3.5.81 \(\int \frac {1}{(1-a^2 x^2)^{3/2} \tanh ^{-1}(a x)} \, dx\) [481]

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 = 0.04, 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 = {6115, 3382} \begin {gather*} \frac {\text {Chi}\left (\tanh ^{-1}(a x)\right )}{a} \end {gather*}

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 3382

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 6115

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 {\text {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 = 0.08, size = 9, normalized size = 1.00 \begin {gather*} \frac {\text {Chi}\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]),x]

[Out]

CoshIntegral[ArcTanh[a*x]]/a

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Maple [A]
time = 0.00, size = 10, normalized size = 1.11

method	result	size
default	\(\frac {\hyperbolicCosineIntegral \left (\arctanh \left (a x \right )\right )}{a}\)	\(10\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Chi(arctanh(a*x))/a

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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),x, algorithm="maxima")

[Out]

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

[Out]

Integral(1/((-(a*x - 1)*(a*x + 1))**(3/2)*atanh(a*x)), 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),x, algorithm="giac")

[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 \begin {gather*} \int \frac {1}{\mathrm {atanh}\left (a\,x\right )\,{\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)*(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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