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

Optimal. Leaf size=9 \[ \frac {\text {Chi}\left (\tanh ^{-1}(a x)\right )}{a} \]

[Out]

Chi(arctanh(a*x))/a

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Rubi [A]  time = 0.06, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5968, 3301} \[ \frac {\text {Chi}\left (\tanh ^{-1}(a x)\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

CoshIntegral[ArcTanh[a*x]]/a

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

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Mathematica [A]  time = 0.02, size = 9, normalized size = 1.00 \[ \frac {\text {Chi}\left (\tanh ^{-1}(a x)\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

CoshIntegral[ArcTanh[a*x]]/a

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {artanh}\left (a x\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.28, size = 10, normalized size = 1.11 \[ \frac {\Chi \left (\arctanh \left (a x \right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Chi(arctanh(a*x))/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.11 \[ \int \frac {1}{\mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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