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

Optimal. Leaf size=70 \[ \frac {1}{4} a^4 x^4 \tanh ^{-1}(a x)+\frac {a^3 x^3}{12}-a^2 x^2 \tanh ^{-1}(a x)-\frac {\text {Li}_2(-a x)}{2}+\frac {\text {Li}_2(a x)}{2}-\frac {3 a x}{4}+\frac {3}{4} \tanh ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6012, 5912, 5916, 321, 206, 302} \[ -\frac {1}{2} \text {PolyLog}(2,-a x)+\frac {1}{2} \text {PolyLog}(2,a x)+\frac {a^3 x^3}{12}+\frac {1}{4} a^4 x^4 \tanh ^{-1}(a x)-a^2 x^2 \tanh ^{-1}(a x)-\frac {3 a x}{4}+\frac {3}{4} \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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

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Mathematica [A]  time = 0.07, size = 73, normalized size = 1.04 \[ \frac {1}{24} \left (6 a^4 x^4 \tanh ^{-1}(a x)+2 a^3 x^3-24 a^2 x^2 \tanh ^{-1}(a x)-12 \text {Li}_2(-a x)+12 \text {Li}_2(a x)-18 a x-9 \log (1-a x)+9 \log (a x+1)\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-18*a*x + 2*a^3*x^3 - 24*a^2*x^2*ArcTanh[a*x] + 6*a^4*x^4*ArcTanh[a*x] - 9*Log[1 - a*x] + 9*Log[1 + a*x] - 12
*PolyLog[2, -(a*x)] + 12*PolyLog[2, a*x])/24

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {artanh}\left (a x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 89, normalized size = 1.27 \[ \frac {a^{4} x^{4} \arctanh \left (a x \right )}{4}-a^{2} x^{2} \arctanh \left (a x \right )+\arctanh \left (a x \right ) \ln \left (a x \right )-\frac {\dilog \left (a x \right )}{2}-\frac {\dilog \left (a x +1\right )}{2}-\frac {\ln \left (a x \right ) \ln \left (a x +1\right )}{2}+\frac {x^{3} a^{3}}{12}-\frac {3 a x}{4}-\frac {3 \ln \left (a x -1\right )}{8}+\frac {3 \ln \left (a x +1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 106, normalized size = 1.51 \[ \frac {1}{24} \, {\left (2 \, a^{2} x^{3} - 18 \, x - \frac {12 \, {\left (\log \left (a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (-a x\right )\right )}}{a} + \frac {12 \, {\left (\log \left (-a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (a x\right )\right )}}{a} + \frac {9 \, \log \left (a x + 1\right )}{a} - \frac {9 \, \log \left (a x - 1\right )}{a}\right )} a + \frac {1}{4} \, {\left (a^{4} x^{4} - 4 \, a^{2} x^{2} + 2 \, \log \left (x^{2}\right )\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(2*a^2*x^3 - 18*x - 12*(log(a*x + 1)*log(x) + dilog(-a*x))/a + 12*(log(-a*x + 1)*log(x) + dilog(a*x))/a +
9*log(a*x + 1)/a - 9*log(a*x - 1)/a)*a + 1/4*(a^4*x^4 - 4*a^2*x^2 + 2*log(x^2))*arctanh(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (a\,x\right )\,{\left (a^2\,x^2-1\right )}^2}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atanh(a*x)*(a^2*x^2 - 1)^2)/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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