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]

1/32*x/(-a^2*x^2+1)^2+15/64*x/(-a^2*x^2+1)+15/64*arctanh(a*x)/a-1/8*arctanh(a*x)/a/(-a^2*x^2+1)^2-3/8*arctanh(
a*x)/a/(-a^2*x^2+1)+1/4*x*arctanh(a*x)^2/(-a^2*x^2+1)^2+3/8*x*arctanh(a*x)^2/(-a^2*x^2+1)+1/8*arctanh(a*x)^3/a

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Rubi [A]  time = 0.11, antiderivative size = 151, normalized size of antiderivative = 1.00, 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 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])

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 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 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 {\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*}

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Mathematica [A]  time = 0.11, 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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fricas [A]  time = 0.54, size = 137, normalized size = 0.91 \[ -\frac {30 \, a^{3} x^{3} - 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 4 \, {\left (3 \, a^{3} x^{3} - 5 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 34 \, a x - {\left (15 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 17\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{128 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\operatorname {artanh}\left (a x\right )^{2}}{{\left (a^{2} x^{2} - 1\right )}^{3}}\,{d x} \]

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)

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maple [C]  time = 0.87, size = 2571, normalized size = 17.03 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(
1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1))*Pi*x^2+3/32*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*
csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(
a^2*x^2-1))*Pi*x^4+3/16*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2
*x^2+1)))^3*Pi*x^2+3/16*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*Pi*x^2+3/8*I*a/
(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*Pi*x^2+3/16*I*a^3/(a*x-1)^2/(a*x+1)^2*
arctanh(a*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*Pi*x^4-3/32*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(
I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*Pi*x^4-3/32*I/a/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)^2*cs
gn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(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)/(1+(a*x+1)^2/(-a^2*x^2
+1)))^2-3/16*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)/(-a^2*x^2+1)^(1/2))^2*csgn(I*(a*x+1)
^2/(a^2*x^2-1))+3/32*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*
x^2+1)))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*Pi*x^4-3/16*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I*(a*x+1)^2
/(a^2*x^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*Pi*x^4-3/32*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I
*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*Pi*x^4+3/32*I/a/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*
x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^
2/(-a^2*x^2+1)))+3/16*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)
^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*Pi*x^2-3/16*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I*(a*x+1)
^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*Pi*x^2+3/8*I*a/(a*x-1)^2/(a*x+1)^2*
arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*Pi*x^2+3/16*I*a/(a*x-1)^2/(a
*x+1)^2*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*Pi*x^2-3/32*I*a^3/(a
*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/
(-a^2*x^2+1)))^2*Pi*x^4-3/32*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*Pi*x^4-3
/16*I*a^3/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*Pi*x^4-3/8*I*a/(a*x-1)^2/(a*
x+1)^2*arctanh(a*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*Pi*x^2-3/16/a*arctanh(a*x)^2*ln(a*x-1)+3/16/a*arcta
nh(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))-1/4*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)
^3*x^2-3/32*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)*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+3/16*I/a/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)^2+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)-1/16/a*arctanh(a*x)^2/(a*x+1)^2-3/16/a*arctanh(a
*x)^2/(a*x+1)-3/8*I*a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*x^2+3/16*I/a/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)^2
*csgn(I/(1+(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))^3-3/16*I/a/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2+3/16*I*a^3/(a*x-1)^
2/(a*x+1)^2*arctanh(a*x)^2*Pi*x^4-3/32*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+(a*x+1)^2/(-a^2*x^2+1)))^3-15/64*a^2/(a*x-1)^2/(a*x+1)^2*x^3+1/8/a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^3-17/64
/a/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)

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maxima [B]  time = 0.34, size = 392, normalized size = 2.60 \[ -\frac {1}{16} \, {\left (\frac {2 \, {\left (3 \, a^{2} x^{3} - 5 \, x\right )}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1} - \frac {3 \, \log \left (a x + 1\right )}{a} + \frac {3 \, \log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right )^{2} - \frac {{\left (30 \, a^{3} x^{3} - 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{3} + 6 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) + 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{3} - 34 \, a x - 3 \, {\left (5 \, a^{4} x^{4} - 10 \, a^{2} x^{2} + 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} + 5\right )} \log \left (a x + 1\right ) + 15 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )\right )} a^{2}}{128 \, {\left (a^{7} x^{4} - 2 \, a^{5} x^{2} + a^{3}\right )}} + \frac {{\left (12 \, a^{2} x^{2} - 3 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 6 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 3 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} - 16\right )} a \operatorname {artanh}\left (a x\right )}{32 \, {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )}} \]

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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mupad [B]  time = 2.06, size = 358, normalized size = 2.37 \[ \frac {\frac {17\,x}{8}-\frac {15\,a^2\,x^3}{8}}{8\,a^4\,x^4-16\,a^2\,x^2+8}-\ln \left (1-a\,x\right )\,\left (\frac {3\,{\ln \left (a\,x+1\right )}^2}{64\,a}-\frac {\frac {7\,x}{2}-3\,a\,x^2+\frac {4}{a}-\frac {5\,a^2\,x^3}{2}}{32\,a^4\,x^4-64\,a^2\,x^2+32}+\frac {\frac {7\,x}{2}+3\,a\,x^2-\frac {4}{a}-\frac {5\,a^2\,x^3}{2}}{32\,a^4\,x^4-64\,a^2\,x^2+32}+\frac {\ln \left (a\,x+1\right )\,\left (10\,x-6\,a^2\,x^3\right )}{32\,a^4\,x^4-64\,a^2\,x^2+32}\right )+{\ln \left (1-a\,x\right )}^2\,\left (\frac {3\,\ln \left (a\,x+1\right )}{64\,a}+\frac {\frac {5\,x}{8}-\frac {3\,a^2\,x^3}{8}}{4\,a^4\,x^4-8\,a^2\,x^2+4}\right )+\frac {{\ln \left (a\,x+1\right )}^3}{64\,a}-\frac {{\ln \left (1-a\,x\right )}^3}{64\,a}-\frac {\ln \left (a\,x+1\right )\,\left (\frac {1}{4\,a^2}-\frac {3\,x^2}{16}\right )}{\frac {1}{a}-2\,a\,x^2+a^3\,x^4}+\frac {{\ln \left (a\,x+1\right )}^2\,\left (\frac {5\,x}{32\,a}-\frac {3\,a\,x^3}{32}\right )}{\frac {1}{a}-2\,a\,x^2+a^3\,x^4}-\frac {\mathrm {atan}\left (a\,x\,1{}\mathrm {i}\right )\,15{}\mathrm {i}}{64\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-atanh(a*x)^2/(a^2*x^2 - 1)^3,x)

[Out]

((17*x)/8 - (15*a^2*x^3)/8)/(8*a^4*x^4 - 16*a^2*x^2 + 8) - log(1 - a*x)*((3*log(a*x + 1)^2)/(64*a) - ((7*x)/2
- 3*a*x^2 + 4/a - (5*a^2*x^3)/2)/(32*a^4*x^4 - 64*a^2*x^2 + 32) + ((7*x)/2 + 3*a*x^2 - 4/a - (5*a^2*x^3)/2)/(3
2*a^4*x^4 - 64*a^2*x^2 + 32) + (log(a*x + 1)*(10*x - 6*a^2*x^3))/(32*a^4*x^4 - 64*a^2*x^2 + 32)) + log(1 - a*x
)^2*((3*log(a*x + 1))/(64*a) + ((5*x)/8 - (3*a^2*x^3)/8)/(4*a^4*x^4 - 8*a^2*x^2 + 4)) + log(a*x + 1)^3/(64*a)
- log(1 - a*x)^3/(64*a) - (atan(a*x*1i)*15i)/(64*a) - (log(a*x + 1)*(1/(4*a^2) - (3*x^2)/16))/(1/a - 2*a*x^2 +
a^3*x^4) + (log(a*x + 1)^2*((5*x)/(32*a) - (3*a*x^3)/32))/(1/a - 2*a*x^2 + a^3*x^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, dx \]

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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