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

Optimal. Leaf size=36 \[ \frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{a^2}-\frac {x}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6032, 6034, 3312, 3301, 5968} \[ \frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{a^2}-\frac {x}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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])

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 32, normalized size = 0.89 \[ \frac {\frac {a x}{\left (a^2 x^2-1\right ) \tanh ^{-1}(a x)}+\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.59, size = 106, normalized size = 2.94 \[ \frac {4 \, a x + {\left ({\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) + {\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{2 \, {\left (a^{4} x^{2} - a^{2}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a*x + ((a^2*x^2 - 1)*log_integral(-(a*x + 1)/(a*x - 1)) + (a^2*x^2 - 1)*log_integral(-(a*x - 1)/(a*x +
1)))*log(-(a*x + 1)/(a*x - 1)))/((a^4*x^2 - a^2)*log(-(a*x + 1)/(a*x - 1)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.20, size = 28, normalized size = 0.78 \[ \frac {-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{2 \arctanh \left (a x \right )}+\Chi \left (2 \arctanh \left (a x \right )\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, x}{{\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right ) - {\left (a^{3} x^{2} - a\right )} \log \left (-a x + 1\right )} - \int -\frac {2 \, {\left (a^{2} x^{2} + 1\right )}}{{\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right ) - {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (-a x + 1\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (a^2\,x^2-1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(atanh(a*x)^2*(a^2*x^2 - 1)^2),x)

[Out]

int(x/(atanh(a*x)^2*(a^2*x^2 - 1)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-a**2*x**2+1)**2/atanh(a*x)**2,x)

[Out]

Integral(x/((a*x - 1)**2*(a*x + 1)**2*atanh(a*x)**2), x)

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