3.164 \(\int x (1-a^2 x^2) \tanh ^{-1}(a x) \, dx\)

Optimal. Leaf size=40 \[ -\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 a^2}-\frac {a x^3}{12}+\frac {x}{4 a} \]

[Out]

1/4*x/a-1/12*a*x^3-1/4*(-a^2*x^2+1)^2*arctanh(a*x)/a^2

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Rubi [A]  time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {5994} \[ -\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 a^2}-\frac {a x^3}{12}+\frac {x}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[x*(1 - a^2*x^2)*ArcTanh[a*x],x]

[Out]

x/(4*a) - (a*x^3)/12 - ((1 - a^2*x^2)^2*ArcTanh[a*x])/(4*a^2)

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 x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx &=-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 a^2}+\frac {\int \left (1-a^2 x^2\right ) \, dx}{4 a}\\ &=\frac {x}{4 a}-\frac {a x^3}{12}-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 69, normalized size = 1.72 \[ -\frac {1}{4} a^2 x^4 \tanh ^{-1}(a x)+\frac {\log (1-a x)}{8 a^2}-\frac {\log (a x+1)}{8 a^2}-\frac {a x^3}{12}+\frac {1}{2} x^2 \tanh ^{-1}(a x)+\frac {x}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(4*a) - (a*x^3)/12 + (x^2*ArcTanh[a*x])/2 - (a^2*x^4*ArcTanh[a*x])/4 + Log[1 - a*x]/(8*a^2) - Log[1 + a*x]/(
8*a^2)

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fricas [A]  time = 0.58, size = 52, normalized size = 1.30 \[ -\frac {2 \, a^{3} x^{3} - 6 \, a x + 3 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{24 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(2*a^3*x^3 - 6*a*x + 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1)))/a^2

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giac [B]  time = 0.29, size = 160, normalized size = 4.00 \[ -\frac {1}{3} \, a {\left (\frac {\frac {3 \, {\left (a x + 1\right )}}{a x - 1} - 1}{a^{3} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{3}} + \frac {6 \, {\left (a x + 1\right )}^{2} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{{\left (a x - 1\right )}^{2} a^{3} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*((3*(a*x + 1)/(a*x - 1) - 1)/(a^3*((a*x + 1)/(a*x - 1) - 1)^3) + 6*(a*x + 1)^2*log(-(a*((a*x + 1)/(a*x
- 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) + 1)/(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) - 1))/((a*
x - 1)^2*a^3*((a*x + 1)/(a*x - 1) - 1)^4))

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maple [A]  time = 0.03, size = 57, normalized size = 1.42 \[ -\frac {a^{2} \arctanh \left (a x \right ) x^{4}}{4}+\frac {\arctanh \left (a x \right ) x^{2}}{2}-\frac {x^{3} a}{12}+\frac {x}{4 a}+\frac {\ln \left (a x -1\right )}{8 a^{2}}-\frac {\ln \left (a x +1\right )}{8 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*arctanh(a*x)*x^4+1/2*arctanh(a*x)*x^2-1/12*x^3*a+1/4*x/a+1/8/a^2*ln(a*x-1)-1/8/a^2*ln(a*x+1)

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maxima [A]  time = 0.30, size = 37, normalized size = 0.92 \[ -\frac {{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )}{4 \, a^{2}} - \frac {a^{2} x^{3} - 3 \, x}{12 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^2*x^2 - 1)^2*arctanh(a*x)/a^2 - 1/12*(a^2*x^3 - 3*x)/a

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mupad [B]  time = 0.86, size = 44, normalized size = 1.10 \[ \frac {x^2\,\mathrm {atanh}\left (a\,x\right )}{2}-\frac {\frac {\mathrm {atanh}\left (a\,x\right )}{4}-\frac {a\,x}{4}}{a^2}-\frac {a\,x^3}{12}-\frac {a^2\,x^4\,\mathrm {atanh}\left (a\,x\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x*atanh(a*x)*(a^2*x^2 - 1),x)

[Out]

(x^2*atanh(a*x))/2 - (atanh(a*x)/4 - (a*x)/4)/a^2 - (a*x^3)/12 - (a^2*x^4*atanh(a*x))/4

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sympy [A]  time = 0.85, size = 46, normalized size = 1.15 \[ \begin {cases} - \frac {a^{2} x^{4} \operatorname {atanh}{\left (a x \right )}}{4} - \frac {a x^{3}}{12} + \frac {x^{2} \operatorname {atanh}{\left (a x \right )}}{2} + \frac {x}{4 a} - \frac {\operatorname {atanh}{\left (a x \right )}}{4 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*x**4*atanh(a*x)/4 - a*x**3/12 + x**2*atanh(a*x)/2 + x/(4*a) - atanh(a*x)/(4*a**2), Ne(a, 0)),
(0, True))

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