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

Optimal. Leaf size=62 \[ \frac {1}{2} a^4 x^2 \tanh ^{-1}(a x)+\frac {a^3 x}{2}+a^2 \text {Li}_2(-a x)-a^2 \text {Li}_2(a x)-\frac {\tanh ^{-1}(a x)}{2 x^2}-\frac {a}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6012, 5916, 325, 206, 5912, 321} \[ a^2 \text {PolyLog}(2,-a x)-a^2 \text {PolyLog}(2,a x)+\frac {1}{2} a^4 x^2 \tanh ^{-1}(a x)+\frac {a^3 x}{2}-\frac {\tanh ^{-1}(a x)}{2 x^2}-\frac {a}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(2*x) + (a^3*x)/2 - ArcTanh[a*x]/(2*x^2) + (a^4*x^2*ArcTanh[a*x])/2 + a^2*PolyLog[2, -(a*x)] - a^2*PolyLog[
2, a*x]

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 5912

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b*PolyLog[2, -(c*x)])/2
, x] + Simp[(b*PolyLog[2, c*x])/2, x]) /; FreeQ[{a, b, c}, x]

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 6012

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

Rubi steps

\begin {align*} \int \frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{x^3} \, dx &=\int \left (\frac {\tanh ^{-1}(a x)}{x^3}-\frac {2 a^2 \tanh ^{-1}(a x)}{x}+a^4 x \tanh ^{-1}(a x)\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int \frac {\tanh ^{-1}(a x)}{x} \, dx\right )+a^4 \int x \tanh ^{-1}(a x) \, dx+\int \frac {\tanh ^{-1}(a x)}{x^3} \, dx\\ &=-\frac {\tanh ^{-1}(a x)}{2 x^2}+\frac {1}{2} a^4 x^2 \tanh ^{-1}(a x)+a^2 \text {Li}_2(-a x)-a^2 \text {Li}_2(a x)+\frac {1}{2} a \int \frac {1}{x^2 \left (1-a^2 x^2\right )} \, dx-\frac {1}{2} a^5 \int \frac {x^2}{1-a^2 x^2} \, dx\\ &=-\frac {a}{2 x}+\frac {a^3 x}{2}-\frac {\tanh ^{-1}(a x)}{2 x^2}+\frac {1}{2} a^4 x^2 \tanh ^{-1}(a x)+a^2 \text {Li}_2(-a x)-a^2 \text {Li}_2(a x)\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 62, normalized size = 1.00 \[ -\frac {-a^4 x^4 \tanh ^{-1}(a x)-a^3 x^3-2 a^2 x^2 \text {Li}_2(-a x)+2 a^2 x^2 \text {Li}_2(a x)+a x+\tanh ^{-1}(a x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {artanh}\left (a x\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 80, normalized size = 1.29 \[ \frac {a^{4} x^{2} \arctanh \left (a x \right )}{2}-2 a^{2} \arctanh \left (a x \right ) \ln \left (a x \right )-\frac {\arctanh \left (a x \right )}{2 x^{2}}+a^{2} \dilog \left (a x \right )+a^{2} \dilog \left (a x +1\right )+a^{2} \ln \left (a x \right ) \ln \left (a x +1\right )+\frac {a^{3} x}{2}-\frac {a}{2 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^4*x^2*arctanh(a*x)-2*a^2*arctanh(a*x)*ln(a*x)-1/2*arctanh(a*x)/x^2+a^2*dilog(a*x)+a^2*dilog(a*x+1)+a^2*l
n(a*x)*ln(a*x+1)+1/2*a^3*x-1/2*a/x

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maxima [A]  time = 0.31, size = 82, normalized size = 1.32 \[ \frac {1}{2} \, {\left (2 \, {\left (\log \left (a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (-a x\right )\right )} a - 2 \, {\left (\log \left (-a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (a x\right )\right )} a + \frac {a^{2} x^{2} - 1}{x}\right )} a + \frac {1}{2} \, {\left (a^{4} x^{2} - 2 \, a^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {atanh}\left (a\,x\right )\,{\left (a^2\,x^2-1\right )}^2}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}{\left (a x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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