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

Optimal. Leaf size=29 \[ \frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{4 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^2} \]

[Out]

SinhIntegral[2*ArcTanh[a*x]]/(4*a^2) + SinhIntegral[4*ArcTanh[a*x]]/(8*a^2)

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Rubi [A]  time = 0.0941661, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6034, 5448, 3298} \[ \frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{4 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

SinhIntegral[2*ArcTanh[a*x]]/(4*a^2) + SinhIntegral[4*ArcTanh[a*x]]/(8*a^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 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 )^3 \tanh ^{-1}(a x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{\sinh (2 x)}{4 x}+\frac{\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^2}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^2}\\ &=\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{4 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^2}\\ \end{align*}

Mathematica [A]  time = 0.107331, size = 24, normalized size = 0.83 \[ \frac{2 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )+\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*SinhIntegral[2*ArcTanh[a*x]] + SinhIntegral[4*ArcTanh[a*x]])/(8*a^2)

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Maple [A]  time = 0.06, size = 24, normalized size = 0.8 \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{4}}+{\frac{{\it Shi} \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(1/4*Shi(2*arctanh(a*x))+1/8*Shi(4*arctanh(a*x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.56244, size = 281, normalized size = 9.69 \begin{align*} \frac{\logintegral \left (\frac{a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - \logintegral \left (\frac{a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + 2 \, \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) - 2 \, \logintegral \left (-\frac{a x - 1}{a x + 1}\right )}{16 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) - log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2
+ 2*a*x + 1)) + 2*log_integral(-(a*x + 1)/(a*x - 1)) - 2*log_integral(-(a*x - 1)/(a*x + 1)))/a^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x}{a^{6} x^{6} \operatorname{atanh}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname{atanh}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname{atanh}{\left (a x \right )} - \operatorname{atanh}{\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)**3/atanh(a*x),x)

[Out]

-Integral(x/(a**6*x**6*atanh(a*x) - 3*a**4*x**4*atanh(a*x) + 3*a**2*x**2*atanh(a*x) - atanh(a*x)), 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 )}^{3} \operatorname{artanh}\left (a x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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