3.3.1 \(\int \frac {(1-a^2 x^2)^2 \tanh ^{-1}(a x)}{x^5} \, dx\) [201]

Optimal. Leaf size=77 \[ -\frac {a}{12 x^3}+\frac {3 a^3}{4 x}-\frac {3}{4} a^4 \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)}{4 x^4}+\frac {a^2 \tanh ^{-1}(a x)}{x^2}-\frac {1}{2} a^4 \text {PolyLog}(2,-a x)+\frac {1}{2} a^4 \text {PolyLog}(2,a x) \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6159, 6037, 331, 212, 6031} \begin {gather*} -\frac {1}{2} a^4 \text {Li}_2(-a x)+\frac {1}{2} a^4 \text {Li}_2(a x)-\frac {3}{4} a^4 \tanh ^{-1}(a x)+\frac {3 a^3}{4 x}+\frac {a^2 \tanh ^{-1}(a x)}{x^2}-\frac {\tanh ^{-1}(a x)}{4 x^4}-\frac {a}{12 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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^5} \, dx &=\int \left (\frac {\tanh ^{-1}(a x)}{x^5}-\frac {2 a^2 \tanh ^{-1}(a x)}{x^3}+\frac {a^4 \tanh ^{-1}(a x)}{x}\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int \frac {\tanh ^{-1}(a x)}{x^3} \, dx\right )+a^4 \int \frac {\tanh ^{-1}(a x)}{x} \, dx+\int \frac {\tanh ^{-1}(a x)}{x^5} \, dx\\ &=-\frac {\tanh ^{-1}(a x)}{4 x^4}+\frac {a^2 \tanh ^{-1}(a x)}{x^2}-\frac {1}{2} a^4 \text {Li}_2(-a x)+\frac {1}{2} a^4 \text {Li}_2(a x)+\frac {1}{4} a \int \frac {1}{x^4 \left (1-a^2 x^2\right )} \, dx-a^3 \int \frac {1}{x^2 \left (1-a^2 x^2\right )} \, dx\\ &=-\frac {a}{12 x^3}+\frac {a^3}{x}-\frac {\tanh ^{-1}(a x)}{4 x^4}+\frac {a^2 \tanh ^{-1}(a x)}{x^2}-\frac {1}{2} a^4 \text {Li}_2(-a x)+\frac {1}{2} a^4 \text {Li}_2(a x)+\frac {1}{4} a^3 \int \frac {1}{x^2 \left (1-a^2 x^2\right )} \, dx-a^5 \int \frac {1}{1-a^2 x^2} \, dx\\ &=-\frac {a}{12 x^3}+\frac {3 a^3}{4 x}-a^4 \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)}{4 x^4}+\frac {a^2 \tanh ^{-1}(a x)}{x^2}-\frac {1}{2} a^4 \text {Li}_2(-a x)+\frac {1}{2} a^4 \text {Li}_2(a x)+\frac {1}{4} a^5 \int \frac {1}{1-a^2 x^2} \, dx\\ &=-\frac {a}{12 x^3}+\frac {3 a^3}{4 x}-\frac {3}{4} a^4 \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)}{4 x^4}+\frac {a^2 \tanh ^{-1}(a x)}{x^2}-\frac {1}{2} a^4 \text {Li}_2(-a x)+\frac {1}{2} a^4 \text {Li}_2(a x)\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 89, normalized size = 1.16 \begin {gather*} -\frac {a}{12 x^3}+\frac {3 a^3}{4 x}-\frac {\tanh ^{-1}(a x)}{4 x^4}+\frac {a^2 \tanh ^{-1}(a x)}{x^2}+\frac {3}{8} a^4 \log (1-a x)-\frac {3}{8} a^4 \log (1+a x)+\frac {1}{2} a^4 (-\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^5,x]

[Out]

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

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Maple [A]
time = 0.30, size = 96, normalized size = 1.25

method	result	size
derivativedivides	\(a^{4} \left (\arctanh \left (a x \right ) \ln \left (a x \right )-\frac {\arctanh \left (a x \right )}{4 a^{4} x^{4}}+\frac {\arctanh \left (a x \right )}{a^{2} x^{2}}-\frac {\dilog \left (a x \right )}{2}-\frac {\dilog \left (a x +1\right )}{2}-\frac {\ln \left (a x \right ) \ln \left (a x +1\right )}{2}-\frac {1}{12 a^{3} x^{3}}+\frac {3}{4 a x}-\frac {3 \ln \left (a x +1\right )}{8}+\frac {3 \ln \left (a x -1\right )}{8}\right )\)	\(96\)
default	\(a^{4} \left (\arctanh \left (a x \right ) \ln \left (a x \right )-\frac {\arctanh \left (a x \right )}{4 a^{4} x^{4}}+\frac {\arctanh \left (a x \right )}{a^{2} x^{2}}-\frac {\dilog \left (a x \right )}{2}-\frac {\dilog \left (a x +1\right )}{2}-\frac {\ln \left (a x \right ) \ln \left (a x +1\right )}{2}-\frac {1}{12 a^{3} x^{3}}+\frac {3}{4 a x}-\frac {3 \ln \left (a x +1\right )}{8}+\frac {3 \ln \left (a x -1\right )}{8}\right )\)	\(96\)
risch	\(-\frac {a}{12 x^{3}}+\frac {3 a^{3}}{4 x}+\frac {3 a^{4} \ln \left (a x \right )}{8}-\frac {3 a^{4} \ln \left (a x +1\right )}{8}-\frac {\ln \left (a x +1\right )}{8 x^{4}}-\frac {a^{4} \dilog \left (a x +1\right )}{2}+\frac {a^{2} \ln \left (a x +1\right )}{2 x^{2}}-\frac {3 a^{4} \ln \left (-a x \right )}{8}+\frac {3 a^{4} \ln \left (-a x +1\right )}{8}+\frac {\ln \left (-a x +1\right )}{8 x^{4}}+\frac {a^{4} \dilog \left (-a x +1\right )}{2}-\frac {a^{2} \ln \left (-a x +1\right )}{2 x^{2}}\)	\(133\)
meijerg	\(-\frac {i a^{4} \left (-\frac {i}{3 x^{3} a^{3}}-\frac {i}{x a}+\frac {4 i \left (\frac {3}{8}-\frac {3 a^{4} x^{4}}{8}\right ) \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 x^{3} a^{3} \sqrt {a^{2} x^{2}}}\right )}{4}-\frac {i a^{4} \left (\frac {2 i a x \polylog \left (2, \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-\frac {2 i a x \polylog \left (2, -\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}\right )}{4}-\frac {i a^{4} \left (\frac {2 i}{x a}+\frac {2 i \left (-a x +1\right ) \left (a x +1\right ) \arctanh \left (a x \right )}{x^{2} a^{2}}\right )}{2}\)	\(183\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 112, normalized size = 1.45 \begin {gather*} -\frac {1}{24} \, {\left (12 \, {\left (\log \left (a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-a x\right )\right )} a^{3} - 12 \, {\left (\log \left (-a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (a x\right )\right )} a^{3} + 9 \, a^{3} \log \left (a x + 1\right ) - 9 \, a^{3} \log \left (a x - 1\right ) - \frac {2 \, {\left (9 \, a^{2} x^{2} - 1\right )}}{x^{3}}\right )} a + \frac {1}{4} \, {\left (2 \, a^{4} \log \left (x^{2}\right ) + \frac {4 \, a^{2} x^{2} - 1}{x^{4}}\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^5,x, algorithm="maxima")

[Out]

-1/24*(12*(log(a*x + 1)*log(x) + dilog(-a*x))*a^3 - 12*(log(-a*x + 1)*log(x) + dilog(a*x))*a^3 + 9*a^3*log(a*x
+ 1) - 9*a^3*log(a*x - 1) - 2*(9*a^2*x^2 - 1)/x^3)*a + 1/4*(2*a^4*log(x^2) + (4*a^2*x^2 - 1)/x^4)*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^5,x, algorithm="fricas")

[Out]

integral((a^4*x^4 - 2*a^2*x^2 + 1)*arctanh(a*x)/x^5, 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^{5}}\, 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**5,x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {atanh}\left (a\,x\right )\,{\left (a^2\,x^2-1\right )}^2}{x^5} \,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^5,x)

[Out]

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

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