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

Optimal. Leaf size=220 \[ \frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {6 i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {6 i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {6 \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^4}+\frac {\tanh ^{-1}(a x)^3}{a^4 \sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{a^4 \sqrt {1-a^2 x^2}}-\frac {6 x}{a^3 \sqrt {1-a^2 x^2}}-\frac {3 x \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}} \]

[Out]

-6*arctan((a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)^2/a^4+6*I*arctanh(a*x)*polylog(2,-I*(a*x+1)/(-a^2*x^2+1)^(1
/2))/a^4-6*I*arctanh(a*x)*polylog(2,I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^4-6*I*polylog(3,-I*(a*x+1)/(-a^2*x^2+1)^(1
/2))/a^4+6*I*polylog(3,I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^4-6*x/a^3/(-a^2*x^2+1)^(1/2)+6*arctanh(a*x)/a^4/(-a^2*x
^2+1)^(1/2)-3*x*arctanh(a*x)^2/a^3/(-a^2*x^2+1)^(1/2)+arctanh(a*x)^3/a^4/(-a^2*x^2+1)^(1/2)+arctanh(a*x)^3*(-a
^2*x^2+1)^(1/2)/a^4

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Rubi [A]  time = 0.41, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6028, 5994, 5952, 4180, 2531, 2282, 6589, 5962, 191} \[ \frac {6 i \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {6 i \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {6 i \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {6 i \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {6 x}{a^3 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^4}+\frac {\tanh ^{-1}(a x)^3}{a^4 \sqrt {1-a^2 x^2}}-\frac {3 x \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{a^4 \sqrt {1-a^2 x^2}}-\frac {6 \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*x)/(a^3*Sqrt[1 - a^2*x^2]) + (6*ArcTanh[a*x])/(a^4*Sqrt[1 - a^2*x^2]) - (3*x*ArcTanh[a*x]^2)/(a^3*Sqrt[1 -
a^2*x^2]) - (6*ArcTan[E^ArcTanh[a*x]]*ArcTanh[a*x]^2)/a^4 + ArcTanh[a*x]^3/(a^4*Sqrt[1 - a^2*x^2]) + (Sqrt[1
- a^2*x^2]*ArcTanh[a*x]^3)/a^4 + ((6*I)*ArcTanh[a*x]*PolyLog[2, (-I)*E^ArcTanh[a*x]])/a^4 - ((6*I)*ArcTanh[a*x
]*PolyLog[2, I*E^ArcTanh[a*x]])/a^4 - ((6*I)*PolyLog[3, (-I)*E^ArcTanh[a*x]])/a^4 + ((6*I)*PolyLog[3, I*E^ArcT
anh[a*x]])/a^4

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 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 5962

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> -Simp[(b*p*(a + b*ArcTa
nh[c*x])^(p - 1))/(c*d*Sqrt[d + e*x^2]), x] + (Dist[b^2*p*(p - 1), Int[(a + b*ArcTanh[c*x])^(p - 2)/(d + e*x^2
)^(3/2), x], x] + Simp[(x*(a + b*ArcTanh[c*x])^p)/(d*Sqrt[d + e*x^2]), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[c^2*d + e, 0] && GtQ[p, 1]

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 6028

Int[((a_.) + ArcTanh[(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*ArcTanh[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*A
rcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegersQ[p, 2*q] && LtQ[q, -1] &&
IGtQ[m, 1] && NeQ[p, -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}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=\frac {\int \frac {x \tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a^2}-\frac {\int \frac {x \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx}{a^2}\\ &=\frac {\tanh ^{-1}(a x)^3}{a^4 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^4}-\frac {3 \int \frac {\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a^3}-\frac {3 \int \frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{a^3}\\ &=\frac {6 \tanh ^{-1}(a x)}{a^4 \sqrt {1-a^2 x^2}}-\frac {3 x \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^3}{a^4 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^4}-\frac {3 \operatorname {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac {6 \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a^3}\\ &=-\frac {6 x}{a^3 \sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{a^4 \sqrt {1-a^2 x^2}}-\frac {3 x \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}-\frac {6 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}+\frac {\tanh ^{-1}(a x)^3}{a^4 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^4}+\frac {(6 i) \operatorname {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac {(6 i) \operatorname {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}\\ &=-\frac {6 x}{a^3 \sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{a^4 \sqrt {1-a^2 x^2}}-\frac {3 x \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}-\frac {6 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}+\frac {\tanh ^{-1}(a x)^3}{a^4 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^4}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {(6 i) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac {(6 i) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}\\ &=-\frac {6 x}{a^3 \sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{a^4 \sqrt {1-a^2 x^2}}-\frac {3 x \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}-\frac {6 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}+\frac {\tanh ^{-1}(a x)^3}{a^4 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^4}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}\\ &=-\frac {6 x}{a^3 \sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{a^4 \sqrt {1-a^2 x^2}}-\frac {3 x \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}-\frac {6 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}+\frac {\tanh ^{-1}(a x)^3}{a^4 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^4}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {6 i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {6 i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}\\ \end {align*}

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Mathematica [A]  time = 0.47, size = 249, normalized size = 1.13 \[ \frac {\frac {6 i \sqrt {1-a^2 x^2} \text {Li}_3\left (-i e^{-\tanh ^{-1}(a x)}\right )-6 i \sqrt {1-a^2 x^2} \text {Li}_3\left (i e^{-\tanh ^{-1}(a x)}\right )-a^2 x^2 \tanh ^{-1}(a x)^3+3 i \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2 \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-3 i \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-6 a x+2 \tanh ^{-1}(a x)^3-3 a x \tanh ^{-1}(a x)^2+6 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}+6 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-6 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )}{a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} x^{3} \operatorname {artanh}\left (a x\right )^{3}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*x^3*arctanh(a*x)^3/(a^4*x^4 - 2*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)^(3/2),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 = 0.51, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \arctanh \left (a x \right )^{3}}{\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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