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

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

[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

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Rubi [A]  time = 0.194307, antiderivative size = 113, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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]

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*}

Mathematica [A]  time = 0.0592525, size = 99, normalized size = 0.88 \[ \frac{a^2 x^2 \left (14 a^2 x^2+32 a^4 x^4 \log (x)-16 a^4 x^4 \log \left (1-a^2 x^2\right )-3\right )+30 \left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)^2-4 a x \left (15 a^4 x^4-10 a^2 x^2+3\right ) \tanh ^{-1}(a x)}{180 x^6} \]

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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Maple [B]  time = 0.06, size = 233, normalized size = 2.1 \begin{align*}{\frac{{a}^{2} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{2\,{x}^{4}}}-{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{6\,{x}^{6}}}-{\frac{{a}^{4} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{2\,{x}^{2}}}-{\frac{{a}^{6}{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{6}}-{\frac{a{\it Artanh} \left ( ax \right ) }{15\,{x}^{5}}}+{\frac{2\,{a}^{3}{\it Artanh} \left ( ax \right ) }{9\,{x}^{3}}}-{\frac{{a}^{5}{\it Artanh} \left ( ax \right ) }{3\,x}}+{\frac{{a}^{6}{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{6}}-{\frac{{a}^{6} \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{24}}+{\frac{{a}^{6}\ln \left ( ax-1 \right ) }{12}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{{a}^{6} \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{24}}-{\frac{{a}^{6}}{12}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{{a}^{6}\ln \left ( ax+1 \right ) }{12}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }-{\frac{4\,{a}^{6}\ln \left ( ax-1 \right ) }{45}}-{\frac{{a}^{2}}{60\,{x}^{4}}}+{\frac{7\,{a}^{4}}{90\,{x}^{2}}}+{\frac{8\,{a}^{6}\ln \left ( ax \right ) }{45}}-{\frac{4\,{a}^{6}\ln \left ( ax+1 \right ) }{45}} \end{align*}

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^2*arctanh(a*x)^2/x^4-1/6*arctanh(a*x)^2/x^6-1/2*a^4*arctanh(a*x)^2/x^2-1/6*a^6*arctanh(a*x)*ln(a*x-1)-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)-1/24*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)-4/45*a^6*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)

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Maxima [A]  time = 0.96366, size = 254, normalized size = 2.25 \begin{align*} \frac{1}{360} \,{\left (64 \, a^{4} \log \left (x\right ) - \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}} \end{align*}

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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Fricas [A]  time = 2.11722, size = 300, normalized size = 2.65 \begin{align*} -\frac{32 \, a^{6} x^{6} \log \left (a^{2} x^{2} - 1\right ) - 64 \, a^{6} x^{6} \log \left (x\right ) - 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}} \end{align*}

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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Sympy [A]  time = 4.79153, size = 148, normalized size = 1.31 \begin{align*} \begin{cases} \frac{8 a^{6} \log{\left (x \right )}}{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} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname{artanh}\left (a x\right )^{2}}{x^{7}}\,{d x} \end{align*}

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]

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