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

Optimal. Leaf size=54 \[ -\frac {1}{4 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^2}{4 a} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {5956, 261} \[ -\frac {1}{4 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^2}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(1 - a^2*x^2)^2,x]

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 44, normalized size = 0.81 \[ \frac {\left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^2-2 a x \tanh ^{-1}(a x)+1}{4 a \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 64, normalized size = 1.19 \[ -\frac {4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 4}{16 \, {\left (a^{3} x^{2} - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 2.01, size = 255, normalized size = 4.72 \[ \frac {1}{8} \, a^{2} {\left (\frac {{\left (a x - 1\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{a - \frac {a {\left (\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1\right )}}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}} - 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{a - \frac {a {\left (\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1\right )}}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}} + 1}\right )}{{\left (a x + 1\right )} a^{4}} + \frac {a x - 1}{{\left (a x + 1\right )} a^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 169, normalized size = 3.13 \[ -\frac {\arctanh \left (a x \right )}{4 a \left (a x -1\right )}-\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{4 a}-\frac {\arctanh \left (a x \right )}{4 a \left (a x +1\right )}+\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{4 a}-\frac {\ln \left (a x -1\right )^{2}}{16 a}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8 a}-\frac {\ln \left (a x +1\right )^{2}}{16 a}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{8 a}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8 a}+\frac {1}{8 a \left (a x -1\right )}-\frac {1}{8 a \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.31, size = 122, normalized size = 2.26 \[ -\frac {1}{4} \, {\left (\frac {2 \, x}{a^{2} x^{2} - 1} - \frac {\log \left (a x + 1\right )}{a} + \frac {\log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right ) - \frac {{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4\right )} a}{16 \, {\left (a^{4} x^{2} - a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*x/(a^2*x^2 - 1) - log(a*x + 1)/a + log(a*x - 1)/a)*arctanh(a*x) - 1/16*((a^2*x^2 - 1)*log(a*x + 1)^2 -
2*(a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) + (a^2*x^2 - 1)*log(a*x - 1)^2 - 4)*a/(a^4*x^2 - a^2)

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mupad [B]  time = 0.97, size = 106, normalized size = 1.96 \[ \frac {{\ln \left (a\,x+1\right )}^2}{16\,a}-\ln \left (1-a\,x\right )\,\left (\frac {\ln \left (a\,x+1\right )}{8\,a}-\frac {x}{2\,\left (2\,a^2\,x^2-2\right )}\right )+\frac {{\ln \left (1-a\,x\right )}^2}{16\,a}+\frac {1}{2\,a\,\left (2\,a^2\,x^2-2\right )}-\frac {x\,\ln \left (a\,x+1\right )}{4\,a\,\left (a\,x^2-\frac {1}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)/(a^2*x^2 - 1)^2,x)

[Out]

log(a*x + 1)^2/(16*a) - log(1 - a*x)*(log(a*x + 1)/(8*a) - x/(2*(2*a^2*x^2 - 2))) + log(1 - a*x)^2/(16*a) + 1/
(2*a*(2*a^2*x^2 - 2)) - (x*log(a*x + 1))/(4*a*(a*x^2 - 1/a))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (a x \right )}}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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