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

Optimal. Leaf size=95 \[ \frac {1-a^2 x^2}{12 a^2}+\frac {\log \left (1-a^2 x^2\right )}{6 a^2}-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}+\frac {x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{6 a}+\frac {x \tanh ^{-1}(a x)}{3 a} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5994, 5942, 5910, 260} \[ \frac {1-a^2 x^2}{12 a^2}+\frac {\log \left (1-a^2 x^2\right )}{6 a^2}-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}+\frac {x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{6 a}+\frac {x \tanh ^{-1}(a x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[x*(1 - a^2*x^2)*ArcTanh[a*x]^2,x]

[Out]

(1 - a^2*x^2)/(12*a^2) + (x*ArcTanh[a*x])/(3*a) + (x*(1 - a^2*x^2)*ArcTanh[a*x])/(6*a) - ((1 - a^2*x^2)^2*ArcT
anh[a*x]^2)/(4*a^2) + Log[1 - a^2*x^2]/(6*a^2)

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 5942

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(b*(d + e*x^2)^q)/(2*c*
q*(2*q + 1)), x] + (Dist[(2*d*q)/(2*q + 1), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTanh[c*x]), x], x] + Simp[(x*(d
+ e*x^2)^q*(a + b*ArcTanh[c*x]))/(2*q + 1), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[q, 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 ) \tanh ^{-1}(a x)^2 \, dx &=-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}+\frac {\int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx}{2 a}\\ &=\frac {1-a^2 x^2}{12 a^2}+\frac {x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{6 a}-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}+\frac {\int \tanh ^{-1}(a x) \, dx}{3 a}\\ &=\frac {1-a^2 x^2}{12 a^2}+\frac {x \tanh ^{-1}(a x)}{3 a}+\frac {x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{6 a}-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}-\frac {1}{3} \int \frac {x}{1-a^2 x^2} \, dx\\ &=\frac {1-a^2 x^2}{12 a^2}+\frac {x \tanh ^{-1}(a x)}{3 a}+\frac {x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{6 a}-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}+\frac {\log \left (1-a^2 x^2\right )}{6 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.69 \[ \frac {\left (6 a x-2 a^3 x^3\right ) \tanh ^{-1}(a x)-a^2 x^2+2 \log \left (1-a^2 x^2\right )-3 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2}{12 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 91, normalized size = 0.96 \[ -\frac {4 \, a^{2} x^{2} + 3 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 4 \, {\left (a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 8 \, \log \left (a^{2} x^{2} - 1\right )}{48 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(4*a^2*x^2 + 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^2 + 4*(a^3*x^3 - 3*a*x)*log(-(a*x + 1
)/(a*x - 1)) - 8*log(a^2*x^2 - 1))/a^2

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giac [B]  time = 0.18, size = 305, normalized size = 3.21 \[ -\frac {1}{3} \, a {\left (\frac {{\left (\frac {3 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{\frac {{\left (a x + 1\right )}^{3} a^{3}}{{\left (a x - 1\right )}^{3}} - \frac {3 \, {\left (a x + 1\right )}^{2} a^{3}}{{\left (a x - 1\right )}^{2}} + \frac {3 \, {\left (a x + 1\right )} a^{3}}{a x - 1} - a^{3}} + \frac {3 \, {\left (a x + 1\right )}^{2} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}}{{\left (\frac {{\left (a x + 1\right )}^{4} a^{3}}{{\left (a x - 1\right )}^{4}} - \frac {4 \, {\left (a x + 1\right )}^{3} a^{3}}{{\left (a x - 1\right )}^{3}} + \frac {6 \, {\left (a x + 1\right )}^{2} a^{3}}{{\left (a x - 1\right )}^{2}} - \frac {4 \, {\left (a x + 1\right )} a^{3}}{a x - 1} + a^{3}\right )} {\left (a x - 1\right )}^{2}} + \frac {a x + 1}{{\left (\frac {{\left (a x + 1\right )}^{2} a^{3}}{{\left (a x - 1\right )}^{2}} - \frac {2 \, {\left (a x + 1\right )} a^{3}}{a x - 1} + a^{3}\right )} {\left (a x - 1\right )}} + \frac {\log \left (-\frac {a x + 1}{a x - 1} + 1\right )}{a^{3}} - \frac {\log \left (-\frac {a x + 1}{a x - 1}\right )}{a^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 185, normalized size = 1.95 \[ -\frac {a^{2} \arctanh \left (a x \right )^{2} x^{4}}{4}+\frac {\arctanh \left (a x \right )^{2} x^{2}}{2}-\frac {a \arctanh \left (a x \right ) x^{3}}{6}+\frac {x \arctanh \left (a x \right )}{2 a}+\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{4 a^{2}}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{4 a^{2}}+\frac {\ln \left (a x -1\right )^{2}}{16 a^{2}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8 a^{2}}+\frac {\ln \left (a x +1\right )^{2}}{16 a^{2}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{8 a^{2}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8 a^{2}}-\frac {x^{2}}{12}+\frac {\ln \left (a x -1\right )}{6 a^{2}}+\frac {\ln \left (a x +1\right )}{6 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*arctanh(a*x)^2*x^4+1/2*arctanh(a*x)^2*x^2-1/6*a*arctanh(a*x)*x^3+1/2*x*arctanh(a*x)/a+1/4/a^2*arctanh
(a*x)*ln(a*x-1)-1/4/a^2*arctanh(a*x)*ln(a*x+1)+1/16/a^2*ln(a*x-1)^2-1/8/a^2*ln(a*x-1)*ln(1/2+1/2*a*x)+1/16/a^2
*ln(a*x+1)^2-1/8/a^2*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/8/a^2*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/12*x^2+1/6/a^2*ln(a
*x-1)+1/6/a^2*ln(a*x+1)

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maxima [A]  time = 0.31, size = 74, normalized size = 0.78 \[ -\frac {{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{2}}{4 \, a^{2}} - \frac {{\left (x^{2} - \frac {2 \, \log \left (a x + 1\right )}{a^{2}} - \frac {2 \, \log \left (a x - 1\right )}{a^{2}}\right )} a + 2 \, {\left (a^{2} x^{3} - 3 \, x\right )} \operatorname {artanh}\left (a x\right )}{12 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.90, size = 77, normalized size = 0.81 \[ \frac {x^2\,{\mathrm {atanh}\left (a\,x\right )}^2}{2}-\frac {{\mathrm {atanh}\left (a\,x\right )}^2}{4\,a^2}-\frac {x^2}{12}+\frac {\ln \left (a^2\,x^2-1\right )}{6\,a^2}+\frac {x\,\mathrm {atanh}\left (a\,x\right )}{2\,a}-\frac {a\,x^3\,\mathrm {atanh}\left (a\,x\right )}{6}-\frac {a^2\,x^4\,{\mathrm {atanh}\left (a\,x\right )}^2}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x*atanh(a*x)^2*(a^2*x^2 - 1),x)

[Out]

(x^2*atanh(a*x)^2)/2 - atanh(a*x)^2/(4*a^2) - x^2/12 + log(a^2*x^2 - 1)/(6*a^2) + (x*atanh(a*x))/(2*a) - (a*x^
3*atanh(a*x))/6 - (a^2*x^4*atanh(a*x)^2)/4

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sympy [A]  time = 1.28, size = 88, normalized size = 0.93 \[ \begin {cases} - \frac {a^{2} x^{4} \operatorname {atanh}^{2}{\left (a x \right )}}{4} - \frac {a x^{3} \operatorname {atanh}{\left (a x \right )}}{6} + \frac {x^{2} \operatorname {atanh}^{2}{\left (a x \right )}}{2} - \frac {x^{2}}{12} + \frac {x \operatorname {atanh}{\left (a x \right )}}{2 a} + \frac {\log {\left (x - \frac {1}{a} \right )}}{3 a^{2}} - \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{4 a^{2}} + \frac {\operatorname {atanh}{\left (a x \right )}}{3 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*x**4*atanh(a*x)**2/4 - a*x**3*atanh(a*x)/6 + x**2*atanh(a*x)**2/2 - x**2/12 + x*atanh(a*x)/(2
*a) + log(x - 1/a)/(3*a**2) - atanh(a*x)**2/(4*a**2) + atanh(a*x)/(3*a**2), Ne(a, 0)), (0, True))

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