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

Optimal. Leaf size=100 \[ -\frac {\tanh ^{-1}(a x)^2}{16 a^3}-\frac {x \tanh ^{-1}(a x)}{8 a^2 \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}+\frac {1}{16 a^3 \left (1-a^2 x^2\right )}-\frac {1}{16 a^3 \left (1-a^2 x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5998, 5956, 261} \[ \frac {1}{16 a^3 \left (1-a^2 x^2\right )}-\frac {1}{16 a^3 \left (1-a^2 x^2\right )^2}-\frac {x \tanh ^{-1}(a x)}{8 a^2 \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {\tanh ^{-1}(a x)^2}{16 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5998

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(
q + 1))/(4*c^3*d*(q + 1)^2), x] + (Dist[1/(2*c^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*c^2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[c^2*d + e, 0] && LtQ[q, -1] && NeQ[q, -5/2]

Rubi steps

\begin {align*} \int \frac {x^2 \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx &=-\frac {1}{16 a^3 \left (1-a^2 x^2\right )^2}+\frac {x \tanh ^{-1}(a x)}{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 )^2} \, dx}{4 a^2}\\ &=-\frac {1}{16 a^3 \left (1-a^2 x^2\right )^2}+\frac {x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {x \tanh ^{-1}(a x)}{8 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^2}{16 a^3}+\frac {\int \frac {x}{\left (1-a^2 x^2\right )^2} \, dx}{8 a}\\ &=-\frac {1}{16 a^3 \left (1-a^2 x^2\right )^2}+\frac {1}{16 a^3 \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {x \tanh ^{-1}(a x)}{8 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^2}{16 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 61, normalized size = 0.61 \[ -\frac {-2 \left (a^3 x^3+a x\right ) \tanh ^{-1}(a x)+a^2 x^2+\left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2}{16 a^3 \left (a^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 95, normalized size = 0.95 \[ -\frac {4 \, a^{2} x^{2} + {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{64 \, {\left (a^{7} x^{4} - 2 \, a^{5} x^{2} + a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(4*a^2*x^2 + (a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^2 - 4*(a^3*x^3 + a*x)*log(-(a*x + 1)/(a
*x - 1)))/(a^7*x^4 - 2*a^5*x^2 + a^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 225, normalized size = 2.25 \[ \frac {\arctanh \left (a x \right )}{16 a^{3} \left (a x -1\right )^{2}}+\frac {\arctanh \left (a x \right )}{16 a^{3} \left (a x -1\right )}+\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{16 a^{3}}-\frac {\arctanh \left (a x \right )}{16 a^{3} \left (a x +1\right )^{2}}+\frac {\arctanh \left (a x \right )}{16 a^{3} \left (a x +1\right )}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{16 a^{3}}+\frac {\ln \left (a x +1\right )^{2}}{64 a^{3}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{32 a^{3}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{32 a^{3}}+\frac {\ln \left (a x -1\right )^{2}}{64 a^{3}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{32 a^{3}}-\frac {1}{64 a^{3} \left (a x -1\right )^{2}}-\frac {1}{64 a^{3} \left (a x -1\right )}-\frac {1}{64 a^{3} \left (a x +1\right )^{2}}+\frac {1}{64 a^{3} \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/a^3*arctanh(a*x)/(a*x-1)^2+1/16/a^3*arctanh(a*x)/(a*x-1)+1/16/a^3*arctanh(a*x)*ln(a*x-1)-1/16/a^3*arctanh
(a*x)/(a*x+1)^2+1/16/a^3*arctanh(a*x)/(a*x+1)-1/16/a^3*arctanh(a*x)*ln(a*x+1)+1/64/a^3*ln(a*x+1)^2+1/32/a^3*ln
(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/32/a^3*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/64/a^3*ln(a*x-1)^2-1/32/a^3*ln(a*x-1)*ln(
1/2+1/2*a*x)-1/64/a^3/(a*x-1)^2-1/64/a^3/(a*x-1)-1/64/a^3/(a*x+1)^2+1/64/a^3/(a*x+1)

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maxima [B]  time = 0.32, size = 179, normalized size = 1.79 \[ \frac {1}{16} \, {\left (\frac {2 \, {\left (a^{2} x^{3} + x\right )}}{a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}} - \frac {\log \left (a x + 1\right )}{a^{3}} + \frac {\log \left (a x - 1\right )}{a^{3}}\right )} \operatorname {artanh}\left (a x\right ) - \frac {{\left (4 \, a^{2} x^{2} - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2}\right )} a}{64 \, {\left (a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*(a^2*x^3 + x)/(a^6*x^4 - 2*a^4*x^2 + a^2) - log(a*x + 1)/a^3 + log(a*x - 1)/a^3)*arctanh(a*x) - 1/64*(
4*a^2*x^2 - (a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2 + 2*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)*log(a*x - 1) -
(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^2)*a/(a^8*x^4 - 2*a^6*x^2 + a^4)

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mupad [B]  time = 1.19, size = 150, normalized size = 1.50 \[ \ln \left (1-a\,x\right )\,\left (\frac {\ln \left (a\,x+1\right )}{32\,a^3}-\frac {\frac {x}{8\,a^2}+\frac {x^3}{8}}{2\,a^4\,x^4-4\,a^2\,x^2+2}\right )-\frac {{\ln \left (a\,x+1\right )}^2}{64\,a^3}-\frac {{\ln \left (1-a\,x\right )}^2}{64\,a^3}-\frac {x^2}{2\,\left (8\,a^5\,x^4-16\,a^3\,x^2+8\,a\right )}+\frac {\ln \left (a\,x+1\right )\,\left (\frac {x}{16\,a^3}+\frac {x^3}{16\,a}\right )}{\frac {1}{a}-2\,a\,x^2+a^3\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1 - a*x)*(log(a*x + 1)/(32*a^3) - (x/(8*a^2) + x^3/8)/(2*a^4*x^4 - 4*a^2*x^2 + 2)) - log(a*x + 1)^2/(64*a^
3) - log(1 - a*x)^2/(64*a^3) - x^2/(2*(8*a - 16*a^3*x^2 + 8*a^5*x^4)) + (log(a*x + 1)*(x/(16*a^3) + x^3/(16*a)
))/(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 \[ - \int \frac {x^{2} \operatorname {atanh}{\left (a x \right )}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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