3.2.65 \(\int (1-a^2 x^2) \tanh ^{-1}(a x) \, dx\) [165]

Optimal. Leaf size=64 \[ \frac {1-a^2 x^2}{6 a}+\frac {2}{3} x \tanh ^{-1}(a x)+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{3 a} \]

[Out]

1/6*(-a^2*x^2+1)/a+2/3*x*arctanh(a*x)+1/3*x*(-a^2*x^2+1)*arctanh(a*x)+1/3*ln(-a^2*x^2+1)/a

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Rubi [A]
time = 0.02, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6089, 6021, 266} \begin {gather*} \frac {1-a^2 x^2}{6 a}+\frac {\log \left (1-a^2 x^2\right )}{3 a}+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {2}{3} x \tanh ^{-1}(a x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)*ArcTanh[a*x],x]

[Out]

(1 - a^2*x^2)/(6*a) + (2*x*ArcTanh[a*x])/3 + (x*(1 - a^2*x^2)*ArcTanh[a*x])/3 + Log[1 - a^2*x^2]/(3*a)

Rule 266

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 6021

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])

Rule 6089

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

Rubi steps

\begin {align*} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx &=\frac {1-a^2 x^2}{6 a}+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {2}{3} \int \tanh ^{-1}(a x) \, dx\\ &=\frac {1-a^2 x^2}{6 a}+\frac {2}{3} x \tanh ^{-1}(a x)+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)-\frac {1}{3} (2 a) \int \frac {x}{1-a^2 x^2} \, dx\\ &=\frac {1-a^2 x^2}{6 a}+\frac {2}{3} x \tanh ^{-1}(a x)+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{3 a}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 47, normalized size = 0.73 \begin {gather*} -\frac {a x^2}{6}+x \tanh ^{-1}(a x)-\frac {1}{3} a^2 x^3 \tanh ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(a*x^2) + x*ArcTanh[a*x] - (a^2*x^3*ArcTanh[a*x])/3 + Log[1 - a^2*x^2]/(3*a)

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Maple [A]
time = 0.08, size = 49, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {-\frac {a^{3} x^{3} \arctanh \left (a x \right )}{3}+a x \arctanh \left (a x \right )-\frac {a^{2} x^{2}}{6}+\frac {\ln \left (a x -1\right )}{3}+\frac {\ln \left (a x +1\right )}{3}}{a}\)	\(49\)
default	\(\frac {-\frac {a^{3} x^{3} \arctanh \left (a x \right )}{3}+a x \arctanh \left (a x \right )-\frac {a^{2} x^{2}}{6}+\frac {\ln \left (a x -1\right )}{3}+\frac {\ln \left (a x +1\right )}{3}}{a}\)	\(49\)
risch	\(\left (-\frac {1}{6} a^{2} x^{3}+\frac {1}{2} x \right ) \ln \left (a x +1\right )+\frac {a^{2} x^{3} \ln \left (-a x +1\right )}{6}-\frac {a \,x^{2}}{6}-\frac {x \ln \left (-a x +1\right )}{2}+\frac {\ln \left (a^{2} x^{2}-1\right )}{3 a}\)	\(67\)
meijerg	\(-\frac {\frac {2 a^{2} x^{2} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (-a^{2} x^{2}+1\right )}{4 a}-\frac {\frac {2 a^{2} x^{2}}{3}-\frac {2 a^{4} x^{4} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{3}}{4 a}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-1/3*a^3*x^3*arctanh(a*x)+a*x*arctanh(a*x)-1/6*a^2*x^2+1/3*ln(a*x-1)+1/3*ln(a*x+1))

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Maxima [A]
time = 0.27, size = 47, normalized size = 0.73 \begin {gather*} -\frac {1}{6} \, {\left (x^{2} - \frac {2 \, \log \left (a x + 1\right )}{a^{2}} - \frac {2 \, \log \left (a x - 1\right )}{a^{2}}\right )} a - \frac {1}{3} \, {\left (a^{2} x^{3} - 3 \, x\right )} \operatorname {artanh}\left (a x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(x^2 - 2*log(a*x + 1)/a^2 - 2*log(a*x - 1)/a^2)*a - 1/3*(a^2*x^3 - 3*x)*arctanh(a*x)

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Fricas [A]
time = 0.36, size = 53, normalized size = 0.83 \begin {gather*} -\frac {a^{2} x^{2} + {\left (a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 2 \, \log \left (a^{2} x^{2} - 1\right )}{6 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(a^2*x^2 + (a^3*x^3 - 3*a*x)*log(-(a*x + 1)/(a*x - 1)) - 2*log(a^2*x^2 - 1))/a

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Sympy [A]
time = 0.20, size = 49, normalized size = 0.77 \begin {gather*} \begin {cases} - \frac {a^{2} x^{3} \operatorname {atanh}{\left (a x \right )}}{3} - \frac {a x^{2}}{6} + x \operatorname {atanh}{\left (a x \right )} + \frac {2 \log {\left (x - \frac {1}{a} \right )}}{3 a} + \frac {2 \operatorname {atanh}{\left (a x \right )}}{3 a} & \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)

[Out]

Piecewise((-a**2*x**3*atanh(a*x)/3 - a*x**2/6 + x*atanh(a*x) + 2*log(x - 1/a)/(3*a) + 2*atanh(a*x)/(3*a), Ne(a
, 0)), (0, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (54) = 108\).
time = 0.42, size = 203, normalized size = 3.17 \begin {gather*} \frac {2}{3} \, a {\left (\frac {\log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{2}} - \frac {\log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{2}} - \frac {{\left (\frac {3 \, {\left (a x + 1\right )}}{a x - 1} - 1\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 )}{a^{2} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{3}} - \frac {a x + 1}{{\left (a x - 1\right )} a^{2} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*a*(log(abs(-a*x - 1)/abs(a*x - 1))/a^2 - log(abs(-(a*x + 1)/(a*x - 1) + 1))/a^2 - (3*(a*x + 1)/(a*x - 1) -
1)*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^2*((a*x + 1)/(a*x - 1) - 1)^3) - (a*x + 1)/((a*x - 1)*a^2*((a*x + 1)/(a*x - 1) - 1)
^2))

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Mupad [B]
time = 0.84, size = 40, normalized size = 0.62 \begin {gather*} x\,\mathrm {atanh}\left (a\,x\right )-\frac {a\,x^2}{6}+\frac {\ln \left (a^2\,x^2-1\right )}{3\,a}-\frac {a^2\,x^3\,\mathrm {atanh}\left (a\,x\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-atanh(a*x)*(a^2*x^2 - 1),x)

[Out]

x*atanh(a*x) - (a*x^2)/6 + log(a^2*x^2 - 1)/(3*a) - (a^2*x^3*atanh(a*x))/3

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