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

Optimal. Leaf size=197 \[ -\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}+2 a \text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-2 a \text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+2 i a \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )-2 i a \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )-2 i a \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )+2 i a \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )-4 a \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)-2 a \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.38, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6014, 6008, 6018, 5952, 4180, 2531, 2282, 6589} \[ 2 a \text {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-2 a \text {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+2 i a \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )-2 i a \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )-2 i a \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )+2 i a \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}-4 a \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)-2 a \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/x) - 2*a*ArcTan[E^ArcTanh[a*x]]*ArcTanh[a*x]^2 - 4*a*ArcTanh[a*x]*ArcTanh
[Sqrt[1 - a*x]/Sqrt[1 + a*x]] + (2*I)*a*ArcTanh[a*x]*PolyLog[2, (-I)*E^ArcTanh[a*x]] - (2*I)*a*ArcTanh[a*x]*Po
lyLog[2, I*E^ArcTanh[a*x]] + 2*a*PolyLog[2, -(Sqrt[1 - a*x]/Sqrt[1 + a*x])] - 2*a*PolyLog[2, Sqrt[1 - a*x]/Sqr
t[1 + a*x]] - (2*I)*a*PolyLog[3, (-I)*E^ArcTanh[a*x]] + (2*I)*a*PolyLog[3, I*E^ArcTanh[a*x]]

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 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 6014

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 6018

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.78, size = 223, normalized size = 1.13 \[ a \left (-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{a x}+2 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-2 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )+2 \text {Li}_2\left (-e^{-\tanh ^{-1}(a x)}\right )-2 \text {Li}_2\left (e^{-\tanh ^{-1}(a x)}\right )+2 i \text {Li}_3\left (-i e^{-\tanh ^{-1}(a x)}\right )-2 i \text {Li}_3\left (i e^{-\tanh ^{-1}(a x)}\right )+i \tanh ^{-1}(a x)^2 \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-i \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}(a x) \log \left (e^{-\tanh ^{-1}(a x)}+1\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{2}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

________________________________________________________________________________________

maple [F]  time = 0.48, size = 0, normalized size = 0.00 \[ \int \frac {\arctanh \left (a x \right )^{2} \sqrt {-a^{2} x^{2}+1}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^2\,\sqrt {1-a^2\,x^2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________