3.5.67 \(\int \frac {\tanh ^{-1}(a x)}{\sqrt {c-a^2 c x^2}} \, dx\) [467]

Optimal. Leaf size=182 \[ -\frac {2 \sqrt {1-a^2 x^2} \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a \sqrt {c-a^2 c x^2}}-\frac {i \sqrt {1-a^2 x^2} \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a \sqrt {c-a^2 c x^2}}+\frac {i \sqrt {1-a^2 x^2} \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6101, 6097} \begin {gather*} -\frac {2 \sqrt {1-a^2 x^2} \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a \sqrt {c-a^2 c x^2}}-\frac {i \sqrt {1-a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a \sqrt {c-a^2 c x^2}}+\frac {i \sqrt {1-a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6097

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 6101

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

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Mathematica [A]
time = 0.08, size = 109, normalized size = 0.60 \begin {gather*} -\frac {i \sqrt {c \left (1-a^2 x^2\right )} \left (\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 )+\text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-\text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )\right )}{a c \sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[c*(1 - a^2*x^2)]*(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]]))/(a*c*Sqrt[1 - a^2*x^2])

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Maple [A]
time = 1.60, size = 302, normalized size = 1.66

method	result	size
default	\(\frac {i \sqrt {-a^{2} x^{2}+1}\, \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}\, \arctanh \left (a x \right ) \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{\left (a x -1\right ) \left (a x +1\right ) a c}-\frac {i \sqrt {-a^{2} x^{2}+1}\, \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}\, \arctanh \left (a x \right ) \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{\left (a x -1\right ) \left (a x +1\right ) a c}+\frac {i \sqrt {-a^{2} x^{2}+1}\, \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}\, \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{\left (a x -1\right ) \left (a x +1\right ) a c}-\frac {i \sqrt {-a^{2} x^{2}+1}\, \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}\, \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{\left (a x -1\right ) \left (a x +1\right ) a c}\)	\(302\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {atanh}\left (a\,x\right )}{\sqrt {c-a^2\,c\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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