3.269 \(\int \frac{\tanh ^{-1}(a x)^2}{(1-a^2 x^2)^2} \, dx\)
Optimal. Leaf size=88 \[ \frac{x}{4 \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{6 a}+\frac{\tanh ^{-1}(a x)}{4 a} \]
[Out]
x/(4*(1 - a^2*x^2)) + ArcTanh[a*x]/(4*a) - ArcTanh[a*x]/(2*a*(1 - a^2*x^2)) + (x*ArcTanh[a*x]^2)/(2*(1 - a^2*x
^2)) + ArcTanh[a*x]^3/(6*a)
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Rubi [A] time = 0.0639225, antiderivative size = 88, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used =
{5956, 5994, 199, 206} \[ \frac{x}{4 \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{6 a}+\frac{\tanh ^{-1}(a x)}{4 a} \]
Antiderivative was successfully verified.
[In]
Int[ArcTanh[a*x]^2/(1 - a^2*x^2)^2,x]
[Out]
x/(4*(1 - a^2*x^2)) + ArcTanh[a*x]/(4*a) - ArcTanh[a*x]/(2*a*(1 - a^2*x^2)) + (x*ArcTanh[a*x]^2)/(2*(1 - a^2*x
^2)) + ArcTanh[a*x]^3/(6*a)
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]
Rule 199
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx &=\frac{x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{6 a}-a \int \frac{x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac{\tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{6 a}+\frac{1}{2} \int \frac{1}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac{x}{4 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{6 a}+\frac{1}{4} \int \frac{1}{1-a^2 x^2} \, dx\\ &=\frac{x}{4 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)}{4 a}-\frac{\tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{6 a}\\ \end{align*}
Mathematica [A] time = 0.11065, size = 93, normalized size = 1.06 \[ \frac{-3 \left (\left (a^2 x^2-1\right ) \log (1-a x)+\left (1-a^2 x^2\right ) \log (a x+1)+2 a x\right )+4 \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^3-12 a x \tanh ^{-1}(a x)^2+12 \tanh ^{-1}(a x)}{24 a \left (a^2 x^2-1\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[ArcTanh[a*x]^2/(1 - a^2*x^2)^2,x]
[Out]
(12*ArcTanh[a*x] - 12*a*x*ArcTanh[a*x]^2 + 4*(-1 + a^2*x^2)*ArcTanh[a*x]^3 - 3*(2*a*x + (-1 + a^2*x^2)*Log[1 -
a*x] + (1 - a^2*x^2)*Log[1 + a*x]))/(24*a*(-1 + a^2*x^2))
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Maple [C] time = 0.413, size = 1695, normalized size = 19.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(arctanh(a*x)^2/(-a^2*x^2+1)^2,x)
[Out]
-1/4/(a*x-1)/(a*x+1)*x-1/4/a*arctanh(a*x)^2/(a*x-1)-1/4/a*arctanh(a*x)^2/(a*x+1)-1/4*I*a/(a*x-1)/(a*x+1)*arcta
nh(a*x)^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^2*Pi*x^2+1/8*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a
*x+1)^2/(-a^2*x^2+1)+1))^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*arctanh(a*x)^2*Pi-1/8*I/a/(a*x-1)/(a*x+1)*csgn(I
*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*arctanh(a*x)^2*Pi+1/8*I/a/(
a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*arctanh(a*x)^2*Pi+1/4*I/a/(a
*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)^2*Pi+1/4*I*a/(a*
x-1)/(a*x+1)*arctanh(a*x)^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3*Pi*x^2-1/8*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2
*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3*Pi*x^2-1/8*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*csgn
(I*(a*x+1)^2/(a^2*x^2-1))^3*Pi*x^2-1/4*I/a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*Pi+1/4*a/(a*x-1)/(a*x+1)*arctanh(a*x
)*x^2+1/6*a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*x^2+1/8*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2
/(-a^2*x^2+1)+1))^3*arctanh(a*x)^2*Pi+1/8*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*arctanh(a*x)^2*P
i+1/4*I/a/(a*x-1)/(a*x+1)*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x)^2*Pi+1/4*I*a/(a*x-1)/(a*x+1)*arcta
nh(a*x)^2*Pi*x^2-1/4*I/a/(a*x-1)/(a*x+1)*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^2*Pi-1/8*I*a/(a*x-1
)/(a*x+1)*arctanh(a*x)^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+
1)+1))^2*Pi*x^2+1/8*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2
-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*Pi*x^2-1/8*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1
/2))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*Pi*x^2-1/4*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*csgn(I*(a*x+1)/(-a^2*x^2+1)
^(1/2))*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*Pi*x^2-1/8*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2
/(-a^2*x^2+1)+1))*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*arctanh(a*x)^2*Pi-1/6/a/(a*
x-1)/(a*x+1)*arctanh(a*x)^3+1/4/a/(a*x-1)/(a*x+1)*arctanh(a*x)-1/4/a*arctanh(a*x)^2*ln(a*x-1)+1/4/a*arctanh(a*
x)^2*ln(a*x+1)-1/2/a*arctanh(a*x)^2*ln((a*x+1)/(-a^2*x^2+1)^(1/2))+1/8*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*csgn
(I/((a*x+1)^2/(-a^2*x^2+1)+1))*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2
+1)+1))*Pi*x^2
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Maxima [B] time = 0.986465, size = 362, normalized size = 4.11 \begin{align*} -\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 )^{2} + \frac{{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{3} - 3 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) -{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{3} - 12 \, a x + 3 \,{\left (2 \, a^{2} x^{2} +{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 2\right )} \log \left (a x + 1\right ) - 6 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )\right )} a^{2}}{48 \,{\left (a^{5} x^{2} - a^{3}\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 \operatorname{artanh}\left (a x\right )}{8 \,{\left (a^{4} x^{2} - a^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(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)^2 + 1/48*((a^2*x^2 - 1)*log(a*x + 1)^3
- 3*(a^2*x^2 - 1)*log(a*x + 1)^2*log(a*x - 1) - (a^2*x^2 - 1)*log(a*x - 1)^3 - 12*a*x + 3*(2*a^2*x^2 + (a^2*x
^2 - 1)*log(a*x - 1)^2 - 2)*log(a*x + 1) - 6*(a^2*x^2 - 1)*log(a*x - 1))*a^2/(a^5*x^2 - a^3) - 1/8*((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*arctanh(a
*x)/(a^4*x^2 - a^2)
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Fricas [A] time = 2.05194, size = 208, normalized size = 2.36 \begin{align*} -\frac{6 \, a x \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} -{\left (a^{2} x^{2} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{3} + 12 \, a x - 6 \,{\left (a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{48 \,{\left (a^{3} x^{2} - a\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctanh(a*x)^2/(-a^2*x^2+1)^2,x, algorithm="fricas")
[Out]
-1/48*(6*a*x*log(-(a*x + 1)/(a*x - 1))^2 - (a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1))^3 + 12*a*x - 6*(a^2*x^2 + 1
)*log(-(a*x + 1)/(a*x - 1)))/(a^3*x^2 - a)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}^{2}{\left (a x \right )}}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(atanh(a*x)**2/(-a**2*x**2+1)**2,x)
[Out]
Integral(atanh(a*x)**2/((a*x - 1)**2*(a*x + 1)**2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (a x\right )^{2}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctanh(a*x)^2/(-a^2*x^2+1)^2,x, algorithm="giac")
[Out]
integrate(arctanh(a*x)^2/(a^2*x^2 - 1)^2, x)