3.4.91 \(\int \frac {\tanh ^{-1}(a x)}{(1-a^2 x^2)^{3/2}} \, dx\) [391]

Optimal. Leaf size=40 \[ -\frac {1}{a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {6105} \begin {gather*} \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {1}{a \sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6105

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-b/(c*d*Sqrt[d + e*x^2]
), x] + Simp[x*((a + b*ArcTanh[c*x])/(d*Sqrt[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0
]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 27, normalized size = 0.68 \begin {gather*} \frac {-1+a x \tanh ^{-1}(a x)}{a \sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.10, size = 38, normalized size = 0.95

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (a x \arctanh \left (a x \right )-1\right )}{a \left (a^{2} x^{2}-1\right )}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 36, normalized size = 0.90 \begin {gather*} \frac {x \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}} - \frac {1}{\sqrt {-a^{2} x^{2} + 1} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 47, normalized size = 1.18 \begin {gather*} -\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - 2\right )}}{2 \, {\left (a^{3} x^{2} - a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 59, normalized size = 1.48 \begin {gather*} -\frac {\sqrt {-a^{2} x^{2} + 1} x \log \left (-\frac {a x + 1}{a x - 1}\right )}{2 \, {\left (a^{2} x^{2} - 1\right )}} - \frac {1}{\sqrt {-a^{2} x^{2} + 1} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\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(atanh(a*x)/(1 - a^2*x^2)^(3/2),x)

[Out]

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

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