3.5.31 \(\int \frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x} \, dx\) [431]

Optimal. Leaf size=100 \[ -\text {ArcSin}(a x)+\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\text {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\text {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \]

[Out]

-arcsin(a*x)-2*arctanh(a*x)*arctanh((-a*x+1)^(1/2)/(a*x+1)^(1/2))+polylog(2,-(-a*x+1)^(1/2)/(a*x+1)^(1/2))-pol
ylog(2,(-a*x+1)^(1/2)/(a*x+1)^(1/2))+(-a^2*x^2+1)^(1/2)*arctanh(a*x)

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6157, 6165, 222} \begin {gather*} \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\text {ArcSin}(a x)+\text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/x,x]

[Out]

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

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 6157

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(
m + 1)*Sqrt[d + e*x^2]*((a + b*ArcTanh[c*x])/(f*(m + 2))), x] + (Dist[d/(m + 2), Int[(f*x)^m*((a + b*ArcTanh[c
*x])/Sqrt[d + e*x^2]), x], x] - Dist[b*c*(d/(f*(m + 2))), Int[(f*x)^(m + 1)/Sqrt[d + e*x^2], x], x]) /; FreeQ[
{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && NeQ[m, -2]

Rule 6165

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.09, size = 91, normalized size = 0.91 \begin {gather*} -2 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right )+\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x) \log \left (1+e^{-\tanh ^{-1}(a x)}\right )+\text {PolyLog}\left (2,-e^{-\tanh ^{-1}(a x)}\right )-\text {PolyLog}\left (2,e^{-\tanh ^{-1}(a x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*ArcTan[Tanh[ArcTanh[a*x]/2]] + Sqrt[1 - a^2*x^2]*ArcTanh[a*x] + ArcTanh[a*x]*Log[1 - E^(-ArcTanh[a*x])] - A
rcTanh[a*x]*Log[1 + E^(-ArcTanh[a*x])] + PolyLog[2, -E^(-ArcTanh[a*x])] - PolyLog[2, E^(-ArcTanh[a*x])]

________________________________________________________________________________________

Maple [A]
time = 1.51, size = 113, normalized size = 1.13

method	result	size
default	\(\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \arctanh \left (a x \right )-2 \arctan \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-\dilog \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-\arctanh \left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-\dilog \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*x^2+1)^(1/2)/x,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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*x^2+1)^(1/2)/x,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________