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

Optimal. Leaf size=35 \[ \frac {\text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{a}-\frac {1}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5966, 6034, 5448, 12, 3298} \[ \frac {\text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{a}-\frac {1}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 30, normalized size = 0.86 \[ \frac {\frac {1}{\left (a^2 x^2-1\right ) \tanh ^{-1}(a x)}+\text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.59, size = 102, normalized size = 2.91 \[ \frac {{\left ({\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 4}{2 \, {\left (a^{3} x^{2} - a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(((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)) + 4)/((a^3*x^2 - a)*log(-(a*x + 1)/(a*x - 1)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.18, size = 36, normalized size = 1.03 \[ \frac {-\frac {1}{2 \arctanh \left (a x \right )}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{2 \arctanh \left (a x \right )}+\Shi \left (2 \arctanh \left (a x \right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-1/2/arctanh(a*x)-1/2/arctanh(a*x)*cosh(2*arctanh(a*x))+Shi(2*arctanh(a*x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a*integrate(-x/((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) + 2/(
(a^3*x^2 - a)*log(a*x + 1) - (a^3*x^2 - a)*log(-a*x + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(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 \[ \int \frac {1}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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