3.380 \(\int \frac {x^3 \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx\)

Optimal. Leaf size=219 \[ -\frac {5 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {5 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {5 i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {5 i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {\sin ^{-1}(a x)}{a^4}+\frac {5 \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3} \]

[Out]

arcsin(a*x)/a^4+5*arctan((a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)^2/a^4-5*I*arctanh(a*x)*polylog(2,-I*(a*x+1)/
(-a^2*x^2+1)^(1/2))/a^4+5*I*arctanh(a*x)*polylog(2,I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^4+5*I*polylog(3,-I*(a*x+1)/
(-a^2*x^2+1)^(1/2))/a^4-5*I*polylog(3,I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^4-arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^4-1/
2*x*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2)/a^3-2/3*arctanh(a*x)^3*(-a^2*x^2+1)^(1/2)/a^4-1/3*x^2*arctanh(a*x)^3*(-a
^2*x^2+1)^(1/2)/a^2

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Rubi [A]  time = 0.57, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6016, 5994, 216, 5952, 4180, 2531, 2282, 6589} \[ -\frac {5 i \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {5 i \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {5 i \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {5 i \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}+\frac {\sin ^{-1}(a x)}{a^4}+\frac {5 \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTanh[a*x]^3)/Sqrt[1 - a^2*x^2],x]

[Out]

ArcSin[a*x]/a^4 - (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/a^4 - (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(2*a^3) + (5*Arc
Tan[E^ArcTanh[a*x]]*ArcTanh[a*x]^2)/a^4 - (2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^3)/(3*a^4) - (x^2*Sqrt[1 - a^2*x^2
]*ArcTanh[a*x]^3)/(3*a^2) - ((5*I)*ArcTanh[a*x]*PolyLog[2, (-I)*E^ArcTanh[a*x]])/a^4 + ((5*I)*ArcTanh[a*x]*Pol
yLog[2, I*E^ArcTanh[a*x]])/a^4 + ((5*I)*PolyLog[3, (-I)*E^ArcTanh[a*x]])/a^4 - ((5*I)*PolyLog[3, I*E^ArcTanh[a
*x]])/a^4

Rule 216

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

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 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 5952

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c*Sqrt[d]), Subs
t[Int[(a + b*x)^p*Sech[x], x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[
p, 0] && GtQ[d, 0]

Rule 5994

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

Rule 6016

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

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^3 \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx &=-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}+\frac {2 \int \frac {x \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx}{3 a^2}+\frac {\int \frac {x^2 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{a}\\ &=-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}+\frac {\int \frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{2 a^3}+\frac {2 \int \frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{a^3}+\frac {\int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{a^2}\\ &=-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}+\frac {\operatorname {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^4}+\frac {2 \operatorname {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{a^3}\\ &=\frac {\sin ^{-1}(a x)}{a^4}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}+\frac {5 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac {i \operatorname {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac {i \operatorname {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac {(4 i) \operatorname {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac {(4 i) \operatorname {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}\\ &=\frac {\sin ^{-1}(a x)}{a^4}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}+\frac {5 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac {5 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {5 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {i \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac {i \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac {(4 i) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac {(4 i) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}\\ &=\frac {\sin ^{-1}(a x)}{a^4}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}+\frac {5 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac {5 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {5 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {(4 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {(4 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}\\ &=\frac {\sin ^{-1}(a x)}{a^4}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}+\frac {5 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac {5 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {5 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {5 i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {5 i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 215, normalized size = 0.98 \[ \frac {\sqrt {1-a^2 x^2} \left (-\frac {3 i \left (10 \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-10 \tanh ^{-1}(a x) \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )+10 \text {Li}_3\left (-i e^{-\tanh ^{-1}(a x)}\right )-10 \text {Li}_3\left (i e^{-\tanh ^{-1}(a x)}\right )+5 \tanh ^{-1}(a x)^2 \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-5 \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+4 i \tan ^{-1}\left (\tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right )\right )}{\sqrt {1-a^2 x^2}}+2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3-3 a x \tanh ^{-1}(a x)^2-6 \left (\tanh ^{-1}(a x)^2+1\right ) \tanh ^{-1}(a x)\right )}{6 a^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*ArcTanh[a*x]^3)/Sqrt[1 - a^2*x^2],x]

[Out]

(Sqrt[1 - a^2*x^2]*(-3*a*x*ArcTanh[a*x]^2 + 2*(1 - a^2*x^2)*ArcTanh[a*x]^3 - 6*ArcTanh[a*x]*(1 + ArcTanh[a*x]^
2) - ((3*I)*((4*I)*ArcTan[Tanh[ArcTanh[a*x]/2]] + 5*ArcTanh[a*x]^2*Log[1 - I/E^ArcTanh[a*x]] - 5*ArcTanh[a*x]^
2*Log[1 + I/E^ArcTanh[a*x]] + 10*ArcTanh[a*x]*PolyLog[2, (-I)/E^ArcTanh[a*x]] - 10*ArcTanh[a*x]*PolyLog[2, I/E
^ArcTanh[a*x]] + 10*PolyLog[3, (-I)/E^ArcTanh[a*x]] - 10*PolyLog[3, I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2]))/(6
*a^4)

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} x^{3} \operatorname {artanh}\left (a x\right )^{3}}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*x^2 + 1)*x^3*arctanh(a*x)^3/(a^2*x^2 - 1), 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(x^3*arctanh(a*x)^3/(-a^2*x^2+1)^(1/2),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 [F]  time = 0.50, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \arctanh \left (a x \right )^{3}}{\sqrt {-a^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctanh(a*x)^3/(-a^2*x^2+1)^(1/2),x)

[Out]

int(x^3*arctanh(a*x)^3/(-a^2*x^2+1)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arctanh(a*x)^3/sqrt(-a^2*x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*atanh(a*x)^3)/(1 - a^2*x^2)^(1/2),x)

[Out]

int((x^3*atanh(a*x)^3)/(1 - a^2*x^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atanh(a*x)**3/(-a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**3*atanh(a*x)**3/sqrt(-(a*x - 1)*(a*x + 1)), x)

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