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

Optimal. Leaf size=138 \[ -\frac {2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{15 a^3}+\frac {2 \tanh ^{-1}(a x)^2}{15 a^3}-\frac {\tanh ^{-1}(a x)}{30 a^3}-\frac {4 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{15 a^3}-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)^2+\frac {x}{30 a^2}-\frac {1}{10} a x^4 \tanh ^{-1}(a x)+\frac {1}{3} x^3 \tanh ^{-1}(a x)^2+\frac {2 x^2 \tanh ^{-1}(a x)}{15 a}-\frac {x^3}{30} \]

[Out]

1/30*x/a^2-1/30*x^3-1/30*arctanh(a*x)/a^3+2/15*x^2*arctanh(a*x)/a-1/10*a*x^4*arctanh(a*x)+2/15*arctanh(a*x)^2/
a^3+1/3*x^3*arctanh(a*x)^2-1/5*a^2*x^5*arctanh(a*x)^2-4/15*arctanh(a*x)*ln(2/(-a*x+1))/a^3-2/15*polylog(2,1-2/
(-a*x+1))/a^3

________________________________________________________________________________________

Rubi [A]  time = 0.41, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6014, 5916, 5980, 321, 206, 5984, 5918, 2402, 2315, 302} \[ -\frac {2 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{15 a^3}-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)^2+\frac {x}{30 a^2}+\frac {2 \tanh ^{-1}(a x)^2}{15 a^3}-\frac {\tanh ^{-1}(a x)}{30 a^3}-\frac {4 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{15 a^3}-\frac {1}{10} a x^4 \tanh ^{-1}(a x)+\frac {1}{3} x^3 \tanh ^{-1}(a x)^2+\frac {2 x^2 \tanh ^{-1}(a x)}{15 a}-\frac {x^3}{30} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(30*a^2) - x^3/30 - ArcTanh[a*x]/(30*a^3) + (2*x^2*ArcTanh[a*x])/(15*a) - (a*x^4*ArcTanh[a*x])/10 + (2*ArcTa
nh[a*x]^2)/(15*a^3) + (x^3*ArcTanh[a*x]^2)/3 - (a^2*x^5*ArcTanh[a*x]^2)/5 - (4*ArcTanh[a*x]*Log[2/(1 - a*x)])/
(15*a^3) - (2*PolyLog[2, 1 - 2/(1 - a*x)])/(15*a^3)

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 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]

Rule 6014

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

Rubi steps

\begin {align*} \int x^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2 \, dx &=-\left (a^2 \int x^4 \tanh ^{-1}(a x)^2 \, dx\right )+\int x^2 \tanh ^{-1}(a x)^2 \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(a x)^2-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)^2-\frac {1}{3} (2 a) \int \frac {x^3 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx+\frac {1}{5} \left (2 a^3\right ) \int \frac {x^5 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(a x)^2-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)^2+\frac {2 \int x \tanh ^{-1}(a x) \, dx}{3 a}-\frac {2 \int \frac {x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{3 a}-\frac {1}{5} (2 a) \int x^3 \tanh ^{-1}(a x) \, dx+\frac {1}{5} (2 a) \int \frac {x^3 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=\frac {x^2 \tanh ^{-1}(a x)}{3 a}-\frac {1}{10} a x^4 \tanh ^{-1}(a x)+\frac {\tanh ^{-1}(a x)^2}{3 a^3}+\frac {1}{3} x^3 \tanh ^{-1}(a x)^2-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)^2-\frac {1}{3} \int \frac {x^2}{1-a^2 x^2} \, dx-\frac {2 \int \frac {\tanh ^{-1}(a x)}{1-a x} \, dx}{3 a^2}-\frac {2 \int x \tanh ^{-1}(a x) \, dx}{5 a}+\frac {2 \int \frac {x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{5 a}+\frac {1}{10} a^2 \int \frac {x^4}{1-a^2 x^2} \, dx\\ &=\frac {x}{3 a^2}+\frac {2 x^2 \tanh ^{-1}(a x)}{15 a}-\frac {1}{10} a x^4 \tanh ^{-1}(a x)+\frac {2 \tanh ^{-1}(a x)^2}{15 a^3}+\frac {1}{3} x^3 \tanh ^{-1}(a x)^2-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)^2-\frac {2 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{3 a^3}+\frac {1}{5} \int \frac {x^2}{1-a^2 x^2} \, dx-\frac {\int \frac {1}{1-a^2 x^2} \, dx}{3 a^2}+\frac {2 \int \frac {\tanh ^{-1}(a x)}{1-a x} \, dx}{5 a^2}+\frac {2 \int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{3 a^2}+\frac {1}{10} a^2 \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 {x}{30 a^2}-\frac {x^3}{30}-\frac {\tanh ^{-1}(a x)}{3 a^3}+\frac {2 x^2 \tanh ^{-1}(a x)}{15 a}-\frac {1}{10} a x^4 \tanh ^{-1}(a x)+\frac {2 \tanh ^{-1}(a x)^2}{15 a^3}+\frac {1}{3} x^3 \tanh ^{-1}(a x)^2-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)^2-\frac {4 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{15 a^3}-\frac {2 \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{3 a^3}+\frac {\int \frac {1}{1-a^2 x^2} \, dx}{10 a^2}+\frac {\int \frac {1}{1-a^2 x^2} \, dx}{5 a^2}-\frac {2 \int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{5 a^2}\\ &=\frac {x}{30 a^2}-\frac {x^3}{30}-\frac {\tanh ^{-1}(a x)}{30 a^3}+\frac {2 x^2 \tanh ^{-1}(a x)}{15 a}-\frac {1}{10} a x^4 \tanh ^{-1}(a x)+\frac {2 \tanh ^{-1}(a x)^2}{15 a^3}+\frac {1}{3} x^3 \tanh ^{-1}(a x)^2-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)^2-\frac {4 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{15 a^3}-\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{3 a^3}+\frac {2 \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{5 a^3}\\ &=\frac {x}{30 a^2}-\frac {x^3}{30}-\frac {\tanh ^{-1}(a x)}{30 a^3}+\frac {2 x^2 \tanh ^{-1}(a x)}{15 a}-\frac {1}{10} a x^4 \tanh ^{-1}(a x)+\frac {2 \tanh ^{-1}(a x)^2}{15 a^3}+\frac {1}{3} x^3 \tanh ^{-1}(a x)^2-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)^2-\frac {4 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{15 a^3}-\frac {2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{15 a^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 95, normalized size = 0.69 \[ -\frac {a^3 x^3+2 \left (3 a^5 x^5-5 a^3 x^3+2\right ) \tanh ^{-1}(a x)^2+\tanh ^{-1}(a x) \left (3 a^4 x^4-4 a^2 x^2+8 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )+1\right )-4 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )-a x}{30 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/30*(-(a*x) + a^3*x^3 + 2*(2 - 5*a^3*x^3 + 3*a^5*x^5)*ArcTanh[a*x]^2 + ArcTanh[a*x]*(1 - 4*a^2*x^2 + 3*a^4*x
^4 + 8*Log[1 + E^(-2*ArcTanh[a*x])]) - 4*PolyLog[2, -E^(-2*ArcTanh[a*x])])/a^3

________________________________________________________________________________________

fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} x^{4} - x^{2}\right )} \operatorname {artanh}\left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 205, normalized size = 1.49 \[ -\frac {a^{2} x^{5} \arctanh \left (a x \right )^{2}}{5}+\frac {x^{3} \arctanh \left (a x \right )^{2}}{3}-\frac {a \,x^{4} \arctanh \left (a x \right )}{10}+\frac {2 x^{2} \arctanh \left (a x \right )}{15 a}+\frac {2 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{15 a^{3}}+\frac {2 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{15 a^{3}}+\frac {\ln \left (a x -1\right )^{2}}{30 a^{3}}-\frac {2 \dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{15 a^{3}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{15 a^{3}}-\frac {\ln \left (a x +1\right )^{2}}{30 a^{3}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{15 a^{3}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{15 a^{3}}-\frac {x^{3}}{30}+\frac {x}{30 a^{2}}+\frac {\ln \left (a x -1\right )}{60 a^{3}}-\frac {\ln \left (a x +1\right )}{60 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*a^2*x^5*arctanh(a*x)^2+1/3*x^3*arctanh(a*x)^2-1/10*a*x^4*arctanh(a*x)+2/15*x^2*arctanh(a*x)/a+2/15/a^3*ar
ctanh(a*x)*ln(a*x-1)+2/15/a^3*arctanh(a*x)*ln(a*x+1)+1/30/a^3*ln(a*x-1)^2-2/15/a^3*dilog(1/2+1/2*a*x)-1/15/a^3
*ln(a*x-1)*ln(1/2+1/2*a*x)-1/30/a^3*ln(a*x+1)^2+1/15/a^3*ln(-1/2*a*x+1/2)*ln(a*x+1)-1/15/a^3*ln(-1/2*a*x+1/2)*
ln(1/2+1/2*a*x)-1/30*x^3+1/30*x/a^2+1/60/a^3*ln(a*x-1)-1/60/a^3*ln(a*x+1)

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 173, normalized size = 1.25 \[ -\frac {1}{60} \, a^{2} {\left (\frac {2 \, a^{3} x^{3} - 2 \, a x + 2 \, \log \left (a x + 1\right )^{2} - 4 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 2 \, \log \left (a x - 1\right )^{2} - \log \left (a x - 1\right )}{a^{5}} + \frac {8 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a^{5}} + \frac {\log \left (a x + 1\right )}{a^{5}}\right )} - \frac {1}{30} \, a {\left (\frac {3 \, a^{2} x^{4} - 4 \, x^{2}}{a^{2}} - \frac {4 \, \log \left (a x + 1\right )}{a^{4}} - \frac {4 \, \log \left (a x - 1\right )}{a^{4}}\right )} \operatorname {artanh}\left (a x\right ) - \frac {1}{15} \, {\left (3 \, a^{2} x^{5} - 5 \, x^{3}\right )} \operatorname {artanh}\left (a x\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*a^2*((2*a^3*x^3 - 2*a*x + 2*log(a*x + 1)^2 - 4*log(a*x + 1)*log(a*x - 1) - 2*log(a*x - 1)^2 - log(a*x -
1))/a^5 + 8*(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) - 1/30*a*((3*a^2
*x^4 - 4*x^2)/a^2 - 4*log(a*x + 1)/a^4 - 4*log(a*x - 1)/a^4)*arctanh(a*x) - 1/15*(3*a^2*x^5 - 5*x^3)*arctanh(a
*x)^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- x^{2} \operatorname {atanh}^{2}{\left (a x \right )}\right )\, dx - \int a^{2} x^{4} \operatorname {atanh}^{2}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-x**2*atanh(a*x)**2, x) - Integral(a**2*x**4*atanh(a*x)**2, x)

________________________________________________________________________________________