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

Optimal. Leaf size=171 \[ \frac {a^2 x^3}{30}+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac {4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}-\frac {8 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{15 a}+\frac {8}{15} x \tanh ^{-1}(a x)^2+\frac {8 \tanh ^{-1}(a x)^2}{15 a}-\frac {16 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{15 a}-\frac {11 x}{30} \]

[Out]

-11/30*x+1/30*a^2*x^3+4/15*(-a^2*x^2+1)*arctanh(a*x)/a+1/10*(-a^2*x^2+1)^2*arctanh(a*x)/a+8/15*arctanh(a*x)^2/
a+8/15*x*arctanh(a*x)^2+4/15*x*(-a^2*x^2+1)*arctanh(a*x)^2+1/5*x*(-a^2*x^2+1)^2*arctanh(a*x)^2-16/15*arctanh(a
*x)*ln(2/(-a*x+1))/a-8/15*polylog(2,1-2/(-a*x+1))/a

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Rubi [A]  time = 0.13, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {5944, 5910, 5984, 5918, 2402, 2315, 8} \[ -\frac {8 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{15 a}+\frac {a^2 x^3}{30}+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac {4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac {8}{15} x \tanh ^{-1}(a x)^2+\frac {8 \tanh ^{-1}(a x)^2}{15 a}-\frac {16 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{15 a}-\frac {11 x}{30} \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)^2*ArcTanh[a*x]^2,x]

[Out]

(-11*x)/30 + (a^2*x^3)/30 + (4*(1 - a^2*x^2)*ArcTanh[a*x])/(15*a) + ((1 - a^2*x^2)^2*ArcTanh[a*x])/(10*a) + (8
*ArcTanh[a*x]^2)/(15*a) + (8*x*ArcTanh[a*x]^2)/15 + (4*x*(1 - a^2*x^2)*ArcTanh[a*x]^2)/15 + (x*(1 - a^2*x^2)^2
*ArcTanh[a*x]^2)/5 - (16*ArcTanh[a*x]*Log[2/(1 - a*x)])/(15*a) - (8*PolyLog[2, 1 - 2/(1 - a*x)])/(15*a)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 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 5944

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(b*p*(d + e*x^2)^q
*(a + b*ArcTanh[c*x])^(p - 1))/(2*c*q*(2*q + 1)), x] + (Dist[(2*d*q)/(2*q + 1), Int[(d + e*x^2)^(q - 1)*(a + b
*ArcTanh[c*x])^p, x], x] - Dist[(b^2*d*p*(p - 1))/(2*q*(2*q + 1)), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTanh[c*x]
)^(p - 2), x], x] + Simp[(x*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^p)/(2*q + 1), x]) /; FreeQ[{a, b, c, d, e}, x]
&& EqQ[c^2*d + e, 0] && GtQ[q, 0] && GtQ[p, 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 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2 \, dx &=\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac {1}{10} \int \left (1-a^2 x^2\right ) \, dx+\frac {4}{5} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2 \, dx\\ &=-\frac {x}{10}+\frac {a^2 x^3}{30}+\frac {4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac {4 \int 1 \, dx}{15}+\frac {8}{15} \int \tanh ^{-1}(a x)^2 \, dx\\ &=-\frac {11 x}{30}+\frac {a^2 x^3}{30}+\frac {4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac {8}{15} x \tanh ^{-1}(a x)^2+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac {1}{15} (16 a) \int \frac {x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=-\frac {11 x}{30}+\frac {a^2 x^3}{30}+\frac {4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac {8 \tanh ^{-1}(a x)^2}{15 a}+\frac {8}{15} x \tanh ^{-1}(a x)^2+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac {16}{15} \int \frac {\tanh ^{-1}(a x)}{1-a x} \, dx\\ &=-\frac {11 x}{30}+\frac {a^2 x^3}{30}+\frac {4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac {8 \tanh ^{-1}(a x)^2}{15 a}+\frac {8}{15} x \tanh ^{-1}(a x)^2+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac {16 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{15 a}+\frac {16}{15} \int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {11 x}{30}+\frac {a^2 x^3}{30}+\frac {4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac {8 \tanh ^{-1}(a x)^2}{15 a}+\frac {8}{15} x \tanh ^{-1}(a x)^2+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac {16 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{15 a}-\frac {16 \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{15 a}\\ &=-\frac {11 x}{30}+\frac {a^2 x^3}{30}+\frac {4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac {8 \tanh ^{-1}(a x)^2}{15 a}+\frac {8}{15} x \tanh ^{-1}(a x)^2+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac {16 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{15 a}-\frac {8 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{15 a}\\ \end {align*}

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Mathematica [A]  time = 0.69, size = 99, normalized size = 0.58 \[ \frac {a x \left (a^2 x^2-11\right )+2 \left (3 a^2 x^2+9 a x+8\right ) (a x-1)^3 \tanh ^{-1}(a x)^2+\tanh ^{-1}(a x) \left (3 a^4 x^4-14 a^2 x^2-32 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )+11\right )+16 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )}{30 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*x*(-11 + a^2*x^2) + 2*(-1 + a*x)^3*(8 + 9*a*x + 3*a^2*x^2)*ArcTanh[a*x]^2 + ArcTanh[a*x]*(11 - 14*a^2*x^2 +
3*a^4*x^4 - 32*Log[1 + E^(-2*ArcTanh[a*x])]) + 16*PolyLog[2, -E^(-2*ArcTanh[a*x])])/(30*a)

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {artanh}\left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 216, normalized size = 1.26 \[ \frac {a^{4} \arctanh \left (a x \right )^{2} x^{5}}{5}-\frac {2 a^{2} \arctanh \left (a x \right )^{2} x^{3}}{3}+x \arctanh \left (a x \right )^{2}+\frac {a^{3} \arctanh \left (a x \right ) x^{4}}{10}-\frac {7 a \arctanh \left (a x \right ) x^{2}}{15}+\frac {8 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{15 a}+\frac {8 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{15 a}+\frac {x^{3} a^{2}}{30}-\frac {11 x}{30}-\frac {11 \ln \left (a x -1\right )}{60 a}+\frac {11 \ln \left (a x +1\right )}{60 a}+\frac {2 \ln \left (a x -1\right )^{2}}{15 a}-\frac {8 \dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{15 a}-\frac {4 \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{15 a}-\frac {2 \ln \left (a x +1\right )^{2}}{15 a}-\frac {4 \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{15 a}+\frac {4 \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{15 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^4*arctanh(a*x)^2*x^5-2/3*a^2*arctanh(a*x)^2*x^3+x*arctanh(a*x)^2+1/10*a^3*arctanh(a*x)*x^4-7/15*a*arctan
h(a*x)*x^2+8/15/a*arctanh(a*x)*ln(a*x-1)+8/15/a*arctanh(a*x)*ln(a*x+1)+1/30*x^3*a^2-11/30*x-11/60/a*ln(a*x-1)+
11/60/a*ln(a*x+1)+2/15/a*ln(a*x-1)^2-8/15/a*dilog(1/2+1/2*a*x)-4/15/a*ln(a*x-1)*ln(1/2+1/2*a*x)-2/15/a*ln(a*x+
1)^2-4/15/a*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)+4/15/a*ln(-1/2*a*x+1/2)*ln(a*x+1)

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maxima [A]  time = 0.33, size = 175, normalized size = 1.02 \[ \frac {1}{60} \, a^{2} {\left (\frac {2 \, a^{3} x^{3} - 22 \, a x - 8 \, \log \left (a x + 1\right )^{2} + 16 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) + 8 \, \log \left (a x - 1\right )^{2} - 11 \, \log \left (a x - 1\right )}{a^{3}} - \frac {32 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a^{3}} + \frac {11 \, \log \left (a x + 1\right )}{a^{3}}\right )} + \frac {1}{30} \, {\left (3 \, a^{2} x^{4} - 14 \, x^{2} + \frac {16 \, \log \left (a x + 1\right )}{a^{2}} + \frac {16 \, \log \left (a x - 1\right )}{a^{2}}\right )} a \operatorname {artanh}\left (a x\right ) + \frac {1}{15} \, {\left (3 \, a^{4} x^{5} - 10 \, a^{2} x^{3} + 15 \, x\right )} \operatorname {artanh}\left (a x\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*a^2*((2*a^3*x^3 - 22*a*x - 8*log(a*x + 1)^2 + 16*log(a*x + 1)*log(a*x - 1) + 8*log(a*x - 1)^2 - 11*log(a*
x - 1))/a^3 - 32*(log(a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))/a^3 + 11*log(a*x + 1)/a^3) + 1/30*(
3*a^2*x^4 - 14*x^2 + 16*log(a*x + 1)/a^2 + 16*log(a*x - 1)/a^2)*a*arctanh(a*x) + 1/15*(3*a^4*x^5 - 10*a^2*x^3
+ 15*x)*arctanh(a*x)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^2*(a^2*x^2 - 1)^2,x)

[Out]

int(atanh(a*x)^2*(a^2*x^2 - 1)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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