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

Optimal. Leaf size=38 \[ \frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^3}-\frac{x^2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)} \]

[Out]

-(x^2/(a*(1 - a^2*x^2)*ArcTanh[a*x])) + SinhIntegral[2*ArcTanh[a*x]]/a^3

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Rubi [A]  time = 0.133315, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {6006, 6034, 5448, 12, 3298} \[ \frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^3}-\frac{x^2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^2/(a*(1 - a^2*x^2)*ArcTanh[a*x])) + SinhIntegral[2*ArcTanh[a*x]]/a^3

Rule 6006

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^m*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^(p + 1))/(b*c*d*(p + 1)), x] - Dist[(f*m)/(b*c*(p + 1)), In
t[(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] && Eq
Q[c^2*d + e, 0] && EqQ[m + 2*q + 2, 0] && LtQ[p, -1]

Rule 6034

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

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{x^2}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx &=-\frac{x^2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{2 \int \frac{x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx}{a}\\ &=-\frac{x^2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{2 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=-\frac{x^2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{2 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=-\frac{x^2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=-\frac{x^2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^3}\\ \end{align*}

Mathematica [A]  time = 0.139602, size = 36, normalized size = 0.95 \[ \frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^3}+\frac{x^2}{a \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^2/(a*(-1 + a^2*x^2)*ArcTanh[a*x]) + SinhIntegral[2*ArcTanh[a*x]]/a^3

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Maple [A]  time = 0.063, size = 36, normalized size = 1. \begin{align*}{\frac{1}{{a}^{3}} \left ({\frac{1}{2\,{\it Artanh} \left ( ax \right ) }}-{\frac{\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{2\,{\it Artanh} \left ( ax \right ) }}+{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^3*(1/2/arctanh(a*x)-1/2/arctanh(a*x)*cosh(2*arctanh(a*x))+Shi(2*arctanh(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \, x^{2}}{{\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right ) -{\left (a^{3} x^{2} - a\right )} \log \left (-a x + 1\right )} - 4 \, \int -\frac{x}{{\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right ) -{\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (-a x + 1\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x^2/((a^3*x^2 - a)*log(a*x + 1) - (a^3*x^2 - a)*log(-a*x + 1)) - 4*integrate(-x/((a^5*x^4 - 2*a^3*x^2 + a)*l
og(a*x + 1) - (a^5*x^4 - 2*a^3*x^2 + a)*log(-a*x + 1)), x)

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Fricas [B]  time = 2.13977, size = 258, normalized size = 6.79 \begin{align*} \frac{4 \, a^{2} x^{2} +{\left ({\left (a^{2} x^{2} - 1\right )} \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) -{\left (a^{2} x^{2} - 1\right )} \logintegral \left (-\frac{a x - 1}{a x + 1}\right )\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{2 \,{\left (a^{5} x^{2} - a^{3}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a^2*x^2 + ((a^2*x^2 - 1)*log_integral(-(a*x + 1)/(a*x - 1)) - (a^2*x^2 - 1)*log_integral(-(a*x - 1)/(a*
x + 1)))*log(-(a*x + 1)/(a*x - 1)))/((a^5*x^2 - a^3)*log(-(a*x + 1)/(a*x - 1)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname{artanh}\left (a x\right )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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