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

Optimal. Leaf size=38 \[ -a \log \left (1-a^2 x^2\right )+a^2 (-x) \tanh ^{-1}(a x)+a \log (x)-\frac {\tanh ^{-1}(a x)}{x} \]

[Out]

-arctanh(a*x)/x-a^2*x*arctanh(a*x)+a*ln(x)-a*ln(-a^2*x^2+1)

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Rubi [A]  time = 0.05, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6014, 5916, 266, 36, 29, 31, 5910, 260} \[ -a \log \left (1-a^2 x^2\right )+a^2 (-x) \tanh ^{-1}(a x)+a \log (x)-\frac {\tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[a*x]/x) - a^2*x*ArcTanh[a*x] + a*Log[x] - a*Log[1 - a^2*x^2]

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 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 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 \frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{x^2} \, dx &=-\left (a^2 \int \tanh ^{-1}(a x) \, dx\right )+\int \frac {\tanh ^{-1}(a x)}{x^2} \, dx\\ &=-\frac {\tanh ^{-1}(a x)}{x}-a^2 x \tanh ^{-1}(a x)+a \int \frac {1}{x \left (1-a^2 x^2\right )} \, dx+a^3 \int \frac {x}{1-a^2 x^2} \, dx\\ &=-\frac {\tanh ^{-1}(a x)}{x}-a^2 x \tanh ^{-1}(a x)-\frac {1}{2} a \log \left (1-a^2 x^2\right )+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {\tanh ^{-1}(a x)}{x}-a^2 x \tanh ^{-1}(a x)-\frac {1}{2} a \log \left (1-a^2 x^2\right )+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} a^3 \operatorname {Subst}\left (\int \frac {1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac {\tanh ^{-1}(a x)}{x}-a^2 x \tanh ^{-1}(a x)+a \log (x)-a \log \left (1-a^2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 38, normalized size = 1.00 \[ -a \log \left (1-a^2 x^2\right )+a^2 (-x) \tanh ^{-1}(a x)+a \log (x)-\frac {\tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[a*x]/x) - a^2*x*ArcTanh[a*x] + a*Log[x] - a*Log[1 - a^2*x^2]

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fricas [A]  time = 0.59, size = 51, normalized size = 1.34 \[ -\frac {2 \, a x \log \left (a^{2} x^{2} - 1\right ) - 2 \, a x \log \relax (x) + {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a*x*log(a^2*x^2 - 1) - 2*a*x*log(x) + (a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1)))/x

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giac [B]  time = 0.38, size = 145, normalized size = 3.82 \[ -a {\left (\frac {2 \, \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{\frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - 1} + \log \left (\frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}}\right ) - \log \left ({\left | \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - 1 \right |}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(2*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)^2/(a*x - 1)^2 - 1) + log((a*x + 1)^2/(a*x - 1)^2) - log(abs((a*x + 1)^2/(
a*x - 1)^2 - 1)))

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maple [A]  time = 0.03, size = 45, normalized size = 1.18 \[ -a^{2} x \arctanh \left (a x \right )-\frac {\arctanh \left (a x \right )}{x}+a \ln \left (a x \right )-a \ln \left (a x -1\right )-a \ln \left (a x +1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*x*arctanh(a*x)-arctanh(a*x)/x+a*ln(a*x)-a*ln(a*x-1)-a*ln(a*x+1)

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maxima [A]  time = 0.31, size = 36, normalized size = 0.95 \[ -a {\left (\log \left (a x + 1\right ) + \log \left (a x - 1\right ) - \log \relax (x)\right )} - {\left (a^{2} x + \frac {1}{x}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(log(a*x + 1) + log(a*x - 1) - log(x)) - (a^2*x + 1/x)*arctanh(a*x)

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mupad [B]  time = 0.81, size = 37, normalized size = 0.97 \[ a\,\ln \relax (x)-a\,\ln \left (a^2\,x^2-1\right )-\frac {\mathrm {atanh}\left (a\,x\right )}{x}-a^2\,x\,\mathrm {atanh}\left (a\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(x) - a*log(a^2*x^2 - 1) - atanh(a*x)/x - a^2*x*atanh(a*x)

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sympy [A]  time = 0.85, size = 41, normalized size = 1.08 \[ \begin {cases} - a^{2} x \operatorname {atanh}{\left (a x \right )} + a \log {\relax (x )} - 2 a \log {\left (x - \frac {1}{a} \right )} - 2 a \operatorname {atanh}{\left (a x \right )} - \frac {\operatorname {atanh}{\left (a x \right )}}{x} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*x*atanh(a*x) + a*log(x) - 2*a*log(x - 1/a) - 2*a*atanh(a*x) - atanh(a*x)/x, Ne(a, 0)), (0, Tr
ue))

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