3.3.9 \(\int \frac {(1-a^2 x^2)^2 \tanh ^{-1}(a x)^2}{x^2} \, dx\) [209]

Optimal. Leaf size=156 \[ \frac {a^2 x}{3}-\frac {1}{3} a \tanh ^{-1}(a x)+\frac {1}{3} a^3 x^2 \tanh ^{-1}(a x)-\frac {2}{3} a \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{x}-2 a^2 x \tanh ^{-1}(a x)^2+\frac {1}{3} a^4 x^3 \tanh ^{-1}(a x)^2+\frac {10}{3} a \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )+2 a \tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )+\frac {5}{3} a \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )-a \text {PolyLog}\left (2,-1+\frac {2}{1+a x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.30, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 13, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {6159, 6021, 6131, 6055, 2449, 2352, 6037, 6135, 6079, 2497, 6127, 327, 212} \begin {gather*} \frac {1}{3} a^4 x^3 \tanh ^{-1}(a x)^2+\frac {1}{3} a^3 x^2 \tanh ^{-1}(a x)+\frac {a^2 x}{3}-2 a^2 x \tanh ^{-1}(a x)^2+\frac {5}{3} a \text {Li}_2\left (1-\frac {2}{1-a x}\right )-a \text {Li}_2\left (\frac {2}{a x+1}-1\right )-\frac {2}{3} a \tanh ^{-1}(a x)^2-\frac {1}{3} a \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)^2}{x}+\frac {10}{3} a \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)+2 a \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 327

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 2352

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

Rule 2449

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 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 6021

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

Rule 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rule 6055

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 6079

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTanh[c*x
])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 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 6127

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 6131

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 6135

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

Rule 6159

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

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Mathematica [A]
time = 0.30, size = 149, normalized size = 0.96 \begin {gather*} \frac {a^2 x^2-a x \tanh ^{-1}(a x)+a^3 x^3 \tanh ^{-1}(a x)-3 \tanh ^{-1}(a x)^2+8 a x \tanh ^{-1}(a x)^2-6 a^2 x^2 \tanh ^{-1}(a x)^2+a^4 x^4 \tanh ^{-1}(a x)^2+6 a x \tanh ^{-1}(a x) \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )+10 a x \tanh ^{-1}(a x) \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )-5 a x \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )-3 a x \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(a x)}\right )}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x^2 - a*x*ArcTanh[a*x] + a^3*x^3*ArcTanh[a*x] - 3*ArcTanh[a*x]^2 + 8*a*x*ArcTanh[a*x]^2 - 6*a^2*x^2*ArcTa
nh[a*x]^2 + a^4*x^4*ArcTanh[a*x]^2 + 6*a*x*ArcTanh[a*x]*Log[1 - E^(-2*ArcTanh[a*x])] + 10*a*x*ArcTanh[a*x]*Log
[1 + E^(-2*ArcTanh[a*x])] - 5*a*x*PolyLog[2, -E^(-2*ArcTanh[a*x])] - 3*a*x*PolyLog[2, E^(-2*ArcTanh[a*x])])/(3
*x)

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Maple [A]
time = 0.58, size = 203, normalized size = 1.30

method	result	size
derivativedivides	\(a \left (\frac {\arctanh \left (a x \right )^{2} a^{3} x^{3}}{3}-2 \arctanh \left (a x \right )^{2} a x -\frac {\arctanh \left (a x \right )^{2}}{a x}+\frac {a^{2} x^{2} \arctanh \left (a x \right )}{3}-\frac {8 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{3}+2 \arctanh \left (a x \right ) \ln \left (a x \right )-\frac {8 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{3}+\frac {a x}{3}+\frac {\ln \left (a x -1\right )}{6}-\frac {\ln \left (a x +1\right )}{6}-\dilog \left (a x \right )-\dilog \left (a x +1\right )-\ln \left (a x \right ) \ln \left (a x +1\right )-\frac {2 \ln \left (a x -1\right )^{2}}{3}+\frac {8 \dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{3}+\frac {4 \ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{3}-\frac {4 \left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{3}+\frac {2 \ln \left (a x +1\right )^{2}}{3}\right )\)	\(203\)
default	\(a \left (\frac {\arctanh \left (a x \right )^{2} a^{3} x^{3}}{3}-2 \arctanh \left (a x \right )^{2} a x -\frac {\arctanh \left (a x \right )^{2}}{a x}+\frac {a^{2} x^{2} \arctanh \left (a x \right )}{3}-\frac {8 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{3}+2 \arctanh \left (a x \right ) \ln \left (a x \right )-\frac {8 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{3}+\frac {a x}{3}+\frac {\ln \left (a x -1\right )}{6}-\frac {\ln \left (a x +1\right )}{6}-\dilog \left (a x \right )-\dilog \left (a x +1\right )-\ln \left (a x \right ) \ln \left (a x +1\right )-\frac {2 \ln \left (a x -1\right )^{2}}{3}+\frac {8 \dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{3}+\frac {4 \ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{3}-\frac {4 \left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{3}+\frac {2 \ln \left (a x +1\right )^{2}}{3}\right )\)	\(203\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(1/3*arctanh(a*x)^2*a^3*x^3-2*arctanh(a*x)^2*a*x-arctanh(a*x)^2/a/x+1/3*a^2*x^2*arctanh(a*x)-8/3*arctanh(a*x
)*ln(a*x+1)+2*arctanh(a*x)*ln(a*x)-8/3*arctanh(a*x)*ln(a*x-1)+1/3*a*x+1/6*ln(a*x-1)-1/6*ln(a*x+1)-dilog(a*x)-d
ilog(a*x+1)-ln(a*x)*ln(a*x+1)-2/3*ln(a*x-1)^2+8/3*dilog(1/2*a*x+1/2)+4/3*ln(a*x-1)*ln(1/2*a*x+1/2)-4/3*(ln(a*x
+1)-ln(1/2*a*x+1/2))*ln(-1/2*a*x+1/2)+2/3*ln(a*x+1)^2)

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Maxima [A]
time = 0.27, size = 200, normalized size = 1.28 \begin {gather*} \frac {1}{6} \, a^{2} {\left (\frac {2 \, {\left (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}\right )}}{a} + \frac {16 \, {\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} - \frac {6 \, {\left (\log \left (a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-a x\right )\right )}}{a} + \frac {6 \, {\left (\log \left (-a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (a x\right )\right )}}{a} - \frac {\log \left (a x + 1\right )}{a} + \frac {\log \left (a x - 1\right )}{a}\right )} + \frac {1}{3} \, {\left (a^{2} x^{2} - 8 \, \log \left (a x + 1\right ) - 8 \, \log \left (a x - 1\right ) + 6 \, \log \left (x\right )\right )} a \operatorname {artanh}\left (a x\right ) + \frac {1}{3} \, {\left (a^{4} x^{3} - 6 \, a^{2} x - \frac {3}{x}\right )} \operatorname {artanh}\left (a x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a^2*(2*(a*x + 2*log(a*x + 1)^2 - 4*log(a*x + 1)*log(a*x - 1) - 2*log(a*x - 1)^2)/a + 16*(log(a*x - 1)*log(
1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))/a - 6*(log(a*x + 1)*log(x) + dilog(-a*x))/a + 6*(log(-a*x + 1)*log(x)
+ dilog(a*x))/a - log(a*x + 1)/a + log(a*x - 1)/a) + 1/3*(a^2*x^2 - 8*log(a*x + 1) - 8*log(a*x - 1) + 6*log(x)
)*a*arctanh(a*x) + 1/3*(a^4*x^3 - 6*a^2*x - 3/x)*arctanh(a*x)^2

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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