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

Optimal. Leaf size=53 \[ \frac{\text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac{\text{Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^2}-\frac{x}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]

[Out]

-(x/(a*(1 - a^2*x^2)^2*ArcTanh[a*x])) + CoshIntegral[2*ArcTanh[a*x]]/(2*a^2) + CoshIntegral[4*ArcTanh[a*x]]/(2
*a^2)

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Rubi [A]  time = 0.245151, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {6032, 6034, 5448, 3301, 5968, 3312} \[ \frac{\text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac{\text{Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^2}-\frac{x}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/(a*(1 - a^2*x^2)^2*ArcTanh[a*x])) + CoshIntegral[2*ArcTanh[a*x]]/(2*a^2) + CoshIntegral[4*ArcTanh[a*x]]/(2
*a^2)

Rule 6032

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

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 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 5968

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

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps

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

Mathematica [A]  time = 0.0994367, size = 75, normalized size = 1.42 \[ \frac{\left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x) \text{Chi}\left (2 \tanh ^{-1}(a x)\right )+\left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x) \text{Chi}\left (4 \tanh ^{-1}(a x)\right )-2 a x}{2 a^2 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*x + (-1 + a^2*x^2)^2*ArcTanh[a*x]*CoshIntegral[2*ArcTanh[a*x]] + (-1 + a^2*x^2)^2*ArcTanh[a*x]*CoshInteg
ral[4*ArcTanh[a*x]])/(2*a^2*(-1 + a^2*x^2)^2*ArcTanh[a*x])

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Maple [A]  time = 0.062, size = 54, normalized size = 1. \begin{align*}{\frac{1}{{a}^{2}} \left ( -{\frac{\sinh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{4\,{\it Artanh} \left ( ax \right ) }}+{\frac{{\it Chi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{2}}-{\frac{\sinh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{8\,{\it Artanh} \left ( ax \right ) }}+{\frac{{\it Chi} \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{2}} \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)^2,x)

[Out]

1/a^2*(-1/4/arctanh(a*x)*sinh(2*arctanh(a*x))+1/2*Chi(2*arctanh(a*x))-1/8/arctanh(a*x)*sinh(4*arctanh(a*x))+1/
2*Chi(4*arctanh(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, 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 )} + \int -\frac{2 \,{\left (3 \, a^{2} x^{2} + 1\right )}}{{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (a x + 1\right ) -{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, 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/(-a^2*x^2+1)^3/arctanh(a*x)^2,x, algorithm="maxima")

[Out]

-2*x/((a^5*x^4 - 2*a^3*x^2 + a)*log(a*x + 1) - (a^5*x^4 - 2*a^3*x^2 + a)*log(-a*x + 1)) + integrate(-2*(3*a^2*
x^2 + 1)/((a^7*x^6 - 3*a^5*x^4 + 3*a^3*x^2 - a)*log(a*x + 1) - (a^7*x^6 - 3*a^5*x^4 + 3*a^3*x^2 - a)*log(-a*x
+ 1)), x)

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Fricas [B]  time = 2.37035, size = 535, normalized size = 10.09 \begin{align*} -\frac{8 \, a x -{\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (\frac{a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) +{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (\frac{a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) +{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) +{\left (a^{4} x^{4} - 2 \, 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 )}{4 \,{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\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/(-a^2*x^2+1)^3/arctanh(a*x)^2,x, algorithm="fricas")

[Out]

-1/4*(8*a*x - ((a^4*x^4 - 2*a^2*x^2 + 1)*log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) + (a^4*x^4
- 2*a^2*x^2 + 1)*log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x + 1)) + (a^4*x^4 - 2*a^2*x^2 + 1)*log_int
egral(-(a*x + 1)/(a*x - 1)) + (a^4*x^4 - 2*a^2*x^2 + 1)*log_integral(-(a*x - 1)/(a*x + 1)))*log(-(a*x + 1)/(a*
x - 1)))/((a^6*x^4 - 2*a^4*x^2 + a^2)*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}{a^{6} x^{6} \operatorname{atanh}^{2}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname{atanh}^{2}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname{atanh}^{2}{\left (a x \right )} - \operatorname{atanh}^{2}{\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)**2,x)

[Out]

-Integral(x/(a**6*x**6*atanh(a*x)**2 - 3*a**4*x**4*atanh(a*x)**2 + 3*a**2*x**2*atanh(a*x)**2 - atanh(a*x)**2),
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 )^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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