3.345 \(\int \frac{\tanh ^{-1}(a x)}{(1-a^2 x^2)^4} \, dx\)

Optimal. Leaf size=134 \[ -\frac{5}{32 a \left (1-a^2 x^2\right )}-\frac{5}{96 a \left (1-a^2 x^2\right )^2}-\frac{1}{36 a \left (1-a^2 x^2\right )^3}+\frac{5 x \tanh ^{-1}(a x)}{16 \left (1-a^2 x^2\right )}+\frac{5 x \tanh ^{-1}(a x)}{24 \left (1-a^2 x^2\right )^2}+\frac{x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac{5 \tanh ^{-1}(a x)^2}{32 a} \]

[Out]

-1/(36*a*(1 - a^2*x^2)^3) - 5/(96*a*(1 - a^2*x^2)^2) - 5/(32*a*(1 - a^2*x^2)) + (x*ArcTanh[a*x])/(6*(1 - a^2*x
^2)^3) + (5*x*ArcTanh[a*x])/(24*(1 - a^2*x^2)^2) + (5*x*ArcTanh[a*x])/(16*(1 - a^2*x^2)) + (5*ArcTanh[a*x]^2)/
(32*a)

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Rubi [A]  time = 0.0738376, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {5960, 5956, 261} \[ -\frac{5}{32 a \left (1-a^2 x^2\right )}-\frac{5}{96 a \left (1-a^2 x^2\right )^2}-\frac{1}{36 a \left (1-a^2 x^2\right )^3}+\frac{5 x \tanh ^{-1}(a x)}{16 \left (1-a^2 x^2\right )}+\frac{5 x \tanh ^{-1}(a x)}{24 \left (1-a^2 x^2\right )^2}+\frac{x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac{5 \tanh ^{-1}(a x)^2}{32 a} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(1 - a^2*x^2)^4,x]

[Out]

-1/(36*a*(1 - a^2*x^2)^3) - 5/(96*a*(1 - a^2*x^2)^2) - 5/(32*a*(1 - a^2*x^2)) + (x*ArcTanh[a*x])/(6*(1 - a^2*x
^2)^3) + (5*x*ArcTanh[a*x])/(24*(1 - a^2*x^2)^2) + (5*x*ArcTanh[a*x])/(16*(1 - a^2*x^2)) + (5*ArcTanh[a*x]^2)/
(32*a)

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 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 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]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^4} \, dx &=-\frac{1}{36 a \left (1-a^2 x^2\right )^3}+\frac{x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac{5}{6} \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx\\ &=-\frac{1}{36 a \left (1-a^2 x^2\right )^3}-\frac{5}{96 a \left (1-a^2 x^2\right )^2}+\frac{x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac{5 x \tanh ^{-1}(a x)}{24 \left (1-a^2 x^2\right )^2}+\frac{5}{8} \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac{1}{36 a \left (1-a^2 x^2\right )^3}-\frac{5}{96 a \left (1-a^2 x^2\right )^2}+\frac{x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac{5 x \tanh ^{-1}(a x)}{24 \left (1-a^2 x^2\right )^2}+\frac{5 x \tanh ^{-1}(a x)}{16 \left (1-a^2 x^2\right )}+\frac{5 \tanh ^{-1}(a x)^2}{32 a}-\frac{1}{16} (5 a) \int \frac{x}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac{1}{36 a \left (1-a^2 x^2\right )^3}-\frac{5}{96 a \left (1-a^2 x^2\right )^2}-\frac{5}{32 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac{5 x \tanh ^{-1}(a x)}{24 \left (1-a^2 x^2\right )^2}+\frac{5 x \tanh ^{-1}(a x)}{16 \left (1-a^2 x^2\right )}+\frac{5 \tanh ^{-1}(a x)^2}{32 a}\\ \end{align*}

Mathematica [A]  time = 0.182143, size = 81, normalized size = 0.6 \[ \frac{45 a^4 x^4-105 a^2 x^2-6 a x \left (15 a^4 x^4-40 a^2 x^2+33\right ) \tanh ^{-1}(a x)+45 \left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)^2+68}{288 a \left (a^2 x^2-1\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a*x]/(1 - a^2*x^2)^4,x]

[Out]

(68 - 105*a^2*x^2 + 45*a^4*x^4 - 6*a*x*(33 - 40*a^2*x^2 + 15*a^4*x^4)*ArcTanh[a*x] + 45*(-1 + a^2*x^2)^3*ArcTa
nh[a*x]^2)/(288*a*(-1 + a^2*x^2)^3)

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Maple [B]  time = 0.059, size = 281, normalized size = 2.1 \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) }{48\,a \left ( ax-1 \right ) ^{3}}}+{\frac{{\it Artanh} \left ( ax \right ) }{16\,a \left ( ax-1 \right ) ^{2}}}-{\frac{5\,{\it Artanh} \left ( ax \right ) }{32\,a \left ( ax-1 \right ) }}-{\frac{5\,{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{32\,a}}-{\frac{{\it Artanh} \left ( ax \right ) }{48\,a \left ( ax+1 \right ) ^{3}}}-{\frac{{\it Artanh} \left ( ax \right ) }{16\,a \left ( ax+1 \right ) ^{2}}}-{\frac{5\,{\it Artanh} \left ( ax \right ) }{32\,a \left ( ax+1 \right ) }}+{\frac{5\,{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{32\,a}}-{\frac{5\, \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{128\,a}}+{\frac{5\,\ln \left ( ax-1 \right ) }{64\,a}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{5\, \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{128\,a}}-{\frac{5}{64\,a}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{5\,\ln \left ( ax+1 \right ) }{64\,a}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{1}{288\,a \left ( ax-1 \right ) ^{3}}}-{\frac{7}{384\,a \left ( ax-1 \right ) ^{2}}}+{\frac{37}{384\,a \left ( ax-1 \right ) }}-{\frac{1}{288\,a \left ( ax+1 \right ) ^{3}}}-{\frac{7}{384\,a \left ( ax+1 \right ) ^{2}}}-{\frac{37}{384\,a \left ( ax+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)/(-a^2*x^2+1)^4,x)

[Out]

-1/48/a*arctanh(a*x)/(a*x-1)^3+1/16/a*arctanh(a*x)/(a*x-1)^2-5/32/a*arctanh(a*x)/(a*x-1)-5/32/a*arctanh(a*x)*l
n(a*x-1)-1/48/a*arctanh(a*x)/(a*x+1)^3-1/16/a*arctanh(a*x)/(a*x+1)^2-5/32/a*arctanh(a*x)/(a*x+1)+5/32/a*arctan
h(a*x)*ln(a*x+1)-5/128/a*ln(a*x-1)^2+5/64/a*ln(a*x-1)*ln(1/2+1/2*a*x)-5/128/a*ln(a*x+1)^2-5/64/a*ln(-1/2*a*x+1
/2)*ln(1/2+1/2*a*x)+5/64/a*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/288/a/(a*x-1)^3-7/384/a/(a*x-1)^2+37/384/a/(a*x-1)-1/2
88/a/(a*x+1)^3-7/384/a/(a*x+1)^2-37/384/a/(a*x+1)

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Maxima [B]  time = 1.00347, size = 324, normalized size = 2.42 \begin{align*} -\frac{1}{96} \,{\left (\frac{2 \,{\left (15 \, a^{4} x^{5} - 40 \, a^{2} x^{3} + 33 \, x\right )}}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1} - \frac{15 \, \log \left (a x + 1\right )}{a} + \frac{15 \, \log \left (a x - 1\right )}{a}\right )} \operatorname{artanh}\left (a x\right ) + \frac{{\left (180 \, a^{4} x^{4} - 420 \, a^{2} x^{2} - 45 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} + 90 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 45 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} + 272\right )} a}{1152 \,{\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^4,x, algorithm="maxima")

[Out]

-1/96*(2*(15*a^4*x^5 - 40*a^2*x^3 + 33*x)/(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1) - 15*log(a*x + 1)/a + 15*log(a
*x - 1)/a)*arctanh(a*x) + 1/1152*(180*a^4*x^4 - 420*a^2*x^2 - 45*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(a*x
+ 1)^2 + 90*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) - 45*(a^6*x^6 - 3*a^4*x^4 + 3*a^2
*x^2 - 1)*log(a*x - 1)^2 + 272)*a/(a^8*x^6 - 3*a^6*x^4 + 3*a^4*x^2 - a^2)

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Fricas [A]  time = 1.94932, size = 293, normalized size = 2.19 \begin{align*} \frac{180 \, a^{4} x^{4} - 420 \, a^{2} x^{2} + 45 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - 12 \,{\left (15 \, a^{5} x^{5} - 40 \, a^{3} x^{3} + 33 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) + 272}{1152 \,{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^4,x, algorithm="fricas")

[Out]

1/1152*(180*a^4*x^4 - 420*a^2*x^2 + 45*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1))^2 - 12*
(15*a^5*x^5 - 40*a^3*x^3 + 33*a*x)*log(-(a*x + 1)/(a*x - 1)) + 272)/(a^7*x^6 - 3*a^5*x^4 + 3*a^3*x^2 - a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}{\left (a x \right )}}{\left (a x - 1\right )^{4} \left (a x + 1\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)/(-a**2*x**2+1)**4,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (a x\right )}{{\left (a^{2} x^{2} - 1\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^4,x, algorithm="giac")

[Out]

integrate(arctanh(a*x)/(a^2*x^2 - 1)^4, x)