3.461 \(\int \sqrt{1-a^2 x^2} \tanh ^{-1}(a x) \, dx\)

Optimal. Leaf size=143 \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a}+\frac{\sqrt{1-a^2 x^2}}{2 a}+\frac{1}{2} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{\tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{a} \]

[Out]

Sqrt[1 - a^2*x^2]/(2*a) + (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/2 - (ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[
a*x])/a - ((I/2)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a + ((I/2)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[
1 + a*x]])/a

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Rubi [A]  time = 0.0540413, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {5942, 5950} \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a}+\frac{\sqrt{1-a^2 x^2}}{2 a}+\frac{1}{2} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{\tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - a^2*x^2]*ArcTanh[a*x],x]

[Out]

Sqrt[1 - a^2*x^2]/(2*a) + (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/2 - (ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[
a*x])/a - ((I/2)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a + ((I/2)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[
1 + a*x]])/a

Rule 5942

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]

Rule 5950

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

Rubi steps

\begin{align*} \int \sqrt{1-a^2 x^2} \tanh ^{-1}(a x) \, dx &=\frac{\sqrt{1-a^2 x^2}}{2 a}+\frac{1}{2} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\frac{1}{2} \int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{\sqrt{1-a^2 x^2}}{2 a}+\frac{1}{2} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{\tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)}{a}-\frac{i \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{2 a}+\frac{i \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.066716, size = 117, normalized size = 0.82 \[ \frac{\sqrt{1-a^2 x^2} \left (-\frac{i \left (\text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-\text{PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )+\tanh ^{-1}(a x) \left (\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )\right )}{\sqrt{1-a^2 x^2}}+a x \tanh ^{-1}(a x)+1\right )}{2 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(1 + a*x*ArcTanh[a*x] - (I*(ArcTanh[a*x]*(Log[1 - I/E^ArcTanh[a*x]] - Log[1 + I/E^ArcTanh[a
*x]]) + PolyLog[2, (-I)/E^ArcTanh[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2]))/(2*a)

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Maple [A]  time = 0.21, size = 152, normalized size = 1.1 \begin{align*}{\frac{ax{\it Artanh} \left ( ax \right ) +1}{2\,a}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{\frac{i}{2}}{\it Artanh} \left ( ax \right ) }{a}\ln \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{i}{2}}{\it Artanh} \left ( ax \right ) }{a}\ln \left ( 1-{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{i}{2}}}{a}{\it dilog} \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{i}{2}}}{a}{\it dilog} \left ( 1-{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-a^{2} x^{2} + 1} \operatorname{artanh}\left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{-a^{2} x^{2} + 1} \operatorname{artanh}\left (a x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*arctanh(a*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-a^{2} x^{2} + 1} \operatorname{artanh}\left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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