3.2.99 \(\int \frac {(1-a^2 x^2)^2 \tanh ^{-1}(a x)}{x^3} \, dx\) [199]

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

[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 = 0.07, 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 = {6159, 6037, 331, 212, 6031, 327} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*a/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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 327

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 331

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 6031

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

Rule 6037

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

Rule 6159

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 = 0.04, size = 61, normalized size = 0.98 \begin {gather*} -\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 {PolyLog}(2,-a x)+\text {PolyLog}(2,a x)) \end {gather*}

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

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Maple [A]
time = 0.29, size = 73, normalized size = 1.18

method	result	size
derivativedivides	\(a^{2} \left (\frac {a^{2} x^{2} \arctanh \left (a x \right )}{2}-\frac {\arctanh \left (a x \right )}{2 a^{2} x^{2}}-2 \arctanh \left (a x \right ) \ln \left (a x \right )+\dilog \left (a x +1\right )+\ln \left (a x \right ) \ln \left (a x +1\right )+\dilog \left (a x \right )+\frac {a x}{2}-\frac {1}{2 a x}\right )\)	\(73\)
default	\(a^{2} \left (\frac {a^{2} x^{2} \arctanh \left (a x \right )}{2}-\frac {\arctanh \left (a x \right )}{2 a^{2} x^{2}}-2 \arctanh \left (a x \right ) \ln \left (a x \right )+\dilog \left (a x +1\right )+\ln \left (a x \right ) \ln \left (a x +1\right )+\dilog \left (a x \right )+\frac {a x}{2}-\frac {1}{2 a x}\right )\)	\(73\)
risch	\(\frac {a^{4} \ln \left (a x +1\right ) x^{2}}{4}+\frac {a^{3} x}{2}+a^{2} \dilog \left (a x +1\right )-\frac {a}{2 x}-\frac {a^{2} \ln \left (a x \right )}{4}-\frac {\ln \left (a x +1\right )}{4 x^{2}}-\frac {a^{4} \ln \left (-a x +1\right ) x^{2}}{4}-a^{2} \dilog \left (-a x +1\right )+\frac {a^{2} \ln \left (-a x \right )}{4}+\frac {\ln \left (-a x +1\right )}{4 x^{2}}\)	\(107\)
meijerg	\(\frac {i a^{2} \left (\frac {2 i}{x a}+\frac {2 i \left (-a x +1\right ) \left (a x +1\right ) \arctanh \left (a x \right )}{x^{2} a^{2}}\right )}{4}+\frac {i a^{2} \left (-2 i x a +2 i \left (-a x +1\right ) \left (a x +1\right ) \arctanh \left (a x \right )\right )}{4}+\frac {i a^{2} \left (\frac {2 i a x \polylog \left (2, \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-\frac {2 i a x \polylog \left (2, -\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}\right )}{2}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(1/2*a^2*x^2*arctanh(a*x)-1/2*arctanh(a*x)/a^2/x^2-2*arctanh(a*x)*ln(a*x)+dilog(a*x+1)+ln(a*x)*ln(a*x+1)+d
ilog(a*x)+1/2*a*x-1/2/a/x)

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Maxima [A]
time = 0.27, size = 82, normalized size = 1.32 \begin {gather*} \frac {1}{2} \, {\left (2 \, {\left (\log \left (a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-a x\right )\right )} a - 2 \, {\left (\log \left (-a x + 1\right ) \log \left (x\right ) + {\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 ) \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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