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

Optimal. Leaf size=186 \[ -\frac{i a \sqrt{1-x^2} \text{PolyLog}\left (2,-\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{2 \sqrt{a-a x^2}}+\frac{i a \sqrt{1-x^2} \text{PolyLog}\left (2,\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{2 \sqrt{a-a x^2}}+\frac{1}{2} \sqrt{a-a x^2}+\frac{1}{2} x \sqrt{a-a x^2} \tanh ^{-1}(x)-\frac{a \sqrt{1-x^2} \tan ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{x+1}}\right ) \tanh ^{-1}(x)}{\sqrt{a-a x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0880305, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5942, 5954, 5950} \[ -\frac{i a \sqrt{1-x^2} \text{PolyLog}\left (2,-\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{2 \sqrt{a-a x^2}}+\frac{i a \sqrt{1-x^2} \text{PolyLog}\left (2,\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{2 \sqrt{a-a x^2}}+\frac{1}{2} \sqrt{a-a x^2}+\frac{1}{2} x \sqrt{a-a x^2} \tanh ^{-1}(x)-\frac{a \sqrt{1-x^2} \tan ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{x+1}}\right ) \tanh ^{-1}(x)}{\sqrt{a-a x^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - a*x^2]*ArcTanh[x],x]

[Out]

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

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 5954

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 - c^2*x^2]/S
qrt[d + e*x^2], Int[(a + b*ArcTanh[c*x])^p/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d
+ e, 0] && IGtQ[p, 0] &&  !GtQ[d, 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{a-a x^2} \tanh ^{-1}(x) \, dx &=\frac{1}{2} \sqrt{a-a x^2}+\frac{1}{2} x \sqrt{a-a x^2} \tanh ^{-1}(x)+\frac{1}{2} a \int \frac{\tanh ^{-1}(x)}{\sqrt{a-a x^2}} \, dx\\ &=\frac{1}{2} \sqrt{a-a x^2}+\frac{1}{2} x \sqrt{a-a x^2} \tanh ^{-1}(x)+\frac{\left (a \sqrt{1-x^2}\right ) \int \frac{\tanh ^{-1}(x)}{\sqrt{1-x^2}} \, dx}{2 \sqrt{a-a x^2}}\\ &=\frac{1}{2} \sqrt{a-a x^2}+\frac{1}{2} x \sqrt{a-a x^2} \tanh ^{-1}(x)-\frac{a \sqrt{1-x^2} \tan ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{1+x}}\right ) \tanh ^{-1}(x)}{\sqrt{a-a x^2}}-\frac{i a \sqrt{1-x^2} \text{Li}_2\left (-\frac{i \sqrt{1-x}}{\sqrt{1+x}}\right )}{2 \sqrt{a-a x^2}}+\frac{i a \sqrt{1-x^2} \text{Li}_2\left (\frac{i \sqrt{1-x}}{\sqrt{1+x}}\right )}{2 \sqrt{a-a x^2}}\\ \end{align*}

Mathematica [A]  time = 0.324162, size = 97, normalized size = 0.52 \[ \frac{1}{2} \sqrt{a \left (1-x^2\right )} \left (-\frac{i \left (\text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(x)}\right )-\text{PolyLog}\left (2,i e^{-\tanh ^{-1}(x)}\right )+\tanh ^{-1}(x) \left (\log \left (1-i e^{-\tanh ^{-1}(x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(x)}\right )\right )\right )}{\sqrt{1-x^2}}+x \tanh ^{-1}(x)+1\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.434, size = 229, normalized size = 1.2 \begin{align*}{\frac{{\it Artanh} \left ( x \right ) x+1}{2}\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}}+{\frac{{\frac{i}{2}}{\it Artanh} \left ( x \right ) }{ \left ( -1+x \right ) \left ( 1+x \right ) }\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}\sqrt{-{x}^{2}+1}\ln \left ( 1+{i \left ( 1+x \right ){\frac{1}{\sqrt{-{x}^{2}+1}}}} \right ) }-{\frac{{\frac{i}{2}}{\it Artanh} \left ( x \right ) }{ \left ( -1+x \right ) \left ( 1+x \right ) }\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}\sqrt{-{x}^{2}+1}\ln \left ( 1-{i \left ( 1+x \right ){\frac{1}{\sqrt{-{x}^{2}+1}}}} \right ) }+{\frac{{\frac{i}{2}}}{ \left ( -1+x \right ) \left ( 1+x \right ) }\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}\sqrt{-{x}^{2}+1}{\it dilog} \left ( 1+{i \left ( 1+x \right ){\frac{1}{\sqrt{-{x}^{2}+1}}}} \right ) }-{\frac{{\frac{i}{2}}}{ \left ( -1+x \right ) \left ( 1+x \right ) }\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}\sqrt{-{x}^{2}+1}{\it dilog} \left ( 1-{i \left ( 1+x \right ){\frac{1}{\sqrt{-{x}^{2}+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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