3.5.29 \(\int x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\) [429]

Optimal. Leaf size=59 \[ \frac {x \sqrt {1-a^2 x^2}}{6 a}+\frac {\text {ArcSin}(a x)}{6 a^2}-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 a^2} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6141, 201, 222} \begin {gather*} \frac {\text {ArcSin}(a x)}{6 a^2}+\frac {x \sqrt {1-a^2 x^2}}{6 a}-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x],x]

[Out]

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

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 6141

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 49, normalized size = 0.83 \begin {gather*} \frac {a x \sqrt {1-a^2 x^2}+\text {ArcSin}(a x)-2 \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{6 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 1.36, size = 99, normalized size = 1.68

method	result	size
default	\(\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (2 a^{2} x^{2} \arctanh \left (a x \right )+a x -2 \arctanh \left (a x \right )\right )}{6 a^{2}}+\frac {i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}+i\right )}{6 a^{2}}-\frac {i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-i\right )}{6 a^{2}}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.46, size = 50, normalized size = 0.85 \begin {gather*} -\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \operatorname {artanh}\left (a x\right )}{3 \, a^{2}} + \frac {\sqrt {-a^{2} x^{2} + 1} x + \frac {\arcsin \left (a x\right )}{a}}{6 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.39, size = 72, normalized size = 1.22 \begin {gather*} \frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x + {\left (a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )\right )} - 2 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{6 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(-(a*x - 1)*(a*x + 1))*atanh(a*x), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x\,\mathrm {atanh}\left (a\,x\right )\,\sqrt {1-a^2\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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