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

Optimal. Leaf size=58 \[ -\frac{x}{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{1}{2 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}+\frac{\text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.23446, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {5966, 6032, 6034, 3312, 3301, 5968} \[ -\frac{x}{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{1}{2 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}+\frac{\text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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])
)

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])

Rubi steps

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

Mathematica [A]  time = 0.0731106, size = 58, normalized size = 1. \[ \frac{2 \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^2 \text{Chi}\left (2 \tanh ^{-1}(a x)\right )+2 a x \tanh ^{-1}(a x)+1}{2 a \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.064, size = 51, normalized size = 0.9 \begin{align*}{\frac{1}{a} \left ( -{\frac{1}{4\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{4\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{\sinh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{2\,{\it Artanh} \left ( ax \right ) }}+{\it Chi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

2*(a*x*log(a*x + 1) - a*x*log(-a*x + 1) + 1)/((a^3*x^2 - a)*log(a*x + 1)^2 - 2*(a^3*x^2 - a)*log(a*x + 1)*log(
-a*x + 1) + (a^3*x^2 - a)*log(-a*x + 1)^2) - integrate(-2*(a^2*x^2 + 1)/((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 = 1.96564, size = 296, normalized size = 5.1 \begin{align*} \frac{4 \, a x \log \left (-\frac{a x + 1}{a x - 1}\right ) +{\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}{2 \,{\left (a^{3} x^{2} - a\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(1/(-a^2*x^2+1)^2/arctanh(a*x)^3,x, algorithm="fricas")

[Out]

1/2*(4*a*x*log(-(a*x + 1)/(a*x - 1)) + ((a^2*x^2 - 1)*log_integral(-(a*x + 1)/(a*x - 1)) + (a^2*x^2 - 1)*log_i
ntegral(-(a*x - 1)/(a*x + 1)))*log(-(a*x + 1)/(a*x - 1))^2 + 4)/((a^3*x^2 - a)*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{1}{\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(1/(-a**2*x**2+1)**2/atanh(a*x)**3,x)

[Out]

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

[Out]

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