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

Optimal. Leaf size=32 \[ \frac {\sin ^{-1}(a x)}{a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5994, 216} \[ \frac {\sin ^{-1}(a x)}{a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 5994

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)
^(q + 1)*(a + b*ArcTanh[c*x])^p)/(2*e*(q + 1)), x] + Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan
h[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0] && NeQ[q, -1]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 29, normalized size = 0.91 \[ \frac {\sin ^{-1}(a x)-\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 58, normalized size = 1.81 \[ -\frac {\sqrt {-a^{2} x^{2} + 1} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 4 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 47, normalized size = 1.47 \[ \frac {\arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{a {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1} \log \left (-\frac {a x + 1}{a x - 1}\right )}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.31, size = 81, normalized size = 2.53 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \arctanh \left (a x \right )}{a^{2}}+\frac {i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}+i\right )}{a^{2}}-\frac {i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-i\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 30, normalized size = 0.94 \[ -\frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )}{a^{2}} + \frac {\arcsin \left (a x\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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