3.5.40 \(\int x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2 \, dx\) [440]

Optimal. Leaf size=254 \[ \frac {x \sqrt {1-a^2 x^2}}{12 a^2}-\frac {\text {ArcSin}(a x)}{6 a^3}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\text {ArcTan}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{4 a^3}-\frac {i \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac {i \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac {i \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}-\frac {i \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{4 a^3} \]

[Out]

-1/6*arcsin(a*x)/a^3+1/4*arctan((a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)^2/a^3-1/4*I*arctanh(a*x)*polylog(2,-I
*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3+1/4*I*arctanh(a*x)*polylog(2,I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3+1/4*I*polylog(
3,-I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3-1/4*I*polylog(3,I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3+1/12*x*(-a^2*x^2+1)^(1/
2)/a^2+1/12*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^3+1/6*x^2*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a-1/8*x*arctanh(a*x)^2
*(-a^2*x^2+1)^(1/2)/a^2+1/4*x^3*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.59, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 29, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6161, 6163, 6141, 222, 6099, 4265, 2611, 2320, 6724, 327} \begin {gather*} -\frac {\text {ArcSin}(a x)}{6 a^3}+\frac {\tanh ^{-1}(a x)^2 \text {ArcTan}\left (e^{\tanh ^{-1}(a x)}\right )}{4 a^3}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac {i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}-\frac {i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac {x \sqrt {1-a^2 x^2}}{12 a^2}+\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2,x]

[Out]

(x*Sqrt[1 - a^2*x^2])/(12*a^2) - ArcSin[a*x]/(6*a^3) + (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(12*a^3) + (x^2*Sqrt[1
- a^2*x^2]*ArcTanh[a*x])/(6*a) - (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(8*a^2) + (x^3*Sqrt[1 - a^2*x^2]*ArcTan
h[a*x]^2)/4 + (ArcTan[E^ArcTanh[a*x]]*ArcTanh[a*x]^2)/(4*a^3) - ((I/4)*ArcTanh[a*x]*PolyLog[2, (-I)*E^ArcTanh[
a*x]])/a^3 + ((I/4)*ArcTanh[a*x]*PolyLog[2, I*E^ArcTanh[a*x]])/a^3 + ((I/4)*PolyLog[3, (-I)*E^ArcTanh[a*x]])/a
^3 - ((I/4)*PolyLog[3, I*E^ArcTanh[a*x]])/a^3

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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 6161

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

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 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2 \, dx &=-\left (a^2 \int \frac {x^4 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\right )+\int \frac {x^2 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^2}+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac {3}{4} \int \frac {x^2 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx+\frac {\int \frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{2 a^2}+\frac {\int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{a}-\frac {1}{2} a \int \frac {x^3 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^3}+\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac {1}{6} \int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx+\frac {\text {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^3}-\frac {3 \int \frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{8 a^2}+\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{a^2}-\frac {\int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{3 a}-\frac {3 \int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{4 a}\\ &=\frac {x \sqrt {1-a^2 x^2}}{12 a^2}+\frac {\sin ^{-1}(a x)}{a^3}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^3}-\frac {i \text {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}+\frac {i \text {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}-\frac {3 \text {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^3}-\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{12 a^2}-\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{3 a^2}-\frac {3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{4 a^2}\\ &=\frac {x \sqrt {1-a^2 x^2}}{12 a^2}-\frac {\sin ^{-1}(a x)}{6 a^3}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{4 a^3}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {(3 i) \text {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^3}-\frac {(3 i) \text {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^3}+\frac {i \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}-\frac {i \text {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=\frac {x \sqrt {1-a^2 x^2}}{12 a^2}-\frac {\sin ^{-1}(a x)}{6 a^3}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{4 a^3}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}-\frac {(3 i) \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^3}+\frac {(3 i) \text {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^3}+\frac {i \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {i \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}\\ &=\frac {x \sqrt {1-a^2 x^2}}{12 a^2}-\frac {\sin ^{-1}(a x)}{6 a^3}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{4 a^3}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac {i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {(3 i) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac {(3 i) \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{4 a^3}\\ &=\frac {x \sqrt {1-a^2 x^2}}{12 a^2}-\frac {\sin ^{-1}(a x)}{6 a^3}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{4 a^3}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac {i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}-\frac {i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}\\ \end {align*}

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Mathematica [A]
time = 0.77, size = 228, normalized size = 0.90 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (6 \tanh ^{-1}(a x)-4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)-6 a x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+a x \left (2+3 \tanh ^{-1}(a x)^2\right )-\frac {i \left (-8 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right )+3 \tanh ^{-1}(a x)^2 \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-3 \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )+6 \text {PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-6 \text {PolyLog}\left (3,i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}\right )}{24 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2,x]

[Out]

(Sqrt[1 - a^2*x^2]*(6*ArcTanh[a*x] - 4*(1 - a^2*x^2)*ArcTanh[a*x] - 6*a*x*(1 - a^2*x^2)*ArcTanh[a*x]^2 + a*x*(
2 + 3*ArcTanh[a*x]^2) - (I*((-8*I)*ArcTan[Tanh[ArcTanh[a*x]/2]] + 3*ArcTanh[a*x]^2*Log[1 - I/E^ArcTanh[a*x]] -
3*ArcTanh[a*x]^2*Log[1 + I/E^ArcTanh[a*x]] + 6*ArcTanh[a*x]*PolyLog[2, (-I)/E^ArcTanh[a*x]] - 6*ArcTanh[a*x]*
PolyLog[2, I/E^ArcTanh[a*x]] + 6*PolyLog[3, (-I)/E^ArcTanh[a*x]] - 6*PolyLog[3, I/E^ArcTanh[a*x]]))/Sqrt[1 - a
^2*x^2]))/(24*a^3)

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Maple [F]
time = 0.73, size = 0, normalized size = 0.00 \[\int x^{2} \arctanh \left (a x \right )^{2} \sqrt {-a^{2} x^{2}+1}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*x^2*arctanh(a*x)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}^{2}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [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^2*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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