3.4.80 \(\int \frac {x^3 \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx\) [380]

Optimal. Leaf size=219 \[ \frac {\text {ArcSin}(a x)}{a^4}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}+\frac {5 \text {ArcTan}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac {5 i \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {5 i \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {5 i \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {5 i \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a^4} \]

[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 = 0.36, 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 = {6163, 6141, 222, 6099, 4265, 2611, 2320, 6724} \begin {gather*} \frac {\text {ArcSin}(a x)}{a^4}+\frac {5 \tanh ^{-1}(a x)^2 \text {ArcTan}\left (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 \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 {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} \end {gather*}

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 222

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 2320

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 2611

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 4265

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 6099

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 6141

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 6163

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(-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 6724

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 {\text {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^4}+\frac {2 \text {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 \text {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac {i \text {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac {(4 i) \text {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac {(4 i) \text {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 \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac {i \text {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac {(4 i) \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac {(4 i) \text {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 \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {i \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac {(4 i) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac {(4 i) \text {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 = 0.62, size = 215, normalized size = 0.98 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (-3 a x \tanh ^{-1}(a x)^2+2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3-6 \tanh ^{-1}(a x) \left (1+\tanh ^{-1}(a x)^2\right )-\frac {3 i \left (4 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )\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 )+10 \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-10 \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )+10 \text {PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-10 \text {PolyLog}\left (3,i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}\right )}{6 a^4} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [F]
time = 0.85, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

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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