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

Optimal. Leaf size=50 \[ \frac {a^3 x^5}{30}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}{6 a^2}-\frac {a x^3}{9}+\frac {x}{6 a} \]

[Out]

1/6*x/a-1/9*a*x^3+1/30*a^3*x^5-1/6*(-a^2*x^2+1)^3*arctanh(a*x)/a^2

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Rubi [A]  time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5994, 194} \[ \frac {a^3 x^5}{30}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}{6 a^2}-\frac {a x^3}{9}+\frac {x}{6 a} \]

Antiderivative was successfully verified.

[In]

Int[x*(1 - a^2*x^2)^2*ArcTanh[a*x],x]

[Out]

x/(6*a) - (a*x^3)/9 + (a^3*x^5)/30 - ((1 - a^2*x^2)^3*ArcTanh[a*x])/(6*a^2)

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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 )^2 \tanh ^{-1}(a x) \, dx &=-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}{6 a^2}+\frac {\int \left (1-a^2 x^2\right )^2 \, dx}{6 a}\\ &=-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}{6 a^2}+\frac {\int \left (1-2 a^2 x^2+a^4 x^4\right ) \, dx}{6 a}\\ &=\frac {x}{6 a}-\frac {a x^3}{9}+\frac {a^3 x^5}{30}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}{6 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 93, normalized size = 1.86 \[ \frac {1}{6} a^4 x^6 \tanh ^{-1}(a x)+\frac {a^3 x^5}{30}-\frac {1}{2} a^2 x^4 \tanh ^{-1}(a x)+\frac {\log (1-a x)}{12 a^2}-\frac {\log (a x+1)}{12 a^2}-\frac {a x^3}{9}+\frac {1}{2} x^2 \tanh ^{-1}(a x)+\frac {x}{6 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(6*a) - (a*x^3)/9 + (a^3*x^5)/30 + (x^2*ArcTanh[a*x])/2 - (a^2*x^4*ArcTanh[a*x])/2 + (a^4*x^6*ArcTanh[a*x])/
6 + Log[1 - a*x]/(12*a^2) - Log[1 + a*x]/(12*a^2)

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fricas [A]  time = 0.60, size = 68, normalized size = 1.36 \[ \frac {6 \, a^{5} x^{5} - 20 \, a^{3} x^{3} + 30 \, a x + 15 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{180 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/180*(6*a^5*x^5 - 20*a^3*x^3 + 30*a*x + 15*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1)))/a
^2

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giac [B]  time = 0.18, size = 176, normalized size = 3.52 \[ \frac {8}{45} \, a {\left (\frac {\frac {10 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - \frac {5 \, {\left (a x + 1\right )}}{a x - 1} + 1}{a^{3} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{5}} + \frac {30 \, {\left (a x + 1\right )}^{3} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{{\left (a x - 1\right )}^{3} a^{3} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{6}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/45*a*((10*(a*x + 1)^2/(a*x - 1)^2 - 5*(a*x + 1)/(a*x - 1) + 1)/(a^3*((a*x + 1)/(a*x - 1) - 1)^5) + 30*(a*x +
1)^3*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)^3*a^3*((a*x + 1)/(a*x - 1) - 1)^6))

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maple [A]  time = 0.03, size = 77, normalized size = 1.54 \[ \frac {a^{4} \arctanh \left (a x \right ) x^{6}}{6}-\frac {a^{2} \arctanh \left (a x \right ) x^{4}}{2}+\frac {\arctanh \left (a x \right ) x^{2}}{2}+\frac {a^{3} x^{5}}{30}-\frac {x^{3} a}{9}+\frac {x}{6 a}+\frac {\ln \left (a x -1\right )}{12 a^{2}}-\frac {\ln \left (a x +1\right )}{12 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a^4*arctanh(a*x)*x^6-1/2*a^2*arctanh(a*x)*x^4+1/2*arctanh(a*x)*x^2+1/30*a^3*x^5-1/9*x^3*a+1/6*x/a+1/12/a^2
*ln(a*x-1)-1/12/a^2*ln(a*x+1)

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maxima [A]  time = 0.30, size = 46, normalized size = 0.92 \[ \frac {{\left (a^{2} x^{2} - 1\right )}^{3} \operatorname {artanh}\left (a x\right )}{6 \, a^{2}} + \frac {3 \, a^{4} x^{5} - 10 \, a^{2} x^{3} + 15 \, x}{90 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(a^2*x^2 - 1)^3*arctanh(a*x)/a^2 + 1/90*(3*a^4*x^5 - 10*a^2*x^3 + 15*x)/a

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mupad [B]  time = 0.92, size = 64, normalized size = 1.28 \[ \frac {x^2\,\mathrm {atanh}\left (a\,x\right )}{2}-\frac {\frac {\mathrm {atanh}\left (a\,x\right )}{6}-\frac {a\,x}{6}}{a^2}-\frac {a\,x^3}{9}+\frac {a^3\,x^5}{30}-\frac {a^2\,x^4\,\mathrm {atanh}\left (a\,x\right )}{2}+\frac {a^4\,x^6\,\mathrm {atanh}\left (a\,x\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(a*x)*(a^2*x^2 - 1)^2,x)

[Out]

(x^2*atanh(a*x))/2 - (atanh(a*x)/6 - (a*x)/6)/a^2 - (a*x^3)/9 + (a^3*x^5)/30 - (a^2*x^4*atanh(a*x))/2 + (a^4*x
^6*atanh(a*x))/6

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sympy [A]  time = 1.52, size = 68, normalized size = 1.36 \[ \begin {cases} \frac {a^{4} x^{6} \operatorname {atanh}{\left (a x \right )}}{6} + \frac {a^{3} x^{5}}{30} - \frac {a^{2} x^{4} \operatorname {atanh}{\left (a x \right )}}{2} - \frac {a x^{3}}{9} + \frac {x^{2} \operatorname {atanh}{\left (a x \right )}}{2} + \frac {x}{6 a} - \frac {\operatorname {atanh}{\left (a x \right )}}{6 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x**6*atanh(a*x)/6 + a**3*x**5/30 - a**2*x**4*atanh(a*x)/2 - a*x**3/9 + x**2*atanh(a*x)/2 + x/(
6*a) - atanh(a*x)/(6*a**2), Ne(a, 0)), (0, True))

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