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

Optimal. Leaf size=82 \[ \frac {1}{4 a^2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^2}{2 a^2 \left (1-a^2 x^2\right )}-\frac {x \tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^2}{4 a^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5994, 5956, 261} \[ \frac {1}{4 a^2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^2}{2 a^2 \left (1-a^2 x^2\right )}-\frac {x \tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^2}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

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Mathematica [A]  time = 0.05, size = 43, normalized size = 0.52 \[ \frac {\left (a^2 x^2+1\right ) \tanh ^{-1}(a x)^2-2 a x \tanh ^{-1}(a x)+1}{4 a^2-4 a^4 x^2} \]

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.62, size = 66, normalized size = 0.80 \[ \frac {4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 4}{16 \, {\left (a^{4} x^{2} - a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(a*x)^2/(-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^4*x^2 - a^2)

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giac [A]  time = 0.17, size = 140, normalized size = 1.71 \[ -\frac {1}{32} \, {\left ({\left (\frac {a x + 1}{{\left (a x - 1\right )} a^{3}} + \frac {a x - 1}{{\left (a x + 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 2 \, {\left (\frac {a x + 1}{{\left (a x - 1\right )} a^{3}} - \frac {a x - 1}{{\left (a x + 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + \frac {2 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{3}} + \frac {2 \, {\left (a x - 1\right )}}{{\left (a x + 1\right )} a^{3}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(((a*x + 1)/((a*x - 1)*a^3) + (a*x - 1)/((a*x + 1)*a^3))*log(-(a*x + 1)/(a*x - 1))^2 - 2*((a*x + 1)/((a*
x - 1)*a^3) - (a*x - 1)/((a*x + 1)*a^3))*log(-(a*x + 1)/(a*x - 1)) + 2*(a*x + 1)/((a*x - 1)*a^3) + 2*(a*x - 1)
/((a*x + 1)*a^3))*a

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maple [B]  time = 0.06, size = 191, normalized size = 2.33 \[ -\frac {\arctanh \left (a x \right )^{2}}{2 a^{2} \left (a^{2} x^{2}-1\right )}+\frac {\arctanh \left (a x \right )}{4 a^{2} \left (a x -1\right )}+\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{4 a^{2}}+\frac {\arctanh \left (a x \right )}{4 a^{2} \left (a x +1\right )}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{4 a^{2}}+\frac {\ln \left (a x -1\right )^{2}}{16 a^{2}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8 a^{2}}+\frac {\ln \left (a x +1\right )^{2}}{16 a^{2}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8 a^{2}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{8 a^{2}}-\frac {1}{8 a^{2} \left (a x -1\right )}+\frac {1}{8 a^{2} \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.33, size = 146, normalized size = 1.78 \[ \frac {{\left (\frac {2 \, x}{a^{2} x^{2} - 1} - \frac {\log \left (a x + 1\right )}{a} + \frac {\log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right )}{4 \, a} + \frac {{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4}{16 \, {\left (a^{4} x^{2} - a^{2}\right )}} - \frac {\operatorname {artanh}\left (a x\right )^{2}}{2 \, {\left (a^{2} x^{2} - 1\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 1.16, size = 198, normalized size = 2.41 \[ \ln \left (1-a\,x\right )\,\left (\frac {\frac {x}{2}-\frac {1}{2\,a}}{4\,a-4\,a^3\,x^2}+\frac {\frac {x}{2}+\frac {1}{2\,a}}{4\,a-4\,a^3\,x^2}+\ln \left (a\,x+1\right )\,\left (\frac {1}{8\,a^2}+\frac {1}{2\,a^2\,\left (2\,a^2\,x^2-2\right )}\right )\right )-{\ln \left (1-a\,x\right )}^2\,\left (\frac {1}{16\,a^2}+\frac {1}{2\,a^2\,\left (4\,a^2\,x^2-4\right )}\right )-\frac {1}{2\,a^2\,\left (2\,a^2\,x^2-2\right )}-{\ln \left (a\,x+1\right )}^2\,\left (\frac {1}{8\,a^3\,\left (a\,x^2-\frac {1}{a}\right )}+\frac {1}{16\,a^2}\right )+\frac {x\,\ln \left (a\,x+1\right )}{4\,a^2\,\left (a\,x^2-\frac {1}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1 - a*x)*((x/2 - 1/(2*a))/(4*a - 4*a^3*x^2) + (x/2 + 1/(2*a))/(4*a - 4*a^3*x^2) + log(a*x + 1)*(1/(8*a^2)
+ 1/(2*a^2*(2*a^2*x^2 - 2)))) - log(1 - a*x)^2*(1/(16*a^2) + 1/(2*a^2*(4*a^2*x^2 - 4))) - 1/(2*a^2*(2*a^2*x^2
- 2)) - log(a*x + 1)^2*(1/(8*a^3*(a*x^2 - 1/a)) + 1/(16*a^2)) + (x*log(a*x + 1))/(4*a^2*(a*x^2 - 1/a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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