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

Optimal. Leaf size=260 \[ \frac{6 i a \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 c x^2+c}}-\frac{6 i a \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 c x^2+c}}-\frac{6 a \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 c x^2+c}}+\frac{6 a \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 c x^2+c}}-\frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^3}{c x}-\frac{6 a \sqrt{a^2 x^2+1} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.372423, antiderivative size = 260, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {4944, 4958, 4956, 4183, 2531, 2282, 6589} \[ \frac{6 i a \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 c x^2+c}}-\frac{6 i a \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 c x^2+c}}-\frac{6 a \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 c x^2+c}}+\frac{6 a \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 c x^2+c}}-\frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^3}{c x}-\frac{6 a \sqrt{a^2 x^2+1} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.325929, size = 174, normalized size = 0.67 \[ -\frac{a \sqrt{a^2 x^2+1} \left (-6 i \tan ^{-1}(a x) \text{PolyLog}\left (2,-e^{i \tan ^{-1}(a x)}\right )+6 i \tan ^{-1}(a x) \text{PolyLog}\left (2,e^{i \tan ^{-1}(a x)}\right )+6 \text{PolyLog}\left (3,-e^{i \tan ^{-1}(a x)}\right )-6 \text{PolyLog}\left (3,e^{i \tan ^{-1}(a x)}\right )+\frac{\sqrt{a^2 x^2+1} \tan ^{-1}(a x)^3}{a x}-3 \tan ^{-1}(a x)^2 \log \left (1-e^{i \tan ^{-1}(a x)}\right )+3 \tan ^{-1}(a x)^2 \log \left (1+e^{i \tan ^{-1}(a x)}\right )\right )}{\sqrt{c \left (a^2 x^2+1\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.38, size = 230, normalized size = 0.9 \begin{align*} -{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{3}}{cx}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-3\,{\frac{a\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}{\sqrt{{a}^{2}{x}^{2}+1}c} \left ( \left ( \arctan \left ( ax \right ) \right ) ^{2}\ln \left ( 1+{\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) - \left ( \arctan \left ( ax \right ) \right ) ^{2}\ln \left ( 1-{\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) -2\,i\arctan \left ( ax \right ){\it polylog} \left ( 2,-{\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) +2\,i\arctan \left ( ax \right ){\it polylog} \left ( 2,{\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) +2\,{\it polylog} \left ( 3,-{\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) -2\,{\it polylog} \left ( 3,{\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(a*x)^3*(c*(a*x-I)*(a*x+I))^(1/2)/c/x-3*a*(arctan(a*x)^2*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))-arctan(a*x)^
2*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2))-2*I*arctan(a*x)*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+2*I*arctan(a*x)*po
lylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))+2*polylog(3,-(1+I*a*x)/(a^2*x^2+1)^(1/2))-2*polylog(3,(1+I*a*x)/(a^2*x^2+
1)^(1/2)))*(c*(a*x-I)*(a*x+I))^(1/2)/(a^2*x^2+1)^(1/2)/c

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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(arctan(a*x)^3/x^2/(a^2*c*x^2+c)^(1/2),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} \arctan \left (a x\right )^{3}}{a^{2} c x^{4} + c x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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