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

Optimal. Leaf size=41 \[ \frac{\text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{2 a}+\frac{\text{Chi}\left (4 \tanh ^{-1}(a x)\right )}{8 a}+\frac{3 \log \left (\tanh ^{-1}(a x)\right )}{8 a} \]

[Out]

CoshIntegral[2*ArcTanh[a*x]]/(2*a) + CoshIntegral[4*ArcTanh[a*x]]/(8*a) + (3*Log[ArcTanh[a*x]])/(8*a)

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Rubi [A]  time = 0.0831106, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {5968, 3312, 3301} \[ \frac{\text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{2 a}+\frac{\text{Chi}\left (4 \tanh ^{-1}(a x)\right )}{8 a}+\frac{3 \log \left (\tanh ^{-1}(a x)\right )}{8 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

CoshIntegral[2*ArcTanh[a*x]]/(2*a) + CoshIntegral[4*ArcTanh[a*x]]/(8*a) + (3*Log[ArcTanh[a*x]])/(8*a)

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])
)

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]

Rubi steps

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

Mathematica [A]  time = 0.0690022, size = 33, normalized size = 0.8 \[ -\frac{-4 \text{Chi}\left (2 \tanh ^{-1}(a x)\right )-\text{Chi}\left (4 \tanh ^{-1}(a x)\right )-3 \log \left (\tanh ^{-1}(a x)\right )}{8 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-4*CoshIntegral[2*ArcTanh[a*x]] - CoshIntegral[4*ArcTanh[a*x]] - 3*Log[ArcTanh[a*x]])/(8*a)

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Maple [A]  time = 0.066, size = 36, normalized size = 0.9 \begin{align*}{\frac{{\it Chi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{2\,a}}+{\frac{{\it Chi} \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{8\,a}}+{\frac{3\,\ln \left ({\it Artanh} \left ( ax \right ) \right ) }{8\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*Chi(2*arctanh(a*x))/a+1/8*Chi(4*arctanh(a*x))/a+3/8*ln(arctanh(a*x))/a

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

[Out]

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

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Fricas [B]  time = 2.27058, size = 325, normalized size = 7.93 \begin{align*} \frac{6 \, \log \left (\log \left (-\frac{a x + 1}{a x - 1}\right )\right ) + \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 ) + 4 \, \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) + 4 \, \logintegral \left (-\frac{a x - 1}{a x + 1}\right )}{16 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(6*log(log(-(a*x + 1)/(a*x - 1))) + log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) + log_integ
ral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x + 1)) + 4*log_integral(-(a*x + 1)/(a*x - 1)) + 4*log_integral(-(a*x
- 1)/(a*x + 1)))/a

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

[Out]

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

[Out]

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