3.3.32 \(\int \frac {\tanh ^{-1}(a x)}{x^2 (1-a^2 x^2)} \, dx\) [232]

Optimal. Leaf size=41 \[ -\frac {\tanh ^{-1}(a x)}{x}+\frac {1}{2} a \tanh ^{-1}(a x)^2+a \log (x)-\frac {1}{2} a \log \left (1-a^2 x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6129, 6037, 272, 36, 29, 31, 6095} \begin {gather*} -\frac {1}{2} a \log \left (1-a^2 x^2\right )+a \log (x)+\frac {1}{2} a \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[a*x]/x) + (a*ArcTanh[a*x]^2)/2 + a*Log[x] - (a*Log[1 - a^2*x^2])/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 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 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 6129

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 41, normalized size = 1.00 \begin {gather*} -\frac {\tanh ^{-1}(a x)}{x}+\frac {1}{2} a \tanh ^{-1}(a x)^2+a \log (x)-\frac {1}{2} a \log \left (1-a^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(119\) vs. \(2(37)=74\).
time = 0.30, size = 120, normalized size = 2.93

method	result	size
risch	\(\frac {a \ln \left (a x +1\right )^{2}}{8}-\frac {\left (a x \ln \left (-a x +1\right )+2\right ) \ln \left (a x +1\right )}{4 x}+\frac {a \ln \left (-a x +1\right )^{2} x +8 a \ln \left (x \right ) x -4 a \ln \left (a^{2} x^{2}-1\right ) x +4 \ln \left (-a x +1\right )}{8 x}\)	\(83\)
derivativedivides	\(a \left (-\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{2}-\frac {\arctanh \left (a x \right )}{a x}+\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{2}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (a x -1\right )^{2}}{8}-\frac {\ln \left (a x -1\right )}{2}+\ln \left (a x \right )-\frac {\ln \left (a x +1\right )}{2}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (a x +1\right )^{2}}{8}\right )\)	\(120\)
default	\(a \left (-\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{2}-\frac {\arctanh \left (a x \right )}{a x}+\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{2}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (a x -1\right )^{2}}{8}-\frac {\ln \left (a x -1\right )}{2}+\ln \left (a x \right )-\frac {\ln \left (a x +1\right )}{2}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (a x +1\right )^{2}}{8}\right )\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(-1/2*arctanh(a*x)*ln(a*x-1)-arctanh(a*x)/a/x+1/2*arctanh(a*x)*ln(a*x+1)+1/4*ln(a*x-1)*ln(1/2*a*x+1/2)-1/8*l
n(a*x-1)^2-1/2*ln(a*x-1)+ln(a*x)-1/2*ln(a*x+1)+1/4*(ln(a*x+1)-ln(1/2*a*x+1/2))*ln(-1/2*a*x+1/2)-1/8*ln(a*x+1)^
2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (37) = 74\).
time = 0.28, size = 82, normalized size = 2.00 \begin {gather*} \frac {1}{8} \, {\left (2 \, {\left (\log \left (a x - 1\right ) - 2\right )} \log \left (a x + 1\right ) - \log \left (a x + 1\right )^{2} - \log \left (a x - 1\right )^{2} - 4 \, \log \left (a x - 1\right ) + 8 \, \log \left (x\right )\right )} a + \frac {1}{2} \, {\left (a \log \left (a x + 1\right ) - a \log \left (a x - 1\right ) - \frac {2}{x}\right )} \operatorname {artanh}\left (a x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(log(a*x - 1) - 2)*log(a*x + 1) - log(a*x + 1)^2 - log(a*x - 1)^2 - 4*log(a*x - 1) + 8*log(x))*a + 1/2*
(a*log(a*x + 1) - a*log(a*x - 1) - 2/x)*arctanh(a*x)

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Fricas [A]
time = 0.38, size = 63, normalized size = 1.54 \begin {gather*} \frac {a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 4 \, a x \log \left (a^{2} x^{2} - 1\right ) + 8 \, a x \log \left (x\right ) - 4 \, \log \left (-\frac {a x + 1}{a x - 1}\right )}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.51, size = 37, normalized size = 0.90 \begin {gather*} \begin {cases} a \log {\left (x \right )} - a \log {\left (x - \frac {1}{a} \right )} + \frac {a \operatorname {atanh}^{2}{\left (a x \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.05, size = 80, normalized size = 1.95 \begin {gather*} \frac {a\,{\ln \left (a\,x+1\right )}^2}{8}+\frac {a\,{\ln \left (1-a\,x\right )}^2}{8}-\frac {\ln \left (a\,x+1\right )}{2\,x}+\frac {\ln \left (1-a\,x\right )}{2\,x}-\frac {a\,\ln \left (a^2\,x^2-1\right )}{2}+a\,\ln \left (x\right )-\frac {a\,\ln \left (a\,x+1\right )\,\ln \left (1-a\,x\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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