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

Optimal. Leaf size=29 \[ -\frac {\log (1-a x)}{a^3}-\frac {x}{a^2}-\frac {x^2}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6150, 43} \[ -\frac {x}{a^2}-\frac {\log (1-a x)}{a^3}-\frac {x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a^2) - x^2/(2*a) - Log[1 - a*x]/a^3

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 29, normalized size = 1.00 \[ -\frac {\log (1-a x)}{a^3}-\frac {x}{a^2}-\frac {x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a^2) - x^2/(2*a) - Log[1 - a*x]/a^3

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fricas [A]  time = 0.54, size = 25, normalized size = 0.86 \[ -\frac {a^{2} x^{2} + 2 \, a x + 2 \, \log \left (a x - 1\right )}{2 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 27, normalized size = 0.93 \[ -\frac {a x^{2} + 2 \, x}{2 \, a^{2}} - \frac {\log \left ({\left | a x - 1 \right |}\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 27, normalized size = 0.93 \[ -\frac {x^{2}}{2 a}-\frac {x}{a^{2}}-\frac {\ln \left (a x -1\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^2/a-x/a^2-1/a^3*ln(a*x-1)

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maxima [A]  time = 0.32, size = 26, normalized size = 0.90 \[ -\frac {a x^{2} + 2 \, x}{2 \, a^{2}} - \frac {\log \left (a x - 1\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.87, size = 23, normalized size = 0.79 \[ -\frac {\ln \left (a\,x-1\right )+a\,x+\frac {a^2\,x^2}{2}}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2*(a*x + 1))/(a^2*x^2 - 1),x)

[Out]

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

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sympy [A]  time = 0.09, size = 22, normalized size = 0.76 \[ - \frac {x^{2}}{2 a} - \frac {x}{a^{2}} - \frac {\log {\left (a x - 1 \right )}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2/(2*a) - x/a**2 - log(a*x - 1)/a**3

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