3.214 \(\int \frac {(1-a^2 x^2)^2 \tanh ^{-1}(a x)^2}{x^7} \, dx\)

Optimal. Leaf size=113 \[ \frac {8}{45} a^6 \log (x)-\frac {a^5 \tanh ^{-1}(a x)}{3 x}+\frac {7 a^4}{90 x^2}+\frac {2 a^3 \tanh ^{-1}(a x)}{9 x^3}-\frac {a^2}{60 x^4}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 x^6}-\frac {4}{45} a^6 \log \left (1-a^2 x^2\right )-\frac {a \tanh ^{-1}(a x)}{15 x^5} \]

[Out]

-1/60*a^2/x^4+7/90*a^4/x^2-1/15*a*arctanh(a*x)/x^5+2/9*a^3*arctanh(a*x)/x^3-1/3*a^5*arctanh(a*x)/x-1/6*(-a^2*x
^2+1)^3*arctanh(a*x)^2/x^6+8/45*a^6*ln(x)-4/45*a^6*ln(-a^2*x^2+1)

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Rubi [A]  time = 0.19, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6008, 6012, 5916, 266, 44, 36, 29, 31} \[ \frac {7 a^4}{90 x^2}-\frac {a^2}{60 x^4}-\frac {4}{45} a^6 \log \left (1-a^2 x^2\right )+\frac {2 a^3 \tanh ^{-1}(a x)}{9 x^3}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 x^6}+\frac {8}{45} a^6 \log (x)-\frac {a^5 \tanh ^{-1}(a x)}{3 x}-\frac {a \tanh ^{-1}(a x)}{15 x^5} \]

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)^2*ArcTanh[a*x]^2)/x^7,x]

[Out]

-a^2/(60*x^4) + (7*a^4)/(90*x^2) - (a*ArcTanh[a*x])/(15*x^5) + (2*a^3*ArcTanh[a*x])/(9*x^3) - (a^5*ArcTanh[a*x
])/(3*x) - ((1 - a^2*x^2)^3*ArcTanh[a*x]^2)/(6*x^6) + (8*a^6*Log[x])/45 - (4*a^6*Log[1 - a^2*x^2])/45

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 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 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 6012

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^7} \, dx &=-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 x^6}+\frac {1}{3} a \int \frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{x^6} \, dx\\ &=-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 x^6}+\frac {1}{3} a \int \left (\frac {\tanh ^{-1}(a x)}{x^6}-\frac {2 a^2 \tanh ^{-1}(a x)}{x^4}+\frac {a^4 \tanh ^{-1}(a x)}{x^2}\right ) \, dx\\ &=-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 x^6}+\frac {1}{3} a \int \frac {\tanh ^{-1}(a x)}{x^6} \, dx-\frac {1}{3} \left (2 a^3\right ) \int \frac {\tanh ^{-1}(a x)}{x^4} \, dx+\frac {1}{3} a^5 \int \frac {\tanh ^{-1}(a x)}{x^2} \, dx\\ &=-\frac {a \tanh ^{-1}(a x)}{15 x^5}+\frac {2 a^3 \tanh ^{-1}(a x)}{9 x^3}-\frac {a^5 \tanh ^{-1}(a x)}{3 x}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 x^6}+\frac {1}{15} a^2 \int \frac {1}{x^5 \left (1-a^2 x^2\right )} \, dx-\frac {1}{9} \left (2 a^4\right ) \int \frac {1}{x^3 \left (1-a^2 x^2\right )} \, dx+\frac {1}{3} a^6 \int \frac {1}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac {a \tanh ^{-1}(a x)}{15 x^5}+\frac {2 a^3 \tanh ^{-1}(a x)}{9 x^3}-\frac {a^5 \tanh ^{-1}(a x)}{3 x}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 x^6}+\frac {1}{30} a^2 \operatorname {Subst}\left (\int \frac {1}{x^3 \left (1-a^2 x\right )} \, dx,x,x^2\right )-\frac {1}{9} a^4 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-a^2 x\right )} \, dx,x,x^2\right )+\frac {1}{6} a^6 \operatorname {Subst}\left (\int \frac {1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {a \tanh ^{-1}(a x)}{15 x^5}+\frac {2 a^3 \tanh ^{-1}(a x)}{9 x^3}-\frac {a^5 \tanh ^{-1}(a x)}{3 x}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 x^6}+\frac {1}{30} a^2 \operatorname {Subst}\left (\int \left (\frac {1}{x^3}+\frac {a^2}{x^2}+\frac {a^4}{x}-\frac {a^6}{-1+a^2 x}\right ) \, dx,x,x^2\right )-\frac {1}{9} a^4 \operatorname {Subst}\left (\int \left (\frac {1}{x^2}+\frac {a^2}{x}-\frac {a^4}{-1+a^2 x}\right ) \, dx,x,x^2\right )+\frac {1}{6} a^6 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{6} a^8 \operatorname {Subst}\left (\int \frac {1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac {a^2}{60 x^4}+\frac {7 a^4}{90 x^2}-\frac {a \tanh ^{-1}(a x)}{15 x^5}+\frac {2 a^3 \tanh ^{-1}(a x)}{9 x^3}-\frac {a^5 \tanh ^{-1}(a x)}{3 x}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 x^6}+\frac {8}{45} a^6 \log (x)-\frac {4}{45} a^6 \log \left (1-a^2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 99, normalized size = 0.88 \[ \frac {30 \left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)^2+a^2 x^2 \left (32 a^4 x^4 \log (x)+14 a^2 x^2-16 a^4 x^4 \log \left (1-a^2 x^2\right )-3\right )-4 a x \left (15 a^4 x^4-10 a^2 x^2+3\right ) \tanh ^{-1}(a x)}{180 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - a^2*x^2)^2*ArcTanh[a*x]^2)/x^7,x]

[Out]

(-4*a*x*(3 - 10*a^2*x^2 + 15*a^4*x^4)*ArcTanh[a*x] + 30*(-1 + a^2*x^2)^3*ArcTanh[a*x]^2 + a^2*x^2*(-3 + 14*a^2
*x^2 + 32*a^4*x^4*Log[x] - 16*a^4*x^4*Log[1 - a^2*x^2]))/(180*x^6)

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fricas [A]  time = 0.59, size = 132, normalized size = 1.17 \[ -\frac {32 \, a^{6} x^{6} \log \left (a^{2} x^{2} - 1\right ) - 64 \, a^{6} x^{6} \log \relax (x) - 28 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 15 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 4 \, {\left (15 \, a^{5} x^{5} - 10 \, a^{3} x^{3} + 3 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{360 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^2*arctanh(a*x)^2/x^7,x, algorithm="fricas")

[Out]

-1/360*(32*a^6*x^6*log(a^2*x^2 - 1) - 64*a^6*x^6*log(x) - 28*a^4*x^4 + 6*a^2*x^2 - 15*(a^6*x^6 - 3*a^4*x^4 + 3
*a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1))^2 + 4*(15*a^5*x^5 - 10*a^3*x^3 + 3*a*x)*log(-(a*x + 1)/(a*x - 1)))/x^6

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giac [B]  time = 0.22, size = 440, normalized size = 3.89 \[ \frac {4}{45} \, {\left (2 \, a^{5} \log \left (-\frac {a x + 1}{a x - 1} - 1\right ) - 2 \, a^{5} \log \left (-\frac {a x + 1}{a x - 1}\right ) + \frac {30 \, {\left (a x + 1\right )}^{3} a^{5} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}}{{\left (a x - 1\right )}^{3} {\left (\frac {{\left (a x + 1\right )}^{6}}{{\left (a x - 1\right )}^{6}} + \frac {6 \, {\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} + \frac {15 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {20 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {15 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {6 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}} + \frac {2 \, {\left (\frac {10 \, {\left (a x + 1\right )}^{2} a^{5}}{{\left (a x - 1\right )}^{2}} + \frac {5 \, {\left (a x + 1\right )} a^{5}}{a x - 1} + a^{5}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{\frac {{\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} + \frac {5 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {10 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {10 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {5 \, {\left (a x + 1\right )}}{a x - 1} + 1} - \frac {\frac {2 \, {\left (a x + 1\right )}^{3} a^{5}}{{\left (a x - 1\right )}^{3}} + \frac {7 \, {\left (a x + 1\right )}^{2} a^{5}}{{\left (a x - 1\right )}^{2}} + \frac {2 \, {\left (a x + 1\right )} a^{5}}{a x - 1}}{\frac {{\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {4 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {6 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {4 \, {\left (a x + 1\right )}}{a x - 1} + 1}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^2*arctanh(a*x)^2/x^7,x, algorithm="giac")

[Out]

4/45*(2*a^5*log(-(a*x + 1)/(a*x - 1) - 1) - 2*a^5*log(-(a*x + 1)/(a*x - 1)) + 30*(a*x + 1)^3*a^5*log(-(a*x + 1
)/(a*x - 1))^2/((a*x - 1)^3*((a*x + 1)^6/(a*x - 1)^6 + 6*(a*x + 1)^5/(a*x - 1)^5 + 15*(a*x + 1)^4/(a*x - 1)^4
+ 20*(a*x + 1)^3/(a*x - 1)^3 + 15*(a*x + 1)^2/(a*x - 1)^2 + 6*(a*x + 1)/(a*x - 1) + 1)) + 2*(10*(a*x + 1)^2*a^
5/(a*x - 1)^2 + 5*(a*x + 1)*a^5/(a*x - 1) + a^5)*log(-(a*x + 1)/(a*x - 1))/((a*x + 1)^5/(a*x - 1)^5 + 5*(a*x +
1)^4/(a*x - 1)^4 + 10*(a*x + 1)^3/(a*x - 1)^3 + 10*(a*x + 1)^2/(a*x - 1)^2 + 5*(a*x + 1)/(a*x - 1) + 1) - (2*
(a*x + 1)^3*a^5/(a*x - 1)^3 + 7*(a*x + 1)^2*a^5/(a*x - 1)^2 + 2*(a*x + 1)*a^5/(a*x - 1))/((a*x + 1)^4/(a*x - 1
)^4 + 4*(a*x + 1)^3/(a*x - 1)^3 + 6*(a*x + 1)^2/(a*x - 1)^2 + 4*(a*x + 1)/(a*x - 1) + 1))*a

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maple [B]  time = 0.07, size = 233, normalized size = 2.06 \[ -\frac {a^{4} \arctanh \left (a x \right )^{2}}{2 x^{2}}+\frac {a^{2} \arctanh \left (a x \right )^{2}}{2 x^{4}}-\frac {\arctanh \left (a x \right )^{2}}{6 x^{6}}-\frac {a \arctanh \left (a x \right )}{15 x^{5}}+\frac {2 a^{3} \arctanh \left (a x \right )}{9 x^{3}}-\frac {a^{5} \arctanh \left (a x \right )}{3 x}-\frac {a^{6} \arctanh \left (a x \right ) \ln \left (a x -1\right )}{6}+\frac {a^{6} \arctanh \left (a x \right ) \ln \left (a x +1\right )}{6}-\frac {a^{6} \ln \left (a x -1\right )^{2}}{24}+\frac {a^{6} \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{12}-\frac {a^{6} \ln \left (a x +1\right )^{2}}{24}-\frac {a^{6} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{12}+\frac {a^{6} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{12}-\frac {a^{2}}{60 x^{4}}+\frac {7 a^{4}}{90 x^{2}}+\frac {8 a^{6} \ln \left (a x \right )}{45}-\frac {4 a^{6} \ln \left (a x -1\right )}{45}-\frac {4 a^{6} \ln \left (a x +1\right )}{45} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)^2*arctanh(a*x)^2/x^7,x)

[Out]

-1/2*a^4*arctanh(a*x)^2/x^2+1/2*a^2*arctanh(a*x)^2/x^4-1/6*arctanh(a*x)^2/x^6-1/15*a*arctanh(a*x)/x^5+2/9*a^3*
arctanh(a*x)/x^3-1/3*a^5*arctanh(a*x)/x-1/6*a^6*arctanh(a*x)*ln(a*x-1)+1/6*a^6*arctanh(a*x)*ln(a*x+1)-1/24*a^6
*ln(a*x-1)^2+1/12*a^6*ln(a*x-1)*ln(1/2+1/2*a*x)-1/24*a^6*ln(a*x+1)^2-1/12*a^6*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)
+1/12*a^6*ln(-1/2*a*x+1/2)*ln(a*x+1)-1/60*a^2/x^4+7/90*a^4/x^2+8/45*a^6*ln(a*x)-4/45*a^6*ln(a*x-1)-4/45*a^6*ln
(a*x+1)

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maxima [A]  time = 0.33, size = 188, normalized size = 1.66 \[ \frac {1}{360} \, {\left (64 \, a^{4} \log \relax (x) - \frac {15 \, a^{4} x^{4} \log \left (a x + 1\right )^{2} + 15 \, a^{4} x^{4} \log \left (a x - 1\right )^{2} + 32 \, a^{4} x^{4} \log \left (a x - 1\right ) - 28 \, a^{2} x^{2} - 2 \, {\left (15 \, a^{4} x^{4} \log \left (a x - 1\right ) - 16 \, a^{4} x^{4}\right )} \log \left (a x + 1\right ) + 6}{x^{4}}\right )} a^{2} + \frac {1}{90} \, {\left (15 \, a^{5} \log \left (a x + 1\right ) - 15 \, a^{5} \log \left (a x - 1\right ) - \frac {2 \, {\left (15 \, a^{4} x^{4} - 10 \, a^{2} x^{2} + 3\right )}}{x^{5}}\right )} a \operatorname {artanh}\left (a x\right ) - \frac {{\left (3 \, a^{4} x^{4} - 3 \, a^{2} x^{2} + 1\right )} \operatorname {artanh}\left (a x\right )^{2}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^2*arctanh(a*x)^2/x^7,x, algorithm="maxima")

[Out]

1/360*(64*a^4*log(x) - (15*a^4*x^4*log(a*x + 1)^2 + 15*a^4*x^4*log(a*x - 1)^2 + 32*a^4*x^4*log(a*x - 1) - 28*a
^2*x^2 - 2*(15*a^4*x^4*log(a*x - 1) - 16*a^4*x^4)*log(a*x + 1) + 6)/x^4)*a^2 + 1/90*(15*a^5*log(a*x + 1) - 15*
a^5*log(a*x - 1) - 2*(15*a^4*x^4 - 10*a^2*x^2 + 3)/x^5)*a*arctanh(a*x) - 1/6*(3*a^4*x^4 - 3*a^2*x^2 + 1)*arcta
nh(a*x)^2/x^6

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mupad [B]  time = 1.58, size = 335, normalized size = 2.96 \[ \frac {8\,a^6\,\ln \relax (x)}{45}-\frac {\frac {3\,a^2}{4}-\frac {7\,a^4\,x^2}{2}}{45\,x^4}-{\ln \left (1-a\,x\right )}^2\,\left (\frac {\frac {a^4\,x^4}{2}-\frac {a^2\,x^2}{2}+\frac {1}{6}}{4\,x^6}-\frac {a^6}{24}\right )-{\ln \left (a\,x+1\right )}^2\,\left (\frac {\frac {a^4\,x^4}{8}-\frac {a^2\,x^2}{8}+\frac {1}{24}}{x^6}-\frac {a^6}{24}\right )-\ln \left (1-a\,x\right )\,\left (\frac {a\,\left (\frac {137\,a^5\,x^5}{2}-30\,a^4\,x^4+15\,a^3\,x^3-10\,a^2\,x^2+\frac {15\,a\,x}{2}-6\right )}{360\,x^5}-\ln \left (a\,x+1\right )\,\left (\frac {\frac {a^4\,x^4}{2}-\frac {a^2\,x^2}{2}+\frac {1}{6}}{2\,x^6}-\frac {a^6}{12}\right )-\frac {a\,\left (137\,a^5\,x^5+60\,a^4\,x^4+30\,a^3\,x^3+20\,a^2\,x^2+15\,a\,x+12\right )}{720\,x^5}+\frac {5\,a^8\,x^2-\frac {15\,a^9\,x^3}{2}}{60\,a^5\,x^5}+\frac {\frac {15\,a^9\,x^3}{2}+5\,a^8\,x^2}{60\,a^5\,x^5}\right )-\frac {4\,a^6\,\ln \left (a^2\,x^2-1\right )}{45}-\frac {a\,\ln \left (a\,x+1\right )\,\left (\frac {a^4\,x^4}{6}-\frac {a^2\,x^2}{9}+\frac {1}{30}\right )}{x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atanh(a*x)^2*(a^2*x^2 - 1)^2)/x^7,x)

[Out]

(8*a^6*log(x))/45 - ((3*a^2)/4 - (7*a^4*x^2)/2)/(45*x^4) - log(1 - a*x)^2*(((a^4*x^4)/2 - (a^2*x^2)/2 + 1/6)/(
4*x^6) - a^6/24) - log(a*x + 1)^2*(((a^4*x^4)/8 - (a^2*x^2)/8 + 1/24)/x^6 - a^6/24) - log(1 - a*x)*((a*((15*a*
x)/2 - 10*a^2*x^2 + 15*a^3*x^3 - 30*a^4*x^4 + (137*a^5*x^5)/2 - 6))/(360*x^5) - log(a*x + 1)*(((a^4*x^4)/2 - (
a^2*x^2)/2 + 1/6)/(2*x^6) - a^6/12) - (a*(15*a*x + 20*a^2*x^2 + 30*a^3*x^3 + 60*a^4*x^4 + 137*a^5*x^5 + 12))/(
720*x^5) + (5*a^8*x^2 - (15*a^9*x^3)/2)/(60*a^5*x^5) + (5*a^8*x^2 + (15*a^9*x^3)/2)/(60*a^5*x^5)) - (4*a^6*log
(a^2*x^2 - 1))/45 - (a*log(a*x + 1)*((a^4*x^4)/6 - (a^2*x^2)/9 + 1/30))/x^5

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sympy [A]  time = 2.60, size = 148, normalized size = 1.31 \[ \begin {cases} \frac {8 a^{6} \log {\relax (x )}}{45} - \frac {8 a^{6} \log {\left (x - \frac {1}{a} \right )}}{45} + \frac {a^{6} \operatorname {atanh}^{2}{\left (a x \right )}}{6} - \frac {8 a^{6} \operatorname {atanh}{\left (a x \right )}}{45} - \frac {a^{5} \operatorname {atanh}{\left (a x \right )}}{3 x} - \frac {a^{4} \operatorname {atanh}^{2}{\left (a x \right )}}{2 x^{2}} + \frac {7 a^{4}}{90 x^{2}} + \frac {2 a^{3} \operatorname {atanh}{\left (a x \right )}}{9 x^{3}} + \frac {a^{2} \operatorname {atanh}^{2}{\left (a x \right )}}{2 x^{4}} - \frac {a^{2}}{60 x^{4}} - \frac {a \operatorname {atanh}{\left (a x \right )}}{15 x^{5}} - \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{6 x^{6}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)**2*atanh(a*x)**2/x**7,x)

[Out]

Piecewise((8*a**6*log(x)/45 - 8*a**6*log(x - 1/a)/45 + a**6*atanh(a*x)**2/6 - 8*a**6*atanh(a*x)/45 - a**5*atan
h(a*x)/(3*x) - a**4*atanh(a*x)**2/(2*x**2) + 7*a**4/(90*x**2) + 2*a**3*atanh(a*x)/(9*x**3) + a**2*atanh(a*x)**
2/(2*x**4) - a**2/(60*x**4) - a*atanh(a*x)/(15*x**5) - atanh(a*x)**2/(6*x**6), Ne(a, 0)), (0, True))

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