3.311 \(\int \frac{\tanh ^{-1}(a x)^2}{(1-a^2 x^2)^3} \, dx\)
Optimal. Leaf size=151 \[ \frac{15 x}{64 \left (1-a^2 x^2\right )}+\frac{x}{32 \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)^2}{8 \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac{3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac{\tanh ^{-1}(a x)^3}{8 a}+\frac{15 \tanh ^{-1}(a x)}{64 a} \]
[Out]
x/(32*(1 - a^2*x^2)^2) + (15*x)/(64*(1 - a^2*x^2)) + (15*ArcTanh[a*x])/(64*a) - ArcTanh[a*x]/(8*a*(1 - a^2*x^2
)^2) - (3*ArcTanh[a*x])/(8*a*(1 - a^2*x^2)) + (x*ArcTanh[a*x]^2)/(4*(1 - a^2*x^2)^2) + (3*x*ArcTanh[a*x]^2)/(8
*(1 - a^2*x^2)) + ArcTanh[a*x]^3/(8*a)
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Rubi [A] time = 0.110423, antiderivative size = 151, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used =
{5964, 5956, 5994, 199, 206} \[ \frac{15 x}{64 \left (1-a^2 x^2\right )}+\frac{x}{32 \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)^2}{8 \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac{3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac{\tanh ^{-1}(a x)^3}{8 a}+\frac{15 \tanh ^{-1}(a x)}{64 a} \]
Antiderivative was successfully verified.
[In]
Int[ArcTanh[a*x]^2/(1 - a^2*x^2)^3,x]
[Out]
x/(32*(1 - a^2*x^2)^2) + (15*x)/(64*(1 - a^2*x^2)) + (15*ArcTanh[a*x])/(64*a) - ArcTanh[a*x]/(8*a*(1 - a^2*x^2
)^2) - (3*ArcTanh[a*x])/(8*a*(1 - a^2*x^2)) + (x*ArcTanh[a*x]^2)/(4*(1 - a^2*x^2)^2) + (3*x*ArcTanh[a*x]^2)/(8
*(1 - a^2*x^2)) + ArcTanh[a*x]^3/(8*a)
Rule 5964
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*p*(d + e*x^2)^(
q + 1)*(a + b*ArcTanh[c*x])^(p - 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])^p, x], x] + Dist[(b^2*p*(p - 1))/(4*(q + 1)^2), Int[(d + e*x^2)^q*(a + b*ArcTanh[c*
x])^(p - 2), x], x] - Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(2*d*(q + 1)), x]) /; FreeQ[{a, b, c
, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && GtQ[p, 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 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 )^3} \, dx &=-\frac{\tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac{x \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}+\frac{1}{8} \int \frac{1}{\left (1-a^2 x^2\right )^3} \, dx+\frac{3}{4} \int \frac{\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac{x}{32 \left (1-a^2 x^2\right )^2}-\frac{\tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac{x \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)^2}{8 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{8 a}+\frac{3}{32} \int \frac{1}{\left (1-a^2 x^2\right )^2} \, dx-\frac{1}{4} (3 a) \int \frac{x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac{x}{32 \left (1-a^2 x^2\right )^2}+\frac{3 x}{64 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac{3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)^2}{8 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{8 a}+\frac{3}{64} \int \frac{1}{1-a^2 x^2} \, dx+\frac{3}{8} \int \frac{1}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac{x}{32 \left (1-a^2 x^2\right )^2}+\frac{15 x}{64 \left (1-a^2 x^2\right )}+\frac{3 \tanh ^{-1}(a x)}{64 a}-\frac{\tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac{3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)^2}{8 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{8 a}+\frac{3}{16} \int \frac{1}{1-a^2 x^2} \, dx\\ &=\frac{x}{32 \left (1-a^2 x^2\right )^2}+\frac{15 x}{64 \left (1-a^2 x^2\right )}+\frac{15 \tanh ^{-1}(a x)}{64 a}-\frac{\tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac{3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)^2}{8 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{8 a}\\ \end{align*}
Mathematica [A] time = 0.107171, size = 127, normalized size = 0.84 \[ \frac{1}{128} \left (-\frac{30 x}{a^2 x^2-1}+\frac{4 x}{\left (a^2 x^2-1\right )^2}-\frac{16 x \left (3 a^2 x^2-5\right ) \tanh ^{-1}(a x)^2}{\left (a^2 x^2-1\right )^2}+\frac{16 \left (3 a^2 x^2-4\right ) \tanh ^{-1}(a x)}{a \left (a^2 x^2-1\right )^2}-\frac{15 \log (1-a x)}{a}+\frac{15 \log (a x+1)}{a}+\frac{16 \tanh ^{-1}(a x)^3}{a}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[ArcTanh[a*x]^2/(1 - a^2*x^2)^3,x]
[Out]
((4*x)/(-1 + a^2*x^2)^2 - (30*x)/(-1 + a^2*x^2) + (16*(-4 + 3*a^2*x^2)*ArcTanh[a*x])/(a*(-1 + a^2*x^2)^2) - (1
6*x*(-5 + 3*a^2*x^2)*ArcTanh[a*x]^2)/(-1 + a^2*x^2)^2 + (16*ArcTanh[a*x]^3)/a - (15*Log[1 - a*x])/a + (15*Log[
1 + a*x])/a)/128
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Maple [C] time = 0.443, size = 2571, normalized size = 17. \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)^3,x)
[Out]
-3/32*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*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*x^4+3/16*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)/(-a^2*x^2+
1)^(1/2))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*x^2+3/8*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)/(-a
^2*x^2+1)^(1/2))*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*x^2-3/16*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*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*x^2+3/16*I*a/(a*x-1)^2/(a*x+1)^
2*arctanh(a*x)^2*Pi*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*x^2-3/32*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*x^4-3/32*I*a^3/(a*x-1
)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3*x^4+3/16*I*a^3/(a*x
-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3*x^4-3/16*I*a^3/(a*x-1)^2/(a*x+1)^2*arct
anh(a*x)^2*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^2*x^4+3/16*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(
a*x+1)^2/(a^2*x^2-1))^3*x^2+3/16*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+
1)^2/(-a^2*x^2+1)+1))^3*x^2-3/8*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3
*x^2+3/8*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^2*x^2-3/32*I/a/(a*x-1)^2
/(a*x+1)^2*Pi*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))-3/16*I/a/(a*x-
1)^2/(a*x+1)^2*Pi*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+3/32*I/a/(
a*x-1)^2/(a*x+1)^2*Pi*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))^2-3/32*I/a/(a*x-1)^2/(a*x+1)^2*Pi*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+3/16*I/a/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)^2+1/8*a^3/(a*x-1)^
2/(a*x+1)^2*arctanh(a*x)^3*x^4+15/64*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)*x^4-1/4*a/(a*x-1)^2/(a*x+1)^2*arctan
h(a*x)^3*x^2-3/32*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)*x^2-3/16*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*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))*x^2+3/32*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*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))*x^4+3/32*I/a/(a*x-1)^2/(a*x+1)^2*Pi*arc
tanh(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))-3/32*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*
csgn(I*(a*x+1)^2/(a^2*x^2-1))*x^4-3/16*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)/(-a^2*x^2+1)
^(1/2))*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*x^4+3/32*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*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*x^4+17/64/(a*x-1)^2/(a*x+1)^2*x+1/16/a
*arctanh(a*x)^2/(a*x-1)^2-3/16/a*arctanh(a*x)^2/(a*x-1)-3/16/a*arctanh(a*x)^2*ln(a*x-1)-1/16/a*arctanh(a*x)^2/
(a*x+1)^2-3/16/a*arctanh(a*x)^2/(a*x+1)+3/16/a*arctanh(a*x)^2*ln(a*x+1)-3/8/a*arctanh(a*x)^2*ln((a*x+1)/(-a^2*
x^2+1)^(1/2))-15/64*a^2/(a*x-1)^2/(a*x+1)^2*x^3-17/64/a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)+1/8/a/(a*x-1)^2/(a*x+
1)^2*arctanh(a*x)^3-3/32*I/a/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3-3/32*I/a/(a
*x-1)^2/(a*x+1)^2*Pi*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+3/16*I/a/(a*x-1
)^2/(a*x+1)^2*Pi*arctanh(a*x)^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3-3/16*I/a/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a
*x)^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^2+3/16*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*x^4-3/8*I*a/(a*x-1
)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*x^2
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Maxima [B] time = 1.02304, size = 529, normalized size = 3.5 \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 )^{2} - \frac{{\left (30 \, a^{3} x^{3} - 2 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{3} + 6 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) + 2 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{3} - 34 \, a x - 3 \,{\left (5 \, a^{4} x^{4} - 10 \, a^{2} x^{2} + 2 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} + 5\right )} \log \left (a x + 1\right ) + 15 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )\right )} a^{2}}{128 \,{\left (a^{7} x^{4} - 2 \, a^{5} x^{2} + a^{3}\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 \operatorname{artanh}\left (a x\right )}{32 \,{\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)^2/(-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)^2 - 1
/128*(30*a^3*x^3 - 2*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^3 + 6*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2*log
(a*x - 1) + 2*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^3 - 34*a*x - 3*(5*a^4*x^4 - 10*a^2*x^2 + 2*(a^4*x^4 - 2*a
^2*x^2 + 1)*log(a*x - 1)^2 + 5)*log(a*x + 1) + 15*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1))*a^2/(a^7*x^4 - 2*a^5
*x^2 + a^3) + 1/32*(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*arctanh(a*x)/(a^6*x^4 - 2*a^4*x^2 +
a^2)
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Fricas [A] time = 1.90494, size = 302, normalized size = 2. \begin{align*} -\frac{30 \, a^{3} x^{3} - 2 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{3} + 4 \,{\left (3 \, a^{3} x^{3} - 5 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - 34 \, a x -{\left (15 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 17\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{128 \,{\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)^2/(-a^2*x^2+1)^3,x, algorithm="fricas")
[Out]
-1/128*(30*a^3*x^3 - 2*(a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^3 + 4*(3*a^3*x^3 - 5*a*x)*log(-(a*x
+ 1)/(a*x - 1))^2 - 34*a*x - (15*a^4*x^4 - 6*a^2*x^2 - 17)*log(-(a*x + 1)/(a*x - 1)))/(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}^{2}{\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)**2/(-a**2*x**2+1)**3,x)
[Out]
-Integral(atanh(a*x)**2/(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 )^{2}}{{\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)^2/(-a^2*x^2+1)^3,x, algorithm="giac")
[Out]
integrate(-arctanh(a*x)^2/(a^2*x^2 - 1)^3, x)