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

Optimal. Leaf size=602 \[ \frac {3 i a^2 c \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {3 i a^2 c \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}+\frac {3 i a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {a^2 c x^2+c}}-\frac {3 i a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {a^2 c x^2+c}}-\frac {3 a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {3 a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {3 i a^2 c \sqrt {a^2 x^2+1} \text {Li}_4\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {3 i a^2 c \sqrt {a^2 x^2+1} \text {Li}_4\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3}{2 x^2}-\frac {3 a \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 x}-\frac {a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.24, antiderivative size = 602, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 11, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {4950, 4962, 4944, 4958, 4954, 4956, 4183, 2531, 6609, 2282, 6589} \[ \frac {3 i a^2 c \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {3 i a^2 c \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}+\frac {3 i a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,-e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {a^2 c x^2+c}}-\frac {3 i a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {a^2 c x^2+c}}-\frac {3 a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (3,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {3 a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (3,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {3 i a^2 c \sqrt {a^2 x^2+1} \text {PolyLog}\left (4,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {3 i a^2 c \sqrt {a^2 x^2+1} \text {PolyLog}\left (4,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3}{2 x^2}-\frac {3 a \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 x}-\frac {a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 a^2 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(2*x) - (Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^3)/(2*x^2) - (a^2*c*Sqrt[1 +
a^2*x^2]*ArcTan[a*x]^3*ArcTanh[E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2] - (6*a^2*c*Sqrt[1 + a^2*x^2]*ArcTan[a*
x]*ArcTanh[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2] + (((3*I)/2)*a^2*c*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)/2)*a^2*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*P
olyLog[2, E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2] + ((3*I)*a^2*c*Sqrt[1 + a^2*x^2]*PolyLog[2, -(Sqrt[1 + I*a*x
]/Sqrt[1 - I*a*x])])/Sqrt[c + a^2*c*x^2] - ((3*I)*a^2*c*Sqrt[1 + a^2*x^2]*PolyLog[2, Sqrt[1 + I*a*x]/Sqrt[1 -
I*a*x]])/Sqrt[c + a^2*c*x^2] - (3*a^2*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[3, -E^(I*ArcTan[a*x])])/Sqrt[c +
a^2*c*x^2] + (3*a^2*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[3, E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2] - ((3*
I)*a^2*c*Sqrt[1 + a^2*x^2]*PolyLog[4, -E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2] + ((3*I)*a^2*c*Sqrt[1 + a^2*x^2
]*PolyLog[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 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 4950

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

Rule 4954

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Simp[(-2*(a + b*ArcTan[c
*x])*ArcTanh[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]])/Sqrt[d], x] + (Simp[(I*b*PolyLog[2, -(Sqrt[1 + I*c*x]/Sqrt[1 -
I*c*x])])/Sqrt[d], x] - Simp[(I*b*PolyLog[2, Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]])/Sqrt[d], x]) /; FreeQ[{a, b, c,
d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 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 4962

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

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 {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{x^3} \, dx &=c \int \frac {\tan ^{-1}(a x)^3}{x^3 \sqrt {c+a^2 c x^2}} \, dx+\left (a^2 c\right ) \int \frac {\tan ^{-1}(a x)^3}{x \sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{2 x^2}+\frac {1}{2} (3 a c) \int \frac {\tan ^{-1}(a x)^2}{x^2 \sqrt {c+a^2 c x^2}} \, dx-\frac {1}{2} \left (a^2 c\right ) \int \frac {\tan ^{-1}(a x)^3}{x \sqrt {c+a^2 c x^2}} \, dx+\frac {\left (a^2 c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)^3}{x \sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {3 a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{2 x^2}+\left (3 a^2 c\right ) \int \frac {\tan ^{-1}(a x)}{x \sqrt {c+a^2 c x^2}} \, dx-\frac {\left (a^2 c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)^3}{x \sqrt {1+a^2 x^2}} \, dx}{2 \sqrt {c+a^2 c x^2}}+\frac {\left (a^2 c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^3 \csc (x) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {3 a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{2 x^2}-\frac {2 a^2 c \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 (a^2 c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^3 \csc (x) \, dx,x,\tan ^{-1}(a x)\right )}{2 \sqrt {c+a^2 c x^2}}+\frac {\left (3 a^2 c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{x \sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}-\frac {\left (3 a^2 c \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 a^2 c \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 {3 a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{2 x^2}-\frac {a^2 c \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 {6 a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i a^2 c \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 a^2 c \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 a^2 c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i a^2 c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (6 i a^2 c \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 a^2 c \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 (3 a^2 c \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 )}{2 \sqrt {c+a^2 c x^2}}-\frac {\left (3 a^2 c \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 )}{2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {3 a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{2 x^2}-\frac {a^2 c \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 {6 a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {c+a^2 c x^2}}-\frac {3 i a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {c+a^2 c x^2}}+\frac {3 i a^2 c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i a^2 c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 a^2 c \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 a^2 c \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 (3 i a^2 c \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 (3 i a^2 c \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 a^2 c \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 a^2 c \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 {3 a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{2 x^2}-\frac {a^2 c \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 {6 a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {c+a^2 c x^2}}-\frac {3 i a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {c+a^2 c x^2}}+\frac {3 i a^2 c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i a^2 c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 a^2 c \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 {3 a^2 c \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 a^2 c \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 a^2 c \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 (3 a^2 c \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 (3 a^2 c \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 {3 a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{2 x^2}-\frac {a^2 c \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 {6 a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {c+a^2 c x^2}}-\frac {3 i a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {c+a^2 c x^2}}+\frac {3 i a^2 c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i a^2 c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 a^2 c \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 {3 a^2 c \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 a^2 c \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 a^2 c \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 {\left (3 i a^2 c \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 (3 i a^2 c \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 {3 a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{2 x^2}-\frac {a^2 c \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 {6 a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {c+a^2 c x^2}}-\frac {3 i a^2 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{2 \sqrt {c+a^2 c x^2}}+\frac {3 i a^2 c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i a^2 c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 a^2 c \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 {3 a^2 c \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 {3 i a^2 c \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 {3 i a^2 c \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 = 5.94, size = 345, normalized size = 0.57 \[ \frac {a^2 \sqrt {c \left (a^2 x^2+1\right )} \left (24 i \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{-i \tan ^{-1}(a x)}\right )+48 \tan ^{-1}(a x) \text {Li}_3\left (e^{-i \tan ^{-1}(a x)}\right )-48 \tan ^{-1}(a x) \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )+24 i \left (\tan ^{-1}(a x)^2+2\right ) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )-48 i \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )-48 i \text {Li}_4\left (e^{-i \tan ^{-1}(a x)}\right )-48 i \text {Li}_4\left (-e^{i \tan ^{-1}(a x)}\right )+2 i \tan ^{-1}(a x)^4-12 \tan \left (\frac {1}{2} \tan ^{-1}(a x)\right ) \tan ^{-1}(a x)^2+8 \tan ^{-1}(a x)^3 \log \left (1-e^{-i \tan ^{-1}(a x)}\right )-8 \tan ^{-1}(a x)^3 \log \left (1+e^{i \tan ^{-1}(a x)}\right )+48 \tan ^{-1}(a x) \log \left (1-e^{i \tan ^{-1}(a x)}\right )-48 \tan ^{-1}(a x) \log \left (1+e^{i \tan ^{-1}(a x)}\right )-12 \tan ^{-1}(a x)^2 \cot \left (\frac {1}{2} \tan ^{-1}(a x)\right )-2 \tan ^{-1}(a x)^3 \csc ^2\left (\frac {1}{2} \tan ^{-1}(a x)\right )+2 \tan ^{-1}(a x)^3 \sec ^2\left (\frac {1}{2} \tan ^{-1}(a x)\right )-i \pi ^4\right )}{16 \sqrt {a^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*Sqrt[c*(1 + a^2*x^2)]*((-I)*Pi^4 + (2*I)*ArcTan[a*x]^4 - 12*ArcTan[a*x]^2*Cot[ArcTan[a*x]/2] - 2*ArcTan[a
*x]^3*Csc[ArcTan[a*x]/2]^2 + 8*ArcTan[a*x]^3*Log[1 - E^((-I)*ArcTan[a*x])] + 48*ArcTan[a*x]*Log[1 - E^(I*ArcTa
n[a*x])] - 48*ArcTan[a*x]*Log[1 + E^(I*ArcTan[a*x])] - 8*ArcTan[a*x]^3*Log[1 + E^(I*ArcTan[a*x])] + (24*I)*Arc
Tan[a*x]^2*PolyLog[2, E^((-I)*ArcTan[a*x])] + (24*I)*(2 + ArcTan[a*x]^2)*PolyLog[2, -E^(I*ArcTan[a*x])] - (48*
I)*PolyLog[2, E^(I*ArcTan[a*x])] + 48*ArcTan[a*x]*PolyLog[3, E^((-I)*ArcTan[a*x])] - 48*ArcTan[a*x]*PolyLog[3,
-E^(I*ArcTan[a*x])] - (48*I)*PolyLog[4, E^((-I)*ArcTan[a*x])] - (48*I)*PolyLog[4, -E^(I*ArcTan[a*x])] + 2*Arc
Tan[a*x]^3*Sec[ArcTan[a*x]/2]^2 - 12*ArcTan[a*x]^2*Tan[ArcTan[a*x]/2]))/(16*Sqrt[1 + a^2*x^2])

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{3}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a^2*c*x^2 + c)*arctan(a*x)^3/x^3, 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(arctan(a*x)^3*(a^2*c*x^2+c)^(1/2)/x^3,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 [A]  time = 1.40, size = 404, normalized size = 0.67 \[ -\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \arctan \left (a x \right )^{2} \left (3 a x +\arctan \left (a x \right )\right )}{2 x^{2}}+\frac {i a^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i \arctan \left (a x \right )^{3} \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-i \arctan \left (a x \right )^{3} \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+6 i \arctan \left (a x \right ) \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+3 \arctan \left (a x \right )^{2} \polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+6 i \arctan \left (a x \right ) \polylog \left (3, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-6 i \arctan \left (a x \right ) \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-3 \arctan \left (a x \right )^{2} \polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-6 i \arctan \left (a x \right ) \polylog \left (3, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+6 \polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-6 \polylog \left (4, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-6 \polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+6 \polylog \left (4, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2 \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*(a*x-I)*(I+a*x))^(1/2)*arctan(a*x)^2*(3*a*x+arctan(a*x))/x^2+1/2*I*a^2*(c*(a*x-I)*(I+a*x))^(1/2)*(I*ar
ctan(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))+6*I*arctan(a*x
)*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))+3*arctan(a*x)^2*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*I*arctan(a*x)*po
lylog(3,-(1+I*a*x)/(a^2*x^2+1)^(1/2))-6*I*arctan(a*x)*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2))-3*arctan(a*x)^2*polylo
g(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*polylog(2,-(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))-6*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*polyl
og(4,(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{3}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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