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

Optimal. Leaf size=109 \[ -\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {\tanh ^{-1}(a x)^2}{2 a^4}-\frac {\tanh ^{-1}(a x)}{4 a^4}-\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^4}+\frac {\tanh ^{-1}(a x)}{2 a^4 \left (1-a^2 x^2\right )}-\frac {x}{4 a^3 \left (1-a^2 x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 109, 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 = {6028, 5984, 5918, 2402, 2315, 5994, 199, 206} \[ -\frac {\text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {x}{4 a^3 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)}{2 a^4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^2}{2 a^4}-\frac {\tanh ^{-1}(a x)}{4 a^4}-\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(4*a^3*(1 - a^2*x^2)) - ArcTanh[a*x]/(4*a^4) + ArcTanh[a*x]/(2*a^4*(1 - a^2*x^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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 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 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 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]

Rule 5994

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

Rule 6028

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

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 64, normalized size = 0.59 \[ -\frac {-4 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )+4 \tanh ^{-1}(a x)^2+\sinh \left (2 \tanh ^{-1}(a x)\right )-2 \tanh ^{-1}(a x) \left (\cosh \left (2 \tanh ^{-1}(a x)\right )-4 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )\right )}{8 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/8*(4*ArcTanh[a*x]^2 - 2*ArcTanh[a*x]*(Cosh[2*ArcTanh[a*x]] - 4*Log[1 + E^(-2*ArcTanh[a*x])]) - 4*PolyLog[2,
-E^(-2*ArcTanh[a*x])] + Sinh[2*ArcTanh[a*x]])/a^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 203, normalized size = 1.86 \[ -\frac {\arctanh \left (a x \right )}{4 a^{4} \left (a x -1\right )}+\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{2 a^{4}}+\frac {\arctanh \left (a x \right )}{4 a^{4} \left (a x +1\right )}+\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{2 a^{4}}+\frac {\ln \left (a x -1\right )^{2}}{8 a^{4}}-\frac {\dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{2 a^{4}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4 a^{4}}-\frac {\ln \left (a x +1\right )^{2}}{8 a^{4}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{4 a^{4}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4 a^{4}}+\frac {1}{8 a^{4} \left (a x -1\right )}+\frac {\ln \left (a x -1\right )}{8 a^{4}}+\frac {1}{8 a^{4} \left (a x +1\right )}-\frac {\ln \left (a x +1\right )}{8 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 177, normalized size = 1.62 \[ -\frac {1}{8} \, a {\left (\frac {{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 2 \, a x - {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )}{a^{7} x^{2} - a^{5}} + \frac {4 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a^{5}} + \frac {\log \left (a x + 1\right )}{a^{5}}\right )} - \frac {1}{2} \, {\left (\frac {1}{a^{6} x^{2} - a^{4}} - \frac {\log \left (a^{2} x^{2} - 1\right )}{a^{4}}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a*(((a^2*x^2 - 1)*log(a*x + 1)^2 - 2*(a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) - (a^2*x^2 - 1)*log(a*x - 1)
^2 - 2*a*x - (a^2*x^2 - 1)*log(a*x - 1))/(a^7*x^2 - a^5) + 4*(log(a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x
+ 1/2))/a^5 + log(a*x + 1)/a^5) - 1/2*(1/(a^6*x^2 - a^4) - log(a^2*x^2 - 1)/a^4)*arctanh(a*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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