3.310 \(\int \frac {x \tanh ^{-1}(a x)^2}{(1-a^2 x^2)^3} \, dx\)

Optimal. Leaf size=125 \[ \frac {3}{32 a^2 \left (1-a^2 x^2\right )}+\frac {1}{32 a^2 \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {3 x \tanh ^{-1}(a x)}{16 a \left (1-a^2 x^2\right )}-\frac {x \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac {3 \tanh ^{-1}(a x)^2}{32 a^2} \]

[Out]

1/32/a^2/(-a^2*x^2+1)^2+3/32/a^2/(-a^2*x^2+1)-1/8*x*arctanh(a*x)/a/(-a^2*x^2+1)^2-3/16*x*arctanh(a*x)/a/(-a^2*
x^2+1)-3/32*arctanh(a*x)^2/a^2+1/4*arctanh(a*x)^2/a^2/(-a^2*x^2+1)^2

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Rubi [A]  time = 0.09, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5994, 5960, 5956, 261} \[ \frac {3}{32 a^2 \left (1-a^2 x^2\right )}+\frac {1}{32 a^2 \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {3 x \tanh ^{-1}(a x)}{16 a \left (1-a^2 x^2\right )}-\frac {x \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac {3 \tanh ^{-1}(a x)^2}{32 a^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcTanh[a*x]^2)/(1 - a^2*x^2)^3,x]

[Out]

1/(32*a^2*(1 - a^2*x^2)^2) + 3/(32*a^2*(1 - a^2*x^2)) - (x*ArcTanh[a*x])/(8*a*(1 - a^2*x^2)^2) - (3*x*ArcTanh[
a*x])/(16*a*(1 - a^2*x^2)) - (3*ArcTanh[a*x]^2)/(32*a^2) + ArcTanh[a*x]^2/(4*a^2*(1 - a^2*x^2)^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 5960

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(q + 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]), x], x] -
Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]))/(2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*
d + e, 0] && LtQ[q, -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 {x \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx &=\frac {\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {\int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx}{2 a}\\ &=\frac {1}{32 a^2 \left (1-a^2 x^2\right )^2}-\frac {x \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {3 \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx}{8 a}\\ &=\frac {1}{32 a^2 \left (1-a^2 x^2\right )^2}-\frac {x \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac {3 x \tanh ^{-1}(a x)}{16 a \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^2}{32 a^2}+\frac {\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}+\frac {3}{16} \int \frac {x}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac {1}{32 a^2 \left (1-a^2 x^2\right )^2}+\frac {3}{32 a^2 \left (1-a^2 x^2\right )}-\frac {x \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac {3 x \tanh ^{-1}(a x)}{16 a \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^2}{32 a^2}+\frac {\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 71, normalized size = 0.57 \[ \frac {-3 a^2 x^2+2 a x \left (3 a^2 x^2-5\right ) \tanh ^{-1}(a x)+\left (-3 a^4 x^4+6 a^2 x^2+5\right ) \tanh ^{-1}(a x)^2+4}{32 a^2 \left (a^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcTanh[a*x]^2)/(1 - a^2*x^2)^3,x]

[Out]

(4 - 3*a^2*x^2 + 2*a*x*(-5 + 3*a^2*x^2)*ArcTanh[a*x] + (5 + 6*a^2*x^2 - 3*a^4*x^4)*ArcTanh[a*x]^2)/(32*a^2*(-1
+ a^2*x^2)^2)

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fricas [A]  time = 0.49, size = 99, normalized size = 0.79 \[ -\frac {12 \, a^{2} x^{2} + {\left (3 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 5\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 4 \, {\left (3 \, a^{3} x^{3} - 5 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 16}{128 \, {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(a*x)^2/(-a^2*x^2+1)^3,x, algorithm="fricas")

[Out]

-1/128*(12*a^2*x^2 + (3*a^4*x^4 - 6*a^2*x^2 - 5)*log(-(a*x + 1)/(a*x - 1))^2 - 4*(3*a^3*x^3 - 5*a*x)*log(-(a*x
+ 1)/(a*x - 1)) - 16)/(a^6*x^4 - 2*a^4*x^2 + a^2)

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giac [B]  time = 0.25, size = 251, normalized size = 2.01 \[ -\frac {1}{512} \, {\left (2 \, {\left (\frac {{\left (a x - 1\right )}^{2} {\left (\frac {4 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )}}{{\left (a x + 1\right )}^{2} a^{3}} - \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2} a^{3}} + \frac {4 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 2 \, {\left (\frac {{\left (a x - 1\right )}^{2} {\left (\frac {8 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )}}{{\left (a x + 1\right )}^{2} a^{3}} + \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2} a^{3}} - \frac {8 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + \frac {{\left (a x - 1\right )}^{2} {\left (\frac {16 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )}}{{\left (a x + 1\right )}^{2} a^{3}} - \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2} a^{3}} + \frac {16 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{3}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(a*x)^2/(-a^2*x^2+1)^3,x, algorithm="giac")

[Out]

-1/512*(2*((a*x - 1)^2*(4*(a*x + 1)/(a*x - 1) - 1)/((a*x + 1)^2*a^3) - (a*x + 1)^2/((a*x - 1)^2*a^3) + 4*(a*x
+ 1)/((a*x - 1)*a^3))*log(-(a*x + 1)/(a*x - 1))^2 + 2*((a*x - 1)^2*(8*(a*x + 1)/(a*x - 1) - 1)/((a*x + 1)^2*a^
3) + (a*x + 1)^2/((a*x - 1)^2*a^3) - 8*(a*x + 1)/((a*x - 1)*a^3))*log(-(a*x + 1)/(a*x - 1)) + (a*x - 1)^2*(16*
(a*x + 1)/(a*x - 1) - 1)/((a*x + 1)^2*a^3) - (a*x + 1)^2/((a*x - 1)^2*a^3) + 16*(a*x + 1)/((a*x - 1)*a^3))*a

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maple [B]  time = 0.07, size = 247, normalized size = 1.98 \[ \frac {\arctanh \left (a x \right )^{2}}{4 a^{2} \left (a^{2} x^{2}-1\right )^{2}}-\frac {\arctanh \left (a x \right )}{32 a^{2} \left (a x -1\right )^{2}}+\frac {3 \arctanh \left (a x \right )}{32 a^{2} \left (a x -1\right )}+\frac {3 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{32 a^{2}}+\frac {\arctanh \left (a x \right )}{32 a^{2} \left (a x +1\right )^{2}}+\frac {3 \arctanh \left (a x \right )}{32 a^{2} \left (a x +1\right )}-\frac {3 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{32 a^{2}}+\frac {3 \ln \left (a x -1\right )^{2}}{128 a^{2}}-\frac {3 \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{64 a^{2}}+\frac {3 \ln \left (a x +1\right )^{2}}{128 a^{2}}+\frac {3 \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{64 a^{2}}-\frac {3 \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{64 a^{2}}+\frac {1}{128 a^{2} \left (a x -1\right )^{2}}-\frac {7}{128 a^{2} \left (a x -1\right )}+\frac {1}{128 a^{2} \left (a x +1\right )^{2}}+\frac {7}{128 a^{2} \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a^2/(a^2*x^2-1)^2*arctanh(a*x)^2-1/32/a^2*arctanh(a*x)/(a*x-1)^2+3/32/a^2*arctanh(a*x)/(a*x-1)+3/32/a^2*ar
ctanh(a*x)*ln(a*x-1)+1/32/a^2*arctanh(a*x)/(a*x+1)^2+3/32/a^2*arctanh(a*x)/(a*x+1)-3/32/a^2*arctanh(a*x)*ln(a*
x+1)+3/128/a^2*ln(a*x-1)^2-3/64/a^2*ln(a*x-1)*ln(1/2+1/2*a*x)+3/128/a^2*ln(a*x+1)^2+3/64/a^2*ln(-1/2*a*x+1/2)*
ln(1/2+1/2*a*x)-3/64/a^2*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/128/a^2/(a*x-1)^2-7/128/a^2/(a*x-1)+1/128/a^2/(a*x+1)^2+
7/128/a^2/(a*x+1)

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maxima [A]  time = 0.33, size = 206, normalized size = 1.65 \[ \frac {{\left (\frac {2 \, {\left (3 \, a^{2} x^{3} - 5 \, x\right )}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1} - \frac {3 \, \log \left (a x + 1\right )}{a} + \frac {3 \, \log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right )}{32 \, a} - \frac {12 \, a^{2} x^{2} - 3 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 6 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 3 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} - 16}{128 \, {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )}} + \frac {\operatorname {artanh}\left (a x\right )^{2}}{4 \, {\left (a^{2} x^{2} - 1\right )}^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(a*x)^2/(-a^2*x^2+1)^3,x, algorithm="maxima")

[Out]

1/32*(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)/a - 1/
128*(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^6*x^4 - 2*a^4*x^2 + a^2) + 1/4*arctanh(a*x)^2/((a
^2*x^2 - 1)^2*a^2)

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mupad [B]  time = 1.48, size = 319, normalized size = 2.55 \[ {\ln \left (a\,x+1\right )}^2\,\left (\frac {1}{16\,a^3\,\left (\frac {1}{a}-2\,a\,x^2+a^3\,x^4\right )}-\frac {3}{128\,a^2}\right )-{\ln \left (1-a\,x\right )}^2\,\left (\frac {3}{128\,a^2}-\frac {1}{4\,a^2\,\left (4\,a^4\,x^4-8\,a^2\,x^2+4\right )}\right )-\ln \left (1-a\,x\right )\,\left (\frac {\frac {1}{4\,a}-\frac {5\,x}{8}+\frac {3\,a^2\,x^3}{8}}{8\,a^5\,x^4-16\,a^3\,x^2+8\,a}-\frac {\frac {5\,x}{8}+\frac {1}{4\,a}-\frac {3\,a^2\,x^3}{8}}{8\,a^5\,x^4-16\,a^3\,x^2+8\,a}+\ln \left (a\,x+1\right )\,\left (\frac {1}{4\,a^2\,\left (2\,a^4\,x^4-4\,a^2\,x^2+2\right )}-\frac {3\,\left (a^4\,x^4-2\,a^2\,x^2+1\right )}{32\,a^2\,\left (2\,a^4\,x^4-4\,a^2\,x^2+2\right )}\right )\right )+\frac {\frac {2}{a^2}-\frac {3\,x^2}{2}}{16\,a^4\,x^4-32\,a^2\,x^2+16}-\frac {\ln \left (a\,x+1\right )\,\left (\frac {5\,x}{32\,a^2}-\frac {3\,x^3}{32}\right )}{\frac {1}{a}-2\,a\,x^2+a^3\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a*x + 1)^2*(1/(16*a^3*(1/a - 2*a*x^2 + a^3*x^4)) - 3/(128*a^2)) - log(1 - a*x)^2*(3/(128*a^2) - 1/(4*a^2*(
4*a^4*x^4 - 8*a^2*x^2 + 4))) - log(1 - a*x)*((1/(4*a) - (5*x)/8 + (3*a^2*x^3)/8)/(8*a - 16*a^3*x^2 + 8*a^5*x^4
) - ((5*x)/8 + 1/(4*a) - (3*a^2*x^3)/8)/(8*a - 16*a^3*x^2 + 8*a^5*x^4) + log(a*x + 1)*(1/(4*a^2*(2*a^4*x^4 - 4
*a^2*x^2 + 2)) - (3*(a^4*x^4 - 2*a^2*x^2 + 1))/(32*a^2*(2*a^4*x^4 - 4*a^2*x^2 + 2)))) + (2/a^2 - (3*x^2)/2)/(1
6*a^4*x^4 - 32*a^2*x^2 + 16) - (log(a*x + 1)*((5*x)/(32*a^2) - (3*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 \[ - \int \frac {x \operatorname {atanh}^{2}{\left (a x \right )}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(a*x)**2/(-a**2*x**2+1)**3,x)

[Out]

-Integral(x*atanh(a*x)**2/(a**6*x**6 - 3*a**4*x**4 + 3*a**2*x**2 - 1), x)

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