3.170 \(\int \frac {(1-a^2 x^2) \tanh ^{-1}(a x)}{x^5} \, dx\)

Optimal. Leaf size=42 \[ \frac {a^3}{4 x}-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 x^4}-\frac {a}{12 x^3} \]

[Out]

-1/12*a/x^3+1/4*a^3/x-1/4*(-a^2*x^2+1)^2*arctanh(a*x)/x^4

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Rubi [A]  time = 0.03, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6008, 14} \[ -\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 x^4}+\frac {a^3}{4 x}-\frac {a}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)*ArcTanh[a*x])/x^5,x]

[Out]

-a/(12*x^3) + a^3/(4*x) - ((1 - a^2*x^2)^2*ArcTanh[a*x])/(4*x^4)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{x^5} \, dx &=-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 x^4}+\frac {1}{4} a \int \frac {1-a^2 x^2}{x^4} \, dx\\ &=-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 x^4}+\frac {1}{4} a \int \left (\frac {1}{x^4}-\frac {a^2}{x^2}\right ) \, dx\\ &=-\frac {a}{12 x^3}+\frac {a^3}{4 x}-\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 x^4}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 71, normalized size = 1.69 \[ \frac {1}{8} a^4 \log (1-a x)-\frac {1}{8} a^4 \log (a x+1)+\frac {a^3}{4 x}+\frac {a^2 \tanh ^{-1}(a x)}{2 x^2}-\frac {\tanh ^{-1}(a x)}{4 x^4}-\frac {a}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - a^2*x^2)*ArcTanh[a*x])/x^5,x]

[Out]

-1/12*a/x^3 + a^3/(4*x) - ArcTanh[a*x]/(4*x^4) + (a^2*ArcTanh[a*x])/(2*x^2) + (a^4*Log[1 - a*x])/8 - (a^4*Log[
1 + a*x])/8

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fricas [A]  time = 1.41, size = 52, normalized size = 1.24 \[ \frac {6 \, a^{3} x^{3} - 2 \, a x - 3 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{24 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^5,x, algorithm="fricas")

[Out]

1/24*(6*a^3*x^3 - 2*a*x - 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1)))/x^4

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giac [B]  time = 0.18, size = 160, normalized size = 3.81 \[ -\frac {1}{3} \, a {\left (\frac {a^{3} {\left (\frac {3 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}}{{\left (\frac {a x + 1}{a x - 1} + 1\right )}^{3}} + \frac {6 \, {\left (a x + 1\right )}^{2} a^{3} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{{\left (a x - 1\right )}^{2} {\left (\frac {a x + 1}{a x - 1} + 1\right )}^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^5,x, algorithm="giac")

[Out]

-1/3*a*(a^3*(3*(a*x + 1)/(a*x - 1) + 1)/((a*x + 1)/(a*x - 1) + 1)^3 + 6*(a*x + 1)^2*a^3*log(-(a*((a*x + 1)/(a*
x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) + 1)/(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) - 1))/((
a*x - 1)^2*((a*x + 1)/(a*x - 1) + 1)^4))

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maple [A]  time = 0.04, size = 59, normalized size = 1.40 \[ \frac {a^{2} \arctanh \left (a x \right )}{2 x^{2}}-\frac {\arctanh \left (a x \right )}{4 x^{4}}+\frac {a^{3}}{4 x}-\frac {a}{12 x^{3}}+\frac {a^{4} \ln \left (a x -1\right )}{8}-\frac {a^{4} \ln \left (a x +1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)*arctanh(a*x)/x^5,x)

[Out]

1/2*a^2*arctanh(a*x)/x^2-1/4*arctanh(a*x)/x^4+1/4*a^3/x-1/12*a/x^3+1/8*a^4*ln(a*x-1)-1/8*a^4*ln(a*x+1)

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maxima [A]  time = 0.30, size = 61, normalized size = 1.45 \[ -\frac {1}{24} \, {\left (3 \, a^{3} \log \left (a x + 1\right ) - 3 \, a^{3} \log \left (a x - 1\right ) - \frac {2 \, {\left (3 \, a^{2} x^{2} - 1\right )}}{x^{3}}\right )} a + \frac {{\left (2 \, a^{2} x^{2} - 1\right )} \operatorname {artanh}\left (a x\right )}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^5,x, algorithm="maxima")

[Out]

-1/24*(3*a^3*log(a*x + 1) - 3*a^3*log(a*x - 1) - 2*(3*a^2*x^2 - 1)/x^3)*a + 1/4*(2*a^2*x^2 - 1)*arctanh(a*x)/x
^4

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mupad [B]  time = 0.84, size = 61, normalized size = 1.45 \[ \frac {a^3}{4\,x}-\frac {\mathrm {atanh}\left (a\,x\right )}{4\,x^4}-\frac {a}{12\,x^3}+\frac {a^5\,\mathrm {atan}\left (\frac {a^2\,x}{\sqrt {-a^2}}\right )}{4\,\sqrt {-a^2}}+\frac {a^2\,\mathrm {atanh}\left (a\,x\right )}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(atanh(a*x)*(a^2*x^2 - 1))/x^5,x)

[Out]

a^3/(4*x) - atanh(a*x)/(4*x^4) - a/(12*x^3) + (a^5*atan((a^2*x)/(-a^2)^(1/2)))/(4*(-a^2)^(1/2)) + (a^2*atanh(a
*x))/(2*x^2)

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sympy [A]  time = 0.89, size = 46, normalized size = 1.10 \[ - \frac {a^{4} \operatorname {atanh}{\left (a x \right )}}{4} + \frac {a^{3}}{4 x} + \frac {a^{2} \operatorname {atanh}{\left (a x \right )}}{2 x^{2}} - \frac {a}{12 x^{3}} - \frac {\operatorname {atanh}{\left (a x \right )}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)*atanh(a*x)/x**5,x)

[Out]

-a**4*atanh(a*x)/4 + a**3/(4*x) + a**2*atanh(a*x)/(2*x**2) - a/(12*x**3) - atanh(a*x)/(4*x**4)

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