3.305 \(\int \frac{\tanh ^{-1}(a x)}{(1-a^2 x^2)^3} \, dx\)

Optimal. Leaf size=94 \[ -\frac{3}{16 a \left (1-a^2 x^2\right )}-\frac{1}{16 a \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac{3 \tanh ^{-1}(a x)^2}{16 a} \]

[Out]

-1/(16*a*(1 - a^2*x^2)^2) - 3/(16*a*(1 - a^2*x^2)) + (x*ArcTanh[a*x])/(4*(1 - a^2*x^2)^2) + (3*x*ArcTanh[a*x])
/(8*(1 - a^2*x^2)) + (3*ArcTanh[a*x]^2)/(16*a)

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Rubi [A]  time = 0.0448151, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {5960, 5956, 261} \[ -\frac{3}{16 a \left (1-a^2 x^2\right )}-\frac{1}{16 a \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac{3 \tanh ^{-1}(a x)^2}{16 a} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(1 - a^2*x^2)^3,x]

[Out]

-1/(16*a*(1 - a^2*x^2)^2) - 3/(16*a*(1 - a^2*x^2)) + (x*ArcTanh[a*x])/(4*(1 - a^2*x^2)^2) + (3*x*ArcTanh[a*x])
/(8*(1 - a^2*x^2)) + (3*ArcTanh[a*x]^2)/(16*a)

Rule 5960

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(q + 1))
/(4*c*d*(q + 1)^2), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x], x] -
Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]))/(2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*
d + e, 0] && LtQ[q, -1] && NeQ[q, -3/2]

Rule 5956

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

Rule 261

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

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx &=-\frac{1}{16 a \left (1-a^2 x^2\right )^2}+\frac{x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac{3}{4} \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac{1}{16 a \left (1-a^2 x^2\right )^2}+\frac{x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac{3 \tanh ^{-1}(a x)^2}{16 a}-\frac{1}{8} (3 a) \int \frac{x}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac{1}{16 a \left (1-a^2 x^2\right )^2}-\frac{3}{16 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac{3 \tanh ^{-1}(a x)^2}{16 a}\\ \end{align*}

Mathematica [A]  time = 0.0634113, size = 65, normalized size = 0.69 \[ \frac{3 a^2 x^2+3 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2+\left (10 a x-6 a^3 x^3\right ) \tanh ^{-1}(a x)-4}{16 a \left (a^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a*x]/(1 - a^2*x^2)^3,x]

[Out]

(-4 + 3*a^2*x^2 + (10*a*x - 6*a^3*x^3)*ArcTanh[a*x] + 3*(-1 + a^2*x^2)^2*ArcTanh[a*x]^2)/(16*a*(-1 + a^2*x^2)^
2)

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Maple [B]  time = 0.061, size = 225, normalized size = 2.4 \begin{align*}{\frac{{\it Artanh} \left ( ax \right ) }{16\,a \left ( ax-1 \right ) ^{2}}}-{\frac{3\,{\it Artanh} \left ( ax \right ) }{16\,a \left ( ax-1 \right ) }}-{\frac{3\,{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{16\,a}}-{\frac{{\it Artanh} \left ( ax \right ) }{16\,a \left ( ax+1 \right ) ^{2}}}-{\frac{3\,{\it Artanh} \left ( ax \right ) }{16\,a \left ( ax+1 \right ) }}+{\frac{3\,{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{16\,a}}-{\frac{3\, \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{64\,a}}+{\frac{3\,\ln \left ( ax-1 \right ) }{32\,a}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{3\, \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{64\,a}}-{\frac{3}{32\,a}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{3\,\ln \left ( ax+1 \right ) }{32\,a}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }-{\frac{1}{64\,a \left ( ax-1 \right ) ^{2}}}+{\frac{7}{64\,a \left ( ax-1 \right ) }}-{\frac{1}{64\,a \left ( ax+1 \right ) ^{2}}}-{\frac{7}{64\,a \left ( ax+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)/(-a^2*x^2+1)^3,x)

[Out]

1/16/a*arctanh(a*x)/(a*x-1)^2-3/16/a*arctanh(a*x)/(a*x-1)-3/16/a*arctanh(a*x)*ln(a*x-1)-1/16/a*arctanh(a*x)/(a
*x+1)^2-3/16/a*arctanh(a*x)/(a*x+1)+3/16/a*arctanh(a*x)*ln(a*x+1)-3/64/a*ln(a*x-1)^2+3/32/a*ln(a*x-1)*ln(1/2+1
/2*a*x)-3/64/a*ln(a*x+1)^2-3/32/a*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)+3/32/a*ln(-1/2*a*x+1/2)*ln(a*x+1)-1/64/a/(a
*x-1)^2+7/64/a/(a*x-1)-1/64/a/(a*x+1)^2-7/64/a/(a*x+1)

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Maxima [B]  time = 0.979753, size = 246, normalized size = 2.62 \begin{align*} -\frac{1}{16} \,{\left (\frac{2 \,{\left (3 \, a^{2} x^{3} - 5 \, x\right )}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1} - \frac{3 \, \log \left (a x + 1\right )}{a} + \frac{3 \, \log \left (a x - 1\right )}{a}\right )} \operatorname{artanh}\left (a x\right ) + \frac{{\left (12 \, a^{2} x^{2} - 3 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 6 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 3 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} - 16\right )} a}{64 \,{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^3,x, algorithm="maxima")

[Out]

-1/16*(2*(3*a^2*x^3 - 5*x)/(a^4*x^4 - 2*a^2*x^2 + 1) - 3*log(a*x + 1)/a + 3*log(a*x - 1)/a)*arctanh(a*x) + 1/6
4*(12*a^2*x^2 - 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2 + 6*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)*log(a*x
- 1) - 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^2 - 16)*a/(a^6*x^4 - 2*a^4*x^2 + a^2)

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Fricas [A]  time = 2.02996, size = 213, normalized size = 2.27 \begin{align*} \frac{12 \, 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 (3 \, a^{3} x^{3} - 5 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - 16}{64 \,{\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^3,x, algorithm="fricas")

[Out]

1/64*(12*a^2*x^2 + 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^2 - 4*(3*a^3*x^3 - 5*a*x)*log(-(a*x +
1)/(a*x - 1)) - 16)/(a^5*x^4 - 2*a^3*x^2 + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\operatorname{atanh}{\left (a x \right )}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)/(-a**2*x**2+1)**3,x)

[Out]

-Integral(atanh(a*x)/(a**6*x**6 - 3*a**4*x**4 + 3*a**2*x**2 - 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\operatorname{artanh}\left (a x\right )}{{\left (a^{2} x^{2} - 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^3,x, algorithm="giac")

[Out]

integrate(-arctanh(a*x)/(a^2*x^2 - 1)^3, x)