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

Optimal. Leaf size=186 \[ -\frac {i a \sqrt {1-x^2} \text {Li}_2\left (-\frac {i \sqrt {1-x}}{\sqrt {x+1}}\right )}{2 \sqrt {a-a x^2}}+\frac {i a \sqrt {1-x^2} \text {Li}_2\left (\frac {i \sqrt {1-x}}{\sqrt {x+1}}\right )}{2 \sqrt {a-a x^2}}+\frac {1}{2} \sqrt {a-a x^2}+\frac {1}{2} x \sqrt {a-a x^2} \tanh ^{-1}(x)-\frac {a \sqrt {1-x^2} \tan ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {x+1}}\right ) \tanh ^{-1}(x)}{\sqrt {a-a x^2}} \]

[Out]

-a*arctan((1-x)^(1/2)/(1+x)^(1/2))*arctanh(x)*(-x^2+1)^(1/2)/(-a*x^2+a)^(1/2)-1/2*I*a*polylog(2,-I*(1-x)^(1/2)
/(1+x)^(1/2))*(-x^2+1)^(1/2)/(-a*x^2+a)^(1/2)+1/2*I*a*polylog(2,I*(1-x)^(1/2)/(1+x)^(1/2))*(-x^2+1)^(1/2)/(-a*
x^2+a)^(1/2)+1/2*(-a*x^2+a)^(1/2)+1/2*x*arctanh(x)*(-a*x^2+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 186, 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 = {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 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]

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]

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 = 0.32, size = 97, normalized size = 0.52 \[ \frac {1}{2} \sqrt {a \left (1-x^2\right )} \left (-\frac {i \left (\text {Li}_2\left (-i e^{-\tanh ^{-1}(x)}\right )-\text {Li}_2\left (i e^{-\tanh ^{-1}(x)}\right )+\tanh ^{-1}(x) \left (\log \left (1-i e^{-\tanh ^{-1}(x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(x)}\right )\right )\right )}{\sqrt {1-x^2}}+x \tanh ^{-1}(x)+1\right ) \]

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

________________________________________________________________________________________

fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-a x^{2} + a} \operatorname {artanh}\relax (x), x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a x^{2} + a} \operatorname {artanh}\relax (x)\,{d x} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.61, size = 229, normalized size = 1.23 \[ \frac {\left (\arctanh \relax (x ) x +1\right ) \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{2}+\frac {i \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}\, \sqrt {-x^{2}+1}\, \arctanh \relax (x ) \ln \left (1+\frac {i \left (1+x \right )}{\sqrt {-x^{2}+1}}\right )}{2 \left (1+x \right ) \left (-1+x \right )}-\frac {i \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}\, \sqrt {-x^{2}+1}\, \arctanh \relax (x ) \ln \left (1-\frac {i \left (1+x \right )}{\sqrt {-x^{2}+1}}\right )}{2 \left (1+x \right ) \left (-1+x \right )}+\frac {i \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}\, \sqrt {-x^{2}+1}\, \dilog \left (1+\frac {i \left (1+x \right )}{\sqrt {-x^{2}+1}}\right )}{2 \left (1+x \right ) \left (-1+x \right )}-\frac {i \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}\, \sqrt {-x^{2}+1}\, \dilog \left (1-\frac {i \left (1+x \right )}{\sqrt {-x^{2}+1}}\right )}{2 \left (1+x \right ) \left (-1+x \right )} \]

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))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a x^{2} + a} \operatorname {artanh}\relax (x)\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {atanh}\relax (x)\,\sqrt {a-a\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(x)*(a - a*x^2)^(1/2),x)

[Out]

int(atanh(x)*(a - a*x^2)^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- a \left (x - 1\right ) \left (x + 1\right )} \operatorname {atanh}{\relax (x )}\, dx \]

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)

________________________________________________________________________________________