22 \tan ^{-1}(a x)}{9 a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{x^3 \tan ^{-1}(a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{x \tan ^{-1}(a x)^2}{a^4 c^2 \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)^2}{a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{\left (2 \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^5 c^2 \sqrt{c+a^2 c x^2}}+\frac{\left (2 \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^5 c^2 \sqrt{c+a^2 c x^2}}\\ &=\frac{2 x^3}{27 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{22 x}{9 a^4 c^2 \sqrt{c+a^2 c x^2}}-\frac{2 x^2 \tan ^{-1}(a x)}{9 a^3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{22 \tan ^{-1}(a x)}{9 a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{x^3 \tan ^{-1}(a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{x \tan ^{-1}(a x)^2}{a^4 c^2 \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)^2}{a^5 c^2 \sqrt{c+a^2 c x^2}}+\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{\left (2 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^5 c^2 \sqrt{c+a^2 c x^2}}+\frac{\left (2 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^5 c^2 \sqrt{c+a^2 c x^2}}\\ &=\frac{2 x^3}{27 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{22 x}{9 a^4 c^2 \sqrt{c+a^2 c x^2}}-\frac{2 x^2 \tan ^{-1}(a x)}{9 a^3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{22 \tan ^{-1}(a x)}{9 a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{x^3 \tan ^{-1}(a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{x \tan ^{-1}(a x)^2}{a^4 c^2 \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)^2}{a^5 c^2 \sqrt{c+a^2 c x^2}}+\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{\left (2 \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^5 c^2 \sqrt{c+a^2 c x^2}}+\frac{\left (2 \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^5 c^2 \sqrt{c+a^2 c x^2}}\\ &=\frac{2 x^3}{27 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{22 x}{9 a^4 c^2 \sqrt{c+a^2 c x^2}}-\frac{2 x^2 \tan ^{-1}(a x)}{9 a^3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{22 \tan ^{-1}(a x)}{9 a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{x^3 \tan ^{-1}(a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{x \tan ^{-1}(a x)^2}{a^4 c^2 \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)^2}{a^5 c^2 \sqrt{c+a^2 c x^2}}+\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{c+a^2 c x^2}}-\frac{2 \sqrt{1+a^2 x^2} \text{Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{c+a^2 c x^2}}+\frac{2 \sqrt{1+a^2 x^2} \text{Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a^5 c^2 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.574489, size = 239, normalized size = 0.54 \[ \frac{\sqrt{c \left (a^2 x^2+1\right )} \left (216 i \tan ^{-1}(a x) \left (\text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )-\text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )\right )-216 \left (\text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )-\text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )\right )-\frac{270 \tan ^{-1}(a x)}{\sqrt{a^2 x^2+1}}-\frac{135 a x \left (\tan ^{-1}(a x)^2-2\right )}{\sqrt{a^2 x^2+1}}+108 \tan ^{-1}(a x)^2 \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right )+\left (9 \tan ^{-1}(a x)^2-2\right ) \sin \left (3 \tan ^{-1}(a x)\right )+6 \tan ^{-1}(a x) \cos \left (3 \tan ^{-1}(a x)\right )\right )}{108 a^5 c^3 \sqrt{a^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*ArcTan[a*x]^2)/(c + a^2*c*x^2)^(5/2),x]

[Out]

(Sqrt[c*(1 + a^2*x^2)]*((-270*ArcTan[a*x])/Sqrt[1 + a^2*x^2] - (135*a*x*(-2 + ArcTan[a*x]^2))/Sqrt[1 + a^2*x^2
] + 6*ArcTan[a*x]*Cos[3*ArcTan[a*x]] + 108*ArcTan[a*x]^2*(Log[1 - I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(I*ArcTan
[a*x])]) + (216*I)*ArcTan[a*x]*(PolyLog[2, (-I)*E^(I*ArcTan[a*x])] - PolyLog[2, I*E^(I*ArcTan[a*x])]) - 216*(P
olyLog[3, (-I)*E^(I*ArcTan[a*x])] - PolyLog[3, I*E^(I*ArcTan[a*x])]) + (-2 + 9*ArcTan[a*x]^2)*Sin[3*ArcTan[a*x
]]))/(108*a^5*c^3*Sqrt[1 + a^2*x^2])

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Maple [F]  time = 0.904, size = 0, normalized size = 0. \begin{align*} \int{{x}^{4} \left ( \arctan \left ( ax \right ) \right ) ^{2} \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arctan(a*x)^2/(a^2*c*x^2+c)^(5/2),x)

[Out]

int(x^4*arctan(a*x)^2/(a^2*c*x^2+c)^(5/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctan(a*x)^2/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a^{2} c x^{2} + c} x^{4} \arctan \left (a x\right )^{2}}{a^{6} c^{3} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} + c^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctan(a*x)^2/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)*x^4*arctan(a*x)^2/(a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 + c^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \operatorname{atan}^{2}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*atan(a*x)**2/(a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(x**4*atan(a*x)**2/(c*(a**2*x**2 + 1))**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctan(a*x)^2/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate(x^4*arctan(a*x)^2/(a^2*c*x^2 + c)^(5/2), x)