3.5.87 \(\int \frac {1}{(1-a^2 x^2)^{3/2} \tanh ^{-1}(a x)^2} \, dx\) [487]

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

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6113, 6181, 3379} \begin {gather*} \frac {\text {Shi}\left (\tanh ^{-1}(a x)\right )}{a}-\frac {1}{a \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3379

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 6113

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 6181

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

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Mathematica [A]
time = 0.04, size = 32, normalized size = 0.91 \begin {gather*} \frac {-\frac {1}{\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}+\text {Shi}\left (\tanh ^{-1}(a x)\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.00, size = 62, normalized size = 1.77

method	result	size
default	\(\frac {\arctanh \left (a x \right ) \hyperbolicSineIntegral \left (\arctanh \left (a x \right )\right ) a^{2} x^{2}-\hyperbolicSineIntegral \left (\arctanh \left (a x \right )\right ) \arctanh \left (a x \right )+\sqrt {-a^{2} x^{2}+1}}{a \arctanh \left (a x \right ) \left (a^{2} x^{2}-1\right )}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)/((a^4*x^4 - 2*a^2*x^2 + 1)*arctanh(a*x)^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((-(a*x - 1)*(a*x + 1))**(3/2)*atanh(a*x)**2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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