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

Optimal. Leaf size=283 \[ -\frac{6 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{6 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{6 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}-\frac{6 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^3}{a^2 c}+\frac{6 i \sqrt{a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^2 \sqrt{a^2 c x^2+c}} \]

[Out]

((6*I)*Sqrt[1 + a^2*x^2]*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^2)/(a^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[c + a^2*c*
x^2]*ArcTan[a*x]^3)/(a^2*c) - ((6*I)*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/(a^2*Sq
rt[c + a^2*c*x^2]) + ((6*I)*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]) + (6*Sqrt[1 + a^2*x^2]*PolyLog[3, (-I)*E^(I*ArcTan[a*x])])/(a^2*Sqrt[c + a^2*c*x^2]) - (6*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 = 0.233566, antiderivative size = 283, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {4930, 4890, 4888, 4181, 2531, 2282, 6589} \[ -\frac{6 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{6 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{6 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}-\frac{6 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt{a^2 c x^2+c}}+\frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^3}{a^2 c}+\frac{6 i \sqrt{a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^2 \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin{align*} \int \frac{x \tan ^{-1}(a x)^3}{\sqrt{c+a^2 c x^2}} \, dx &=\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^2 c}-\frac{3 \int \frac{\tan ^{-1}(a x)^2}{\sqrt{c+a^2 c x^2}} \, dx}{a}\\ &=\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^2 c}-\frac{\left (3 \sqrt{1+a^2 x^2}\right ) \int \frac{\tan ^{-1}(a x)^2}{\sqrt{1+a^2 x^2}} \, dx}{a \sqrt{c+a^2 c x^2}}\\ &=\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^2 c}-\frac{\left (3 \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int x^2 \sec (x) \, dx,x,\tan ^{-1}(a x)\right )}{a^2 \sqrt{c+a^2 c x^2}}\\ &=\frac{6 i \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{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^2 c}+\frac{\left (6 \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 (6 \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{6 i \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{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^2 c}-\frac{6 i \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{6 i \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 (6 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^2 \sqrt{c+a^2 c x^2}}-\frac{\left (6 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^2 \sqrt{c+a^2 c x^2}}\\ &=\frac{6 i \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{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^2 c}-\frac{6 i \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{6 i \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 (6 \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 (6 \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{6 i \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{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^3}{a^2 c}-\frac{6 i \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{6 i \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{6 \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{6 \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 = 0.275091, size = 168, normalized size = 0.59 \[ \frac{\sqrt{c \left (a^2 x^2+1\right )} \left (\tan ^{-1}(a x)^3-\frac{3 \left (2 i \tan ^{-1}(a x) \left (\text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )-\text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )\right )-2 \text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )+2 \text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )+\tan ^{-1}(a x)^2 \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right )\right )}{\sqrt{a^2 x^2+1}}\right )}{a^2 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 1.898, size = 0, normalized size = 0. \begin{align*} \int{x \left ( \arctan \left ( ax \right ) \right ) ^{3}{\frac{1}{\sqrt{{a}^{2}c{x}^{2}+c}}}}\, dx \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]

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

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

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

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

[Out]

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