3.202 \(\int \frac {(1-a^2 x^2)^2 \tanh ^{-1}(a x)}{x^6} \, dx\)

Optimal. Leaf size=83 \[ \frac {8}{15} a^5 \log (x)-\frac {a^4 \tanh ^{-1}(a x)}{x}+\frac {7 a^3}{30 x^2}+\frac {2 a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac {4}{15} a^5 \log \left (1-a^2 x^2\right )-\frac {\tanh ^{-1}(a x)}{5 x^5}-\frac {a}{20 x^4} \]

[Out]

-1/20*a/x^4+7/30*a^3/x^2-1/5*arctanh(a*x)/x^5+2/3*a^2*arctanh(a*x)/x^3-a^4*arctanh(a*x)/x+8/15*a^5*ln(x)-4/15*
a^5*ln(-a^2*x^2+1)

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Rubi [A]  time = 0.14, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6012, 5916, 266, 44, 36, 29, 31} \[ \frac {7 a^3}{30 x^2}-\frac {4}{15} a^5 \log \left (1-a^2 x^2\right )+\frac {2 a^2 \tanh ^{-1}(a x)}{3 x^3}+\frac {8}{15} a^5 \log (x)-\frac {a^4 \tanh ^{-1}(a x)}{x}-\frac {a}{20 x^4}-\frac {\tanh ^{-1}(a x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)^2*ArcTanh[a*x])/x^6,x]

[Out]

-a/(20*x^4) + (7*a^3)/(30*x^2) - ArcTanh[a*x]/(5*x^5) + (2*a^2*ArcTanh[a*x])/(3*x^3) - (a^4*ArcTanh[a*x])/x +
(8*a^5*Log[x])/15 - (4*a^5*Log[1 - a^2*x^2])/15

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 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)}{x^6} \, dx &=\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\\ &=-\left (\left (2 a^2\right ) \int \frac {\tanh ^{-1}(a x)}{x^4} \, dx\right )+a^4 \int \frac {\tanh ^{-1}(a x)}{x^2} \, dx+\int \frac {\tanh ^{-1}(a x)}{x^6} \, dx\\ &=-\frac {\tanh ^{-1}(a x)}{5 x^5}+\frac {2 a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac {a^4 \tanh ^{-1}(a x)}{x}+\frac {1}{5} a \int \frac {1}{x^5 \left (1-a^2 x^2\right )} \, dx-\frac {1}{3} \left (2 a^3\right ) \int \frac {1}{x^3 \left (1-a^2 x^2\right )} \, dx+a^5 \int \frac {1}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac {\tanh ^{-1}(a x)}{5 x^5}+\frac {2 a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac {a^4 \tanh ^{-1}(a x)}{x}+\frac {1}{10} a \operatorname {Subst}\left (\int \frac {1}{x^3 \left (1-a^2 x\right )} \, dx,x,x^2\right )-\frac {1}{3} a^3 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-a^2 x\right )} \, dx,x,x^2\right )+\frac {1}{2} a^5 \operatorname {Subst}\left (\int \frac {1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {\tanh ^{-1}(a x)}{5 x^5}+\frac {2 a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac {a^4 \tanh ^{-1}(a x)}{x}+\frac {1}{10} a \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}{3} a^3 \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}{2} a^5 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} a^7 \operatorname {Subst}\left (\int \frac {1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac {a}{20 x^4}+\frac {7 a^3}{30 x^2}-\frac {\tanh ^{-1}(a x)}{5 x^5}+\frac {2 a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac {a^4 \tanh ^{-1}(a x)}{x}+\frac {8}{15} a^5 \log (x)-\frac {4}{15} a^5 \log \left (1-a^2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 1.00 \[ \frac {8}{15} a^5 \log (x)-\frac {a^4 \tanh ^{-1}(a x)}{x}+\frac {7 a^3}{30 x^2}+\frac {2 a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac {4}{15} a^5 \log \left (1-a^2 x^2\right )-\frac {\tanh ^{-1}(a x)}{5 x^5}-\frac {a}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - a^2*x^2)^2*ArcTanh[a*x])/x^6,x]

[Out]

-1/20*a/x^4 + (7*a^3)/(30*x^2) - ArcTanh[a*x]/(5*x^5) + (2*a^2*ArcTanh[a*x])/(3*x^3) - (a^4*ArcTanh[a*x])/x +
(8*a^5*Log[x])/15 - (4*a^5*Log[1 - a^2*x^2])/15

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fricas [A]  time = 0.55, size = 81, normalized size = 0.98 \[ -\frac {16 \, a^{5} x^{5} \log \left (a^{2} x^{2} - 1\right ) - 32 \, a^{5} x^{5} \log \relax (x) - 14 \, a^{3} x^{3} + 3 \, a x + 2 \, {\left (15 \, a^{4} x^{4} - 10 \, a^{2} x^{2} + 3\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{60 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^2*arctanh(a*x)/x^6,x, algorithm="fricas")

[Out]

-1/60*(16*a^5*x^5*log(a^2*x^2 - 1) - 32*a^5*x^5*log(x) - 14*a^3*x^3 + 3*a*x + 2*(15*a^4*x^4 - 10*a^2*x^2 + 3)*
log(-(a*x + 1)/(a*x - 1)))/x^5

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giac [B]  time = 0.18, size = 265, normalized size = 3.19 \[ -\frac {4}{15} \, {\left (2 \, a^{4} \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right ) - 2 \, a^{4} \log \left ({\left | -\frac {a x + 1}{a x - 1} - 1 \right |}\right ) + \frac {\frac {2 \, {\left (a x + 1\right )}^{3} a^{4}}{{\left (a x - 1\right )}^{3}} + \frac {7 \, {\left (a x + 1\right )}^{2} a^{4}}{{\left (a x - 1\right )}^{2}} + \frac {2 \, {\left (a x + 1\right )} a^{4}}{a x - 1}}{{\left (\frac {a x + 1}{a x - 1} + 1\right )}^{4}} - \frac {2 \, {\left (\frac {10 \, {\left (a x + 1\right )}^{2} a^{4}}{{\left (a x - 1\right )}^{2}} + \frac {5 \, {\left (a x + 1\right )} a^{4}}{a x - 1} + a^{4}\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{{\left (\frac {a x + 1}{a x - 1} + 1\right )}^{5}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^2*arctanh(a*x)/x^6,x, algorithm="giac")

[Out]

-4/15*(2*a^4*log(abs(-a*x - 1)/abs(a*x - 1)) - 2*a^4*log(abs(-(a*x + 1)/(a*x - 1) - 1)) + (2*(a*x + 1)^3*a^4/(
a*x - 1)^3 + 7*(a*x + 1)^2*a^4/(a*x - 1)^2 + 2*(a*x + 1)*a^4/(a*x - 1))/((a*x + 1)/(a*x - 1) + 1)^4 - 2*(10*(a
*x + 1)^2*a^4/(a*x - 1)^2 + 5*(a*x + 1)*a^4/(a*x - 1) + a^4)*log(-(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a
*x - 1) - a) + 1)/(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) - 1))/((a*x + 1)/(a*x - 1) + 1)^5)*
a

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maple [A]  time = 0.04, size = 80, normalized size = 0.96 \[ -\frac {a^{4} \arctanh \left (a x \right )}{x}+\frac {2 a^{2} \arctanh \left (a x \right )}{3 x^{3}}-\frac {\arctanh \left (a x \right )}{5 x^{5}}-\frac {a}{20 x^{4}}+\frac {7 a^{3}}{30 x^{2}}+\frac {8 a^{5} \ln \left (a x \right )}{15}-\frac {4 a^{5} \ln \left (a x -1\right )}{15}-\frac {4 a^{5} \ln \left (a x +1\right )}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)^2*arctanh(a*x)/x^6,x)

[Out]

-a^4*arctanh(a*x)/x+2/3*a^2*arctanh(a*x)/x^3-1/5*arctanh(a*x)/x^5-1/20*a/x^4+7/30*a^3/x^2+8/15*a^5*ln(a*x)-4/1
5*a^5*ln(a*x-1)-4/15*a^5*ln(a*x+1)

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maxima [A]  time = 0.31, size = 71, normalized size = 0.86 \[ -\frac {1}{60} \, {\left (16 \, a^{4} \log \left (a^{2} x^{2} - 1\right ) - 16 \, a^{4} \log \left (x^{2}\right ) - \frac {14 \, a^{2} x^{2} - 3}{x^{4}}\right )} a - \frac {{\left (15 \, a^{4} x^{4} - 10 \, a^{2} x^{2} + 3\right )} \operatorname {artanh}\left (a x\right )}{15 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^2*arctanh(a*x)/x^6,x, algorithm="maxima")

[Out]

-1/60*(16*a^4*log(a^2*x^2 - 1) - 16*a^4*log(x^2) - (14*a^2*x^2 - 3)/x^4)*a - 1/15*(15*a^4*x^4 - 10*a^2*x^2 + 3
)*arctanh(a*x)/x^5

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mupad [B]  time = 0.89, size = 70, normalized size = 0.84 \[ \frac {8\,a^5\,\ln \relax (x)}{15}-\frac {a}{20\,x^4}-\frac {\mathrm {atanh}\left (a\,x\right )}{5\,x^5}-\frac {4\,a^5\,\ln \left (a^2\,x^2-1\right )}{15}+\frac {7\,a^3}{30\,x^2}+\frac {2\,a^2\,\mathrm {atanh}\left (a\,x\right )}{3\,x^3}-\frac {a^4\,\mathrm {atanh}\left (a\,x\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atanh(a*x)*(a^2*x^2 - 1)^2)/x^6,x)

[Out]

(8*a^5*log(x))/15 - a/(20*x^4) - atanh(a*x)/(5*x^5) - (4*a^5*log(a^2*x^2 - 1))/15 + (7*a^3)/(30*x^2) + (2*a^2*
atanh(a*x))/(3*x^3) - (a^4*atanh(a*x))/x

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sympy [A]  time = 1.93, size = 88, normalized size = 1.06 \[ \begin {cases} \frac {8 a^{5} \log {\relax (x )}}{15} - \frac {8 a^{5} \log {\left (x - \frac {1}{a} \right )}}{15} - \frac {8 a^{5} \operatorname {atanh}{\left (a x \right )}}{15} - \frac {a^{4} \operatorname {atanh}{\left (a x \right )}}{x} + \frac {7 a^{3}}{30 x^{2}} + \frac {2 a^{2} \operatorname {atanh}{\left (a x \right )}}{3 x^{3}} - \frac {a}{20 x^{4}} - \frac {\operatorname {atanh}{\left (a x \right )}}{5 x^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)**2*atanh(a*x)/x**6,x)

[Out]

Piecewise((8*a**5*log(x)/15 - 8*a**5*log(x - 1/a)/15 - 8*a**5*atanh(a*x)/15 - a**4*atanh(a*x)/x + 7*a**3/(30*x
**2) + 2*a**2*atanh(a*x)/(3*x**3) - a/(20*x**4) - atanh(a*x)/(5*x**5), Ne(a, 0)), (0, True))

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