3.2.71 \(\int \frac {(1-a^2 x^2) \tanh ^{-1}(a x)}{x^6} \, dx\) [171]

Optimal. Leaf size=71 \[ -\frac {a}{20 x^4}+\frac {a^3}{15 x^2}-\frac {\tanh ^{-1}(a x)}{5 x^5}+\frac {a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac {2}{15} a^5 \log (x)+\frac {1}{15} a^5 \log \left (1-a^2 x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6161, 6037, 272, 46} \begin {gather*} -\frac {2}{15} a^5 \log (x)+\frac {a^3}{15 x^2}+\frac {a^2 \tanh ^{-1}(a x)}{3 x^3}+\frac {1}{15} a^5 \log \left (1-a^2 x^2\right )-\frac {\tanh ^{-1}(a x)}{5 x^5}-\frac {a}{20 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 272

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 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 6161

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

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Mathematica [A]
time = 0.02, size = 71, normalized size = 1.00 \begin {gather*} -\frac {a}{20 x^4}+\frac {a^3}{15 x^2}-\frac {\tanh ^{-1}(a x)}{5 x^5}+\frac {a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac {2}{15} a^5 \log (x)+\frac {1}{15} a^5 \log \left (1-a^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.21, size = 68, normalized size = 0.96

method	result	size
derivativedivides	\(a^{5} \left (\frac {\arctanh \left (a x \right )}{3 a^{3} x^{3}}-\frac {\arctanh \left (a x \right )}{5 a^{5} x^{5}}+\frac {\ln \left (a x +1\right )}{15}+\frac {\ln \left (a x -1\right )}{15}-\frac {1}{20 a^{4} x^{4}}+\frac {1}{15 a^{2} x^{2}}-\frac {2 \ln \left (a x \right )}{15}\right )\)	\(68\)
default	\(a^{5} \left (\frac {\arctanh \left (a x \right )}{3 a^{3} x^{3}}-\frac {\arctanh \left (a x \right )}{5 a^{5} x^{5}}+\frac {\ln \left (a x +1\right )}{15}+\frac {\ln \left (a x -1\right )}{15}-\frac {1}{20 a^{4} x^{4}}+\frac {1}{15 a^{2} x^{2}}-\frac {2 \ln \left (a x \right )}{15}\right )\)	\(68\)
risch	\(\frac {\left (5 a^{2} x^{2}-3\right ) \ln \left (a x +1\right )}{30 x^{5}}-\frac {8 \ln \left (x \right ) a^{5} x^{5}-4 \ln \left (-a^{2} x^{2}+1\right ) a^{5} x^{5}-4 a^{3} x^{3}+10 x^{2} \ln \left (-a x +1\right ) a^{2}+3 a x -6 \ln \left (-a x +1\right )}{60 x^{5}}\)	\(94\)
meijerg	\(\frac {a^{5} \left (\frac {\frac {4}{25} a^{4} x^{4}+\frac {4}{15} a^{2} x^{2}+\frac {4}{5}}{a^{4} x^{4}}+\frac {\frac {2 \ln \left (1-\sqrt {a^{2} x^{2}}\right )}{5}-\frac {2 \ln \left (1+\sqrt {a^{2} x^{2}}\right )}{5}}{a^{4} x^{4} \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{5}-\frac {4}{25}+\frac {4 \ln \left (x \right )}{5}+\frac {4 \ln \left (i a \right )}{5}-\frac {1}{a^{4} x^{4}}-\frac {2}{3 a^{2} x^{2}}\right )}{4}+\frac {a^{5} \left (-\frac {2 \left (10 a^{2} x^{2}+30\right )}{45 a^{2} x^{2}}-\frac {2 \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 a^{2} x^{2} \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{3}+\frac {4}{9}-\frac {4 \ln \left (x \right )}{3}-\frac {4 \ln \left (i a \right )}{3}+\frac {2}{a^{2} x^{2}}\right )}{4}\)	\(224\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 62, normalized size = 0.87 \begin {gather*} \frac {1}{60} \, {\left (4 \, a^{4} \log \left (a^{2} x^{2} - 1\right ) - 4 \, a^{4} \log \left (x^{2}\right ) + \frac {4 \, a^{2} x^{2} - 3}{x^{4}}\right )} a + \frac {{\left (5 \, a^{2} x^{2} - 3\right )} \operatorname {artanh}\left (a x\right )}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 73, normalized size = 1.03 \begin {gather*} \frac {4 \, a^{5} x^{5} \log \left (a^{2} x^{2} - 1\right ) - 8 \, a^{5} x^{5} \log \left (x\right ) + 4 \, a^{3} x^{3} - 3 \, a x + 2 \, {\left (5 \, a^{2} x^{2} - 3\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(4*a^5*x^5*log(a^2*x^2 - 1) - 8*a^5*x^5*log(x) + 4*a^3*x^3 - 3*a*x + 2*(5*a^2*x^2 - 3)*log(-(a*x + 1)/(a*
x - 1)))/x^5

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Sympy [A]
time = 0.55, size = 75, normalized size = 1.06 \begin {gather*} \begin {cases} - \frac {2 a^{5} \log {\left (x \right )}}{15} + \frac {2 a^{5} \log {\left (x - \frac {1}{a} \right )}}{15} + \frac {2 a^{5} \operatorname {atanh}{\left (a x \right )}}{15} + \frac {a^{3}}{15 x^{2}} + \frac {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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*a**5*log(x)/15 + 2*a**5*log(x - 1/a)/15 + 2*a**5*atanh(a*x)/15 + a**3/(15*x**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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (59) = 118\).
time = 0.40, size = 281, normalized size = 3.96 \begin {gather*} \frac {2}{15} \, {\left (a^{4} \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right ) - a^{4} \log \left ({\left | -\frac {a x + 1}{a x - 1} - 1 \right |}\right ) + \frac {\frac {{\left (a x + 1\right )}^{3} a^{4}}{{\left (a x - 1\right )}^{3}} - \frac {4 \, {\left (a x + 1\right )}^{2} a^{4}}{{\left (a x - 1\right )}^{2}} + \frac {{\left (a x + 1\right )} a^{4}}{a x - 1}}{{\left (\frac {a x + 1}{a x - 1} + 1\right )}^{4}} - \frac {{\left (\frac {15 \, {\left (a x + 1\right )}^{3} a^{4}}{{\left (a x - 1\right )}^{3}} - \frac {5 \, {\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(a^4*log(abs(-a*x - 1)/abs(a*x - 1)) - a^4*log(abs(-(a*x + 1)/(a*x - 1) - 1)) + ((a*x + 1)^3*a^4/(a*x - 1
)^3 - 4*(a*x + 1)^2*a^4/(a*x - 1)^2 + (a*x + 1)*a^4/(a*x - 1))/((a*x + 1)/(a*x - 1) + 1)^4 - (15*(a*x + 1)^3*a
^4/(a*x - 1)^3 - 5*(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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Mupad [B]
time = 0.88, size = 59, normalized size = 0.83 \begin {gather*} \frac {a^5\,\ln \left (a^2\,x^2-1\right )}{15}-\frac {\frac {\mathrm {atanh}\left (a\,x\right )}{5}+\frac {a\,x}{20}-\frac {a^3\,x^3}{15}-\frac {a^2\,x^2\,\mathrm {atanh}\left (a\,x\right )}{3}}{x^5}-\frac {2\,a^5\,\ln \left (x\right )}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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