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

Optimal. Leaf size=68 \[ \frac{\text{Shi}\left (\tanh ^{-1}(a x)\right )}{2 a^2}-\frac{x}{2 a \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}-\frac{1}{2 a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.19596, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6006, 5966, 6034, 3298} \[ \frac{\text{Shi}\left (\tanh ^{-1}(a x)\right )}{2 a^2}-\frac{x}{2 a \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}-\frac{1}{2 a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5966

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((d + e*x^2)^(q + 1
)*(a + b*ArcTanh[c*x])^(p + 1))/(b*c*d*(p + 1)), x] + Dist[(2*c*(q + 1))/(b*(p + 1)), Int[x*(d + e*x^2)^q*(a +
b*ArcTanh[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && 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 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 )^{3/2} \tanh ^{-1}(a x)^3} \, dx &=-\frac{x}{2 a \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}+\frac{\int \frac{1}{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)^2} \, dx}{2 a}\\ &=-\frac{x}{2 a \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}-\frac{1}{2 a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}+\frac{1}{2} \int \frac{x}{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)} \, dx\\ &=-\frac{x}{2 a \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}-\frac{1}{2 a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^2}\\ &=-\frac{x}{2 a \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}-\frac{1}{2 a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}+\frac{\text{Shi}\left (\tanh ^{-1}(a x)\right )}{2 a^2}\\ \end{align*}

Mathematica [A]  time = 0.122611, size = 43, normalized size = 0.63 \[ \frac{\text{Shi}\left (\tanh ^{-1}(a x)\right )-\frac{a x+\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.248, size = 154, normalized size = 2.3 \begin{align*}{\frac{1}{4\,{a}^{2} \left ( ax-1 \right ) \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{1}{4\,{a}^{2} \left ( ax-1 \right ){\it Artanh} \left ( ax \right ) }\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}-{\frac{{\it Ei} \left ( 1,-{\it Artanh} \left ( ax \right ) \right ) }{4\,{a}^{2}}}+{\frac{1}{4\,{a}^{2} \left ( ax+1 \right ) \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}-{\frac{1}{4\,{a}^{2} \left ( ax+1 \right ){\it Artanh} \left ( ax \right ) }\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{{\it Ei} \left ( 1,{\it Artanh} \left ( ax \right ) \right ) }{4\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-(a*x-1)*(a*x+1))^(1/2)/a^2/(a*x-1)/arctanh(a*x)^2+1/4*(-(a*x-1)*(a*x+1))^(1/2)/a^2/(a*x-1)/arctanh(a*x)-
1/4*Ei(1,-arctanh(a*x))/a^2+1/4*(-(a*x-1)*(a*x+1))^(1/2)/a^2/(a*x+1)/arctanh(a*x)^2-1/4*(-(a*x-1)*(a*x+1))^(1/
2)/a^2/(a*x+1)/arctanh(a*x)+1/4*Ei(1,arctanh(a*x))/a^2

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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 )}^{\frac{3}{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)^(3/2)/arctanh(a*x)^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a^{2} x^{2} + 1} x}{{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname{artanh}\left (a x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*x/((a^4*x^4 - 2*a^2*x^2 + 1)*arctanh(a*x)^3), x)

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

[Out]

Integral(x/((-(a*x - 1)*(a*x + 1))**(3/2)*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 )}^{\frac{3}{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)^(3/2)/arctanh(a*x)^3,x, algorithm="giac")

[Out]

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