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

Optimal. Leaf size=87 \[ \frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {\tanh ^{-1}(a x)^2}{2 a^4}+\frac {\tanh ^{-1}(a x)}{2 a^4}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^4}-\frac {x}{2 a^3}-\frac {x^2 \tanh ^{-1}(a x)}{2 a^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5980, 5916, 321, 206, 5984, 5918, 2402, 2315} \[ \frac {\text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {x^2 \tanh ^{-1}(a x)}{2 a^2}-\frac {x}{2 a^3}-\frac {\tanh ^{-1}(a x)^2}{2 a^4}+\frac {\tanh ^{-1}(a x)}{2 a^4}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 5918

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

Rule 5980

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

Rule 5984

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 60, normalized size = 0.69 \[ \frac {\tanh ^{-1}(a x) \left (-a^2 x^2+2 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )+1\right )-\text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )-a x+\tanh ^{-1}(a x)^2}{2 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(a*x) + ArcTanh[a*x]^2 + ArcTanh[a*x]*(1 - a^2*x^2 + 2*Log[1 + E^(-2*ArcTanh[a*x])]) - PolyLog[2, -E^(-2*Arc
Tanh[a*x])])/(2*a^4)

________________________________________________________________________________________

fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {x^{3} \operatorname {artanh}\left (a x\right )}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.06, size = 165, normalized size = 1.90 \[ -\frac {x^{2} \arctanh \left (a x \right )}{2 a^{2}}-\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{2 a^{4}}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{2 a^{4}}-\frac {x}{2 a^{3}}-\frac {\ln \left (a x -1\right )}{4 a^{4}}+\frac {\ln \left (a x +1\right )}{4 a^{4}}-\frac {\ln \left (a x -1\right )^{2}}{8 a^{4}}+\frac {\dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{2 a^{4}}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4 a^{4}}+\frac {\ln \left (a x +1\right )^{2}}{8 a^{4}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{4 a^{4}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 120, normalized size = 1.38 \[ -\frac {1}{8} \, a {\left (\frac {4 \, a x - \log \left (a x + 1\right )^{2} + 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) + \log \left (a x - 1\right )^{2} + 2 \, \log \left (a x - 1\right )}{a^{5}} - \frac {4 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a^{5}} - \frac {2 \, \log \left (a x + 1\right )}{a^{5}}\right )} - \frac {1}{2} \, {\left (\frac {x^{2}}{a^{2}} + \frac {\log \left (a^{2} x^{2} - 1\right )}{a^{4}}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a*((4*a*x - log(a*x + 1)^2 + 2*log(a*x + 1)*log(a*x - 1) + log(a*x - 1)^2 + 2*log(a*x - 1))/a^5 - 4*(log(
a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))/a^5 - 2*log(a*x + 1)/a^5) - 1/2*(x^2/a^2 + log(a^2*x^2 -
1)/a^4)*arctanh(a*x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {x^3\,\mathrm {atanh}\left (a\,x\right )}{a^2\,x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________