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]

1/4*x/(-a^2*x^2+1)+1/4*arctanh(a*x)/a-1/2*arctanh(a*x)/a/(-a^2*x^2+1)+1/2*x*arctanh(a*x)^2/(-a^2*x^2+1)+1/6*ar
ctanh(a*x)^3/a

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, 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 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 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 )^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 = 0.09, 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))

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 95, normalized size = 1.08 \[ -\frac {6 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - {\left (a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 12 \, a x - 6 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{48 \, {\left (a^{3} x^{2} - a\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 2.08, size = 88, normalized size = 1.00 \[ \frac {1}{16} \, a^{2} {\left (\frac {{\left (a x - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}}{{\left (a x + 1\right )} a^{4}} + \frac {2 \, {\left (a x - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{{\left (a x + 1\right )} a^{4}} + \frac {2 \, {\left (a x - 1\right )}}{{\left (a x + 1\right )} a^{4}}\right )} \]

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]

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

________________________________________________________________________________________

maple [C]  time = 0.81, size = 1695, normalized size = 19.26 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*x^2+1/4*
I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*x^2-1/8*I*a/(a*x-1)/(a*x+1)*arctanh
(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*x^2-1/4*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(
-a^2*x^2+1)))^2*x^2+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+1/8*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*csgn(I/(1
+(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)/(1+(a*x+1)^2/
(-a^2*x^2+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)*arctanh(a*x)^2*Pi*csg
n(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*x^2-1/8*I*a/(a*x-1)/(a*x
+1)*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*x^2-1/4*I*a/(a*x-1)/(
a*x+1)*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*x^2-1/8*I/a/(a*x-1
)/(a*x+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))*csgn(I/(1+(a*
x+1)^2/(-a^2*x^2+1)))*arctanh(a*x)^2*Pi-1/8*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)
/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*x^2-1/6/a/(a*x-1)/(a*x+1)*arctanh(a*x)^3+1/4
/a/(a*x-1)/(a*x+1)*arctanh(a*x)+1/8*I*a/(a*x-1)/(a*x+1)*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)/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*x^2-1/4/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))-1/4*I/a/(a*x-1)/
(a*x+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)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*a
rctanh(a*x)^2*Pi-1/4*I/a/(a*x-1)/(a*x+1)*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+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*Pi+1/4*I/a/(a*x-1)/(a*x+1)*csgn(I/(1+(a*x+1)^2/(-a^2*
x^2+1)))^2*arctanh(a*x)^2*Pi+1/4*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*Pi*x^2-1/4/(a*x-1)/(a*x+1)*x+1/6*a/(a*x-1)
/(a*x+1)*arctanh(a*x)^3*x^2+1/4*a/(a*x-1)/(a*x+1)*arctanh(a*x)*x^2-1/4/a*arctanh(a*x)^2/(a*x-1)-1/4/a*arctanh(
a*x)^2/(a*x+1)

________________________________________________________________________________________

maxima [B]  time = 0.34, size = 268, normalized size = 3.05 \[ -\frac {1}{4} \, {\left (\frac {2 \, x}{a^{2} x^{2} - 1} - \frac {\log \left (a x + 1\right )}{a} + \frac {\log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right )^{2} + \frac {{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{3} - 3 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) - {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{3} - 12 \, a x + 3 \, {\left (2 \, a^{2} x^{2} + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 2\right )} \log \left (a x + 1\right ) - 6 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )\right )} a^{2}}{48 \, {\left (a^{5} x^{2} - a^{3}\right )}} - \frac {{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4\right )} a \operatorname {artanh}\left (a x\right )}{8 \, {\left (a^{4} x^{2} - a^{2}\right )}} \]

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)

________________________________________________________________________________________

mupad [B]  time = 1.50, size = 213, normalized size = 2.42 \[ \frac {{\ln \left (a\,x+1\right )}^3}{48\,a}-\frac {\ln \left (a\,x+1\right )}{4\,\left (a-a^3\,x^2\right )}-\frac {{\ln \left (1-a\,x\right )}^3}{48\,a}-\frac {x}{4\,a^2\,x^2-4}+\frac {\ln \left (1-a\,x\right )}{4\,a-4\,a^3\,x^2}+\frac {\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2}{16\,a}-\frac {{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )}{16\,a}-\frac {x\,{\ln \left (a\,x+1\right )}^2}{8\,\left (a^2\,x^2-1\right )}-\frac {x\,{\ln \left (1-a\,x\right )}^2}{2\,\left (4\,a^2\,x^2-4\right )}+\frac {x\,\ln \left (a\,x+1\right )\,\ln \left (1-a\,x\right )}{4\,a^2\,x^2-4}-\frac {\mathrm {atan}\left (a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^2/(a^2*x^2 - 1)^2,x)

[Out]

log(a*x + 1)^3/(48*a) - log(a*x + 1)/(4*(a - a^3*x^2)) - log(1 - a*x)^3/(48*a) - x/(4*a^2*x^2 - 4) - (atan(a*x
*1i)*1i)/(4*a) + log(1 - a*x)/(4*a - 4*a^3*x^2) + (log(a*x + 1)*log(1 - a*x)^2)/(16*a) - (log(a*x + 1)^2*log(1
- a*x))/(16*a) - (x*log(a*x + 1)^2)/(8*(a^2*x^2 - 1)) - (x*log(1 - a*x)^2)/(2*(4*a^2*x^2 - 4)) + (x*log(a*x +
1)*log(1 - a*x))/(4*a^2*x^2 - 4)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)

________________________________________________________________________________________