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

Optimal. Leaf size=72 \[ \frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^2}-\frac{x}{2 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{a^2 x^2+1}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.112101, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5996, 6034, 5448, 12, 3298} \[ \frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^2}-\frac{x}{2 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{a^2 x^2+1}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5996

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*(x_))/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(x*(a + b*ArcTa
nh[c*x])^(p + 1))/(b*c*d*(p + 1)*(d + e*x^2)), x] + (Dist[4/(b^2*(p + 1)*(p + 2)), Int[(x*(a + b*ArcTanh[c*x])
^(p + 2))/(d + e*x^2)^2, x], x] + Simp[((1 + c^2*x^2)*(a + b*ArcTanh[c*x])^(p + 2))/(b^2*e*(p + 1)*(p + 2)*(d
+ e*x^2)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[p, -1] && NeQ[p, -2]

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

Mathematica [A]  time = 0.0614714, size = 66, normalized size = 0.92 \[ \frac{2 \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^2 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )+\left (a^2 x^2+1\right ) \tanh ^{-1}(a x)+a x}{2 a^2 \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.06, size = 43, normalized size = 0.6 \begin{align*}{\frac{1}{{a}^{2}} \left ( -{\frac{\sinh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{4\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\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/(-a^2*x^2+1)^2/arctanh(a*x)^3,x)

[Out]

1/a^2*(-1/4/arctanh(a*x)^2*sinh(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 \, a x +{\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) -{\left (a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )}{{\left (a^{4} x^{2} - a^{2}\right )} \log \left (a x + 1\right )^{2} - 2 \,{\left (a^{4} x^{2} - a^{2}\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right ) +{\left (a^{4} x^{2} - a^{2}\right )} \log \left (-a x + 1\right )^{2}} - 4 \, \int -\frac{x}{{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) -{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*x + (a^2*x^2 + 1)*log(a*x + 1) - (a^2*x^2 + 1)*log(-a*x + 1))/((a^4*x^2 - a^2)*log(a*x + 1)^2 - 2*(a^4*x^
2 - a^2)*log(a*x + 1)*log(-a*x + 1) + (a^4*x^2 - a^2)*log(-a*x + 1)^2) - 4*integrate(-x/((a^4*x^4 - 2*a^2*x^2
+ 1)*log(a*x + 1) - (a^4*x^4 - 2*a^2*x^2 + 1)*log(-a*x + 1)), x)

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Fricas [B]  time = 1.95548, size = 317, normalized size = 4.4 \begin{align*} \frac{{\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} + 4 \, a x + 2 \,{\left (a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{2 \,{\left (a^{4} x^{2} - a^{2}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname{atanh}^{3}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname{artanh}\left (a x\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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