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

Optimal. Leaf size=144 \[ -\frac {i \sqrt {1-x^2} \text {Li}_2\left (-\frac {i \sqrt {1-x}}{\sqrt {x+1}}\right )}{\sqrt {a-a x^2}}+\frac {i \sqrt {1-x^2} \text {Li}_2\left (\frac {i \sqrt {1-x}}{\sqrt {x+1}}\right )}{\sqrt {a-a x^2}}-\frac {2 \sqrt {1-x^2} \tan ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {x+1}}\right ) \tanh ^{-1}(x)}{\sqrt {a-a x^2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5954, 5950} \[ -\frac {i \sqrt {1-x^2} \text {PolyLog}\left (2,-\frac {i \sqrt {1-x}}{\sqrt {x+1}}\right )}{\sqrt {a-a x^2}}+\frac {i \sqrt {1-x^2} \text {PolyLog}\left (2,\frac {i \sqrt {1-x}}{\sqrt {x+1}}\right )}{\sqrt {a-a x^2}}-\frac {2 \sqrt {1-x^2} \tan ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {x+1}}\right ) \tanh ^{-1}(x)}{\sqrt {a-a x^2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[x]/Sqrt[a - a*x^2],x]

[Out]

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

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

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Mathematica [A]  time = 0.11, size = 90, normalized size = 0.62 \[ -\frac {i \sqrt {a \left (1-x^2\right )} \left (\text {Li}_2\left (-i e^{-\tanh ^{-1}(x)}\right )-\text {Li}_2\left (i e^{-\tanh ^{-1}(x)}\right )+\tanh ^{-1}(x) \left (\log \left (1-i e^{-\tanh ^{-1}(x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(x)}\right )\right )\right )}{a \sqrt {1-x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a x^{2} + a} \operatorname {artanh}\relax (x)}{a x^{2} - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.50, size = 210, normalized size = 1.46 \[ \frac {i \ln \left (1+\frac {i \left (1+x \right )}{\sqrt {-x^{2}+1}}\right ) \arctanh \relax (x ) \sqrt {-x^{2}+1}\, \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{a \left (x^{2}-1\right )}-\frac {i \ln \left (1-\frac {i \left (1+x \right )}{\sqrt {-x^{2}+1}}\right ) \arctanh \relax (x ) \sqrt {-x^{2}+1}\, \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{a \left (x^{2}-1\right )}+\frac {i \dilog \left (1+\frac {i \left (1+x \right )}{\sqrt {-x^{2}+1}}\right ) \sqrt {-x^{2}+1}\, \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{a \left (x^{2}-1\right )}-\frac {i \dilog \left (1-\frac {i \left (1+x \right )}{\sqrt {-x^{2}+1}}\right ) \sqrt {-x^{2}+1}\, \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{a \left (x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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