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

Optimal. Leaf size=116 \[ -\frac {\tanh ^{-1}(a x)^2}{12 a^4}+\frac {x \tanh ^{-1}(a x)}{6 a^3}-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)^2-\frac {x^2}{180 a^2}+\frac {7 \log \left (1-a^2 x^2\right )}{90 a^4}-\frac {1}{15} a x^5 \tanh ^{-1}(a x)+\frac {1}{4} x^4 \tanh ^{-1}(a x)^2+\frac {x^3 \tanh ^{-1}(a x)}{18 a}-\frac {x^4}{60} \]

[Out]

-1/180*x^2/a^2-1/60*x^4+1/6*x*arctanh(a*x)/a^3+1/18*x^3*arctanh(a*x)/a-1/15*a*x^5*arctanh(a*x)-1/12*arctanh(a*
x)^2/a^4+1/4*x^4*arctanh(a*x)^2-1/6*a^2*x^6*arctanh(a*x)^2+7/90*ln(-a^2*x^2+1)/a^4

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Rubi [A]  time = 0.44, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6014, 5916, 5980, 266, 43, 5910, 260, 5948} \[ -\frac {x^2}{180 a^2}+\frac {7 \log \left (1-a^2 x^2\right )}{90 a^4}-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)^2+\frac {x \tanh ^{-1}(a x)}{6 a^3}-\frac {\tanh ^{-1}(a x)^2}{12 a^4}-\frac {1}{15} a x^5 \tanh ^{-1}(a x)+\frac {1}{4} x^4 \tanh ^{-1}(a x)^2+\frac {x^3 \tanh ^{-1}(a x)}{18 a}-\frac {x^4}{60} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^2/(180*a^2) - x^4/60 + (x*ArcTanh[a*x])/(6*a^3) + (x^3*ArcTanh[a*x])/(18*a) - (a*x^5*ArcTanh[a*x])/15 - Arc
Tanh[a*x]^2/(12*a^4) + (x^4*ArcTanh[a*x]^2)/4 - (a^2*x^6*ArcTanh[a*x]^2)/6 + (7*Log[1 - a^2*x^2])/(90*a^4)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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 5948

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

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 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^3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2 \, dx &=-\left (a^2 \int x^5 \tanh ^{-1}(a x)^2 \, dx\right )+\int x^3 \tanh ^{-1}(a x)^2 \, dx\\ &=\frac {1}{4} x^4 \tanh ^{-1}(a x)^2-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)^2-\frac {1}{2} a \int \frac {x^4 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx+\frac {1}{3} a^3 \int \frac {x^6 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=\frac {1}{4} x^4 \tanh ^{-1}(a x)^2-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)^2+\frac {\int x^2 \tanh ^{-1}(a x) \, dx}{2 a}-\frac {\int \frac {x^2 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{2 a}-\frac {1}{3} a \int x^4 \tanh ^{-1}(a x) \, dx+\frac {1}{3} a \int \frac {x^4 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=\frac {x^3 \tanh ^{-1}(a x)}{6 a}-\frac {1}{15} a x^5 \tanh ^{-1}(a x)+\frac {1}{4} x^4 \tanh ^{-1}(a x)^2-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)^2-\frac {1}{6} \int \frac {x^3}{1-a^2 x^2} \, dx+\frac {\int \tanh ^{-1}(a x) \, dx}{2 a^3}-\frac {\int \frac {\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{2 a^3}-\frac {\int x^2 \tanh ^{-1}(a x) \, dx}{3 a}+\frac {\int \frac {x^2 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{3 a}+\frac {1}{15} a^2 \int \frac {x^5}{1-a^2 x^2} \, dx\\ &=\frac {x \tanh ^{-1}(a x)}{2 a^3}+\frac {x^3 \tanh ^{-1}(a x)}{18 a}-\frac {1}{15} a x^5 \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)^2}{4 a^4}+\frac {1}{4} x^4 \tanh ^{-1}(a x)^2-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)^2-\frac {1}{12} \operatorname {Subst}\left (\int \frac {x}{1-a^2 x} \, dx,x,x^2\right )+\frac {1}{9} \int \frac {x^3}{1-a^2 x^2} \, dx-\frac {\int \tanh ^{-1}(a x) \, dx}{3 a^3}+\frac {\int \frac {\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{3 a^3}-\frac {\int \frac {x}{1-a^2 x^2} \, dx}{2 a^2}+\frac {1}{30} a^2 \operatorname {Subst}\left (\int \frac {x^2}{1-a^2 x} \, dx,x,x^2\right )\\ &=\frac {x \tanh ^{-1}(a x)}{6 a^3}+\frac {x^3 \tanh ^{-1}(a x)}{18 a}-\frac {1}{15} a x^5 \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)^2}{12 a^4}+\frac {1}{4} x^4 \tanh ^{-1}(a x)^2-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)^2+\frac {\log \left (1-a^2 x^2\right )}{4 a^4}+\frac {1}{18} \operatorname {Subst}\left (\int \frac {x}{1-a^2 x} \, dx,x,x^2\right )-\frac {1}{12} \operatorname {Subst}\left (\int \left (-\frac {1}{a^2}-\frac {1}{a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )+\frac {\int \frac {x}{1-a^2 x^2} \, dx}{3 a^2}+\frac {1}{30} a^2 \operatorname {Subst}\left (\int \left (-\frac {1}{a^4}-\frac {x}{a^2}-\frac {1}{a^4 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {x^2}{20 a^2}-\frac {x^4}{60}+\frac {x \tanh ^{-1}(a x)}{6 a^3}+\frac {x^3 \tanh ^{-1}(a x)}{18 a}-\frac {1}{15} a x^5 \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)^2}{12 a^4}+\frac {1}{4} x^4 \tanh ^{-1}(a x)^2-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)^2+\frac {2 \log \left (1-a^2 x^2\right )}{15 a^4}+\frac {1}{18} \operatorname {Subst}\left (\int \left (-\frac {1}{a^2}-\frac {1}{a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac {x^2}{180 a^2}-\frac {x^4}{60}+\frac {x \tanh ^{-1}(a x)}{6 a^3}+\frac {x^3 \tanh ^{-1}(a x)}{18 a}-\frac {1}{15} a x^5 \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)^2}{12 a^4}+\frac {1}{4} x^4 \tanh ^{-1}(a x)^2-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)^2+\frac {7 \log \left (1-a^2 x^2\right )}{90 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 88, normalized size = 0.76 \[ -\frac {3 a^4 x^4+a^2 x^2-14 \log \left (1-a^2 x^2\right )+15 \left (2 a^6 x^6-3 a^4 x^4+1\right ) \tanh ^{-1}(a x)^2+2 a x \left (6 a^4 x^4-5 a^2 x^2-15\right ) \tanh ^{-1}(a x)}{180 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/180*(a^2*x^2 + 3*a^4*x^4 + 2*a*x*(-15 - 5*a^2*x^2 + 6*a^4*x^4)*ArcTanh[a*x] + 15*(1 - 3*a^4*x^4 + 2*a^6*x^6
)*ArcTanh[a*x]^2 - 14*Log[1 - a^2*x^2])/a^4

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fricas [A]  time = 0.62, size = 109, normalized size = 0.94 \[ -\frac {12 \, a^{4} x^{4} + 4 \, a^{2} x^{2} + 15 \, {\left (2 \, a^{6} x^{6} - 3 \, a^{4} x^{4} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 4 \, {\left (6 \, a^{5} x^{5} - 5 \, a^{3} x^{3} - 15 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 56 \, \log \left (a^{2} x^{2} - 1\right )}{720 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/720*(12*a^4*x^4 + 4*a^2*x^2 + 15*(2*a^6*x^6 - 3*a^4*x^4 + 1)*log(-(a*x + 1)/(a*x - 1))^2 + 4*(6*a^5*x^5 - 5
*a^3*x^3 - 15*a*x)*log(-(a*x + 1)/(a*x - 1)) - 56*log(a^2*x^2 - 1))/a^4

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giac [B]  time = 0.17, size = 522, normalized size = 4.50 \[ -\frac {1}{45} \, {\left (\frac {15 \, {\left (\frac {3 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {2 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {3 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}}{\frac {{\left (a x + 1\right )}^{6} a^{5}}{{\left (a x - 1\right )}^{6}} - \frac {6 \, {\left (a x + 1\right )}^{5} a^{5}}{{\left (a x - 1\right )}^{5}} + \frac {15 \, {\left (a x + 1\right )}^{4} a^{5}}{{\left (a x - 1\right )}^{4}} - \frac {20 \, {\left (a x + 1\right )}^{3} a^{5}}{{\left (a x - 1\right )}^{3}} + \frac {15 \, {\left (a x + 1\right )}^{2} a^{5}}{{\left (a x - 1\right )}^{2}} - \frac {6 \, {\left (a x + 1\right )} a^{5}}{a x - 1} + a^{5}} + \frac {{\left (\frac {45 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {25 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {35 \, {\left (a x + 1\right )}}{a x - 1} - 7\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{\frac {{\left (a x + 1\right )}^{5} a^{5}}{{\left (a x - 1\right )}^{5}} - \frac {5 \, {\left (a x + 1\right )}^{4} a^{5}}{{\left (a x - 1\right )}^{4}} + \frac {10 \, {\left (a x + 1\right )}^{3} a^{5}}{{\left (a x - 1\right )}^{3}} - \frac {10 \, {\left (a x + 1\right )}^{2} a^{5}}{{\left (a x - 1\right )}^{2}} + \frac {5 \, {\left (a x + 1\right )} a^{5}}{a x - 1} - a^{5}} + \frac {\frac {7 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {2 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {7 \, {\left (a x + 1\right )}}{a x - 1}}{\frac {{\left (a x + 1\right )}^{4} a^{5}}{{\left (a x - 1\right )}^{4}} - \frac {4 \, {\left (a x + 1\right )}^{3} a^{5}}{{\left (a x - 1\right )}^{3}} + \frac {6 \, {\left (a x + 1\right )}^{2} a^{5}}{{\left (a x - 1\right )}^{2}} - \frac {4 \, {\left (a x + 1\right )} a^{5}}{a x - 1} + a^{5}} + \frac {7 \, \log \left (-\frac {a x + 1}{a x - 1} + 1\right )}{a^{5}} - \frac {7 \, \log \left (-\frac {a x + 1}{a x - 1}\right )}{a^{5}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/45*(15*(3*(a*x + 1)^4/(a*x - 1)^4 + 2*(a*x + 1)^3/(a*x - 1)^3 + 3*(a*x + 1)^2/(a*x - 1)^2)*log(-(a*x + 1)/(
a*x - 1))^2/((a*x + 1)^6*a^5/(a*x - 1)^6 - 6*(a*x + 1)^5*a^5/(a*x - 1)^5 + 15*(a*x + 1)^4*a^5/(a*x - 1)^4 - 20
*(a*x + 1)^3*a^5/(a*x - 1)^3 + 15*(a*x + 1)^2*a^5/(a*x - 1)^2 - 6*(a*x + 1)*a^5/(a*x - 1) + a^5) + (45*(a*x +
1)^3/(a*x - 1)^3 - 25*(a*x + 1)^2/(a*x - 1)^2 + 35*(a*x + 1)/(a*x - 1) - 7)*log(-(a*x + 1)/(a*x - 1))/((a*x +
1)^5*a^5/(a*x - 1)^5 - 5*(a*x + 1)^4*a^5/(a*x - 1)^4 + 10*(a*x + 1)^3*a^5/(a*x - 1)^3 - 10*(a*x + 1)^2*a^5/(a*
x - 1)^2 + 5*(a*x + 1)*a^5/(a*x - 1) - a^5) + (7*(a*x + 1)^3/(a*x - 1)^3 - 2*(a*x + 1)^2/(a*x - 1)^2 + 7*(a*x
+ 1)/(a*x - 1))/((a*x + 1)^4*a^5/(a*x - 1)^4 - 4*(a*x + 1)^3*a^5/(a*x - 1)^3 + 6*(a*x + 1)^2*a^5/(a*x - 1)^2 -
4*(a*x + 1)*a^5/(a*x - 1) + a^5) + 7*log(-(a*x + 1)/(a*x - 1) + 1)/a^5 - 7*log(-(a*x + 1)/(a*x - 1))/a^5)*a

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maple [B]  time = 0.06, size = 205, normalized size = 1.77 \[ -\frac {a^{2} x^{6} \arctanh \left (a x \right )^{2}}{6}+\frac {x^{4} \arctanh \left (a x \right )^{2}}{4}-\frac {a \,x^{5} \arctanh \left (a x \right )}{15}+\frac {x^{3} \arctanh \left (a x \right )}{18 a}+\frac {x \arctanh \left (a x \right )}{6 a^{3}}+\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{12 a^{4}}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{12 a^{4}}+\frac {\ln \left (a x -1\right )^{2}}{48 a^{4}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{24 a^{4}}+\frac {\ln \left (a x +1\right )^{2}}{48 a^{4}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{24 a^{4}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{24 a^{4}}-\frac {x^{4}}{60}-\frac {x^{2}}{180 a^{2}}+\frac {7 \ln \left (a x -1\right )}{90 a^{4}}+\frac {7 \ln \left (a x +1\right )}{90 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a^2*x^6*arctanh(a*x)^2+1/4*x^4*arctanh(a*x)^2-1/15*a*x^5*arctanh(a*x)+1/18*x^3*arctanh(a*x)/a+1/6*x*arcta
nh(a*x)/a^3+1/12/a^4*arctanh(a*x)*ln(a*x-1)-1/12/a^4*arctanh(a*x)*ln(a*x+1)+1/48/a^4*ln(a*x-1)^2-1/24/a^4*ln(a
*x-1)*ln(1/2+1/2*a*x)+1/48/a^4*ln(a*x+1)^2+1/24/a^4*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/24/a^4*ln(-1/2*a*x+1/2)
*ln(a*x+1)-1/60*x^4-1/180*x^2/a^2+7/90/a^4*ln(a*x-1)+7/90/a^4*ln(a*x+1)

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maxima [A]  time = 0.33, size = 146, normalized size = 1.26 \[ -\frac {1}{180} \, a {\left (\frac {2 \, {\left (6 \, a^{4} x^{5} - 5 \, a^{2} x^{3} - 15 \, x\right )}}{a^{4}} + \frac {15 \, \log \left (a x + 1\right )}{a^{5}} - \frac {15 \, \log \left (a x - 1\right )}{a^{5}}\right )} \operatorname {artanh}\left (a x\right ) - \frac {1}{12} \, {\left (2 \, a^{2} x^{6} - 3 \, x^{4}\right )} \operatorname {artanh}\left (a x\right )^{2} - \frac {12 \, a^{4} x^{4} + 4 \, a^{2} x^{2} + 2 \, {\left (15 \, \log \left (a x - 1\right ) - 28\right )} \log \left (a x + 1\right ) - 15 \, \log \left (a x + 1\right )^{2} - 15 \, \log \left (a x - 1\right )^{2} - 56 \, \log \left (a x - 1\right )}{720 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/180*a*(2*(6*a^4*x^5 - 5*a^2*x^3 - 15*x)/a^4 + 15*log(a*x + 1)/a^5 - 15*log(a*x - 1)/a^5)*arctanh(a*x) - 1/1
2*(2*a^2*x^6 - 3*x^4)*arctanh(a*x)^2 - 1/720*(12*a^4*x^4 + 4*a^2*x^2 + 2*(15*log(a*x - 1) - 28)*log(a*x + 1) -
15*log(a*x + 1)^2 - 15*log(a*x - 1)^2 - 56*log(a*x - 1))/a^4

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mupad [B]  time = 0.98, size = 101, normalized size = 0.87 \[ -\frac {a^2\,x^2-14\,\ln \left (a^2\,x^2-1\right )+3\,a^4\,x^4+15\,{\mathrm {atanh}\left (a\,x\right )}^2-10\,a^3\,x^3\,\mathrm {atanh}\left (a\,x\right )+12\,a^5\,x^5\,\mathrm {atanh}\left (a\,x\right )-30\,a\,x\,\mathrm {atanh}\left (a\,x\right )-45\,a^4\,x^4\,{\mathrm {atanh}\left (a\,x\right )}^2+30\,a^6\,x^6\,{\mathrm {atanh}\left (a\,x\right )}^2}{180\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*x^2 - 14*log(a^2*x^2 - 1) + 3*a^4*x^4 + 15*atanh(a*x)^2 - 10*a^3*x^3*atanh(a*x) + 12*a^5*x^5*atanh(a*x)
- 30*a*x*atanh(a*x) - 45*a^4*x^4*atanh(a*x)^2 + 30*a^6*x^6*atanh(a*x)^2)/(180*a^4)

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sympy [A]  time = 2.14, size = 114, normalized size = 0.98 \[ \begin {cases} - \frac {a^{2} x^{6} \operatorname {atanh}^{2}{\left (a x \right )}}{6} - \frac {a x^{5} \operatorname {atanh}{\left (a x \right )}}{15} + \frac {x^{4} \operatorname {atanh}^{2}{\left (a x \right )}}{4} - \frac {x^{4}}{60} + \frac {x^{3} \operatorname {atanh}{\left (a x \right )}}{18 a} - \frac {x^{2}}{180 a^{2}} + \frac {x \operatorname {atanh}{\left (a x \right )}}{6 a^{3}} + \frac {7 \log {\left (x - \frac {1}{a} \right )}}{45 a^{4}} - \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{12 a^{4}} + \frac {7 \operatorname {atanh}{\left (a x \right )}}{45 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*x**6*atanh(a*x)**2/6 - a*x**5*atanh(a*x)/15 + x**4*atanh(a*x)**2/4 - x**4/60 + x**3*atanh(a*x
)/(18*a) - x**2/(180*a**2) + x*atanh(a*x)/(6*a**3) + 7*log(x - 1/a)/(45*a**4) - atanh(a*x)**2/(12*a**4) + 7*at
anh(a*x)/(45*a**4), Ne(a, 0)), (0, True))

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