3.4.89 \(\int \frac {x^2 \tanh ^{-1}(a x)}{(1-a^2 x^2)^{3/2}} \, dx\) [389]

Optimal. Leaf size=137 \[ -\frac {1}{a^3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)}{a^2 \sqrt {1-a^2 x^2}}+\frac {2 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a^3}+\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}-\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3} \]

[Out]

2*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)/a^3+I*polylog(2,-I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^3-I*pol
ylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^3-1/a^3/(-a^2*x^2+1)^(1/2)+x*arctanh(a*x)/a^2/(-a^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6145, 6097} \begin {gather*} \frac {2 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a^3}+\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a^3}-\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a^3}+\frac {x \tanh ^{-1}(a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {1}{a^3 \sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*ArcTanh[a*x])/(1 - a^2*x^2)^(3/2),x]

[Out]

-(1/(a^3*Sqrt[1 - a^2*x^2])) + (x*ArcTanh[a*x])/(a^2*Sqrt[1 - a^2*x^2]) + (2*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x
]]*ArcTanh[a*x])/a^3 + (I*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^3 - (I*PolyLog[2, (I*Sqrt[1 - a*x]
)/Sqrt[1 + a*x]])/a^3

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 6145

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(-b)*((d + e*x^2)
^(q + 1)/(4*c^3*d*(q + 1)^2)), x] + (Dist[1/(2*c^2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x
], x] - Simp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])/(2*c^2*d*(q + 1))), x]) /; FreeQ[{a, b, c, d, e}, x]
&& EqQ[c^2*d + e, 0] && LtQ[q, -1] && NeQ[q, -5/2]

Rubi steps

\begin {align*} \int \frac {x^2 \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=-\frac {1}{a^3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {\int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{a^2}\\ &=-\frac {1}{a^3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)}{a^2 \sqrt {1-a^2 x^2}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a^3}+\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}-\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.20, size = 121, normalized size = 0.88 \begin {gather*} \frac {i \left (\frac {i}{\sqrt {1-a^2 x^2}}-\frac {i a x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}+\tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\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^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*ArcTanh[a*x])/(1 - a^2*x^2)^(3/2),x]

[Out]

(I*(I/Sqrt[1 - a^2*x^2] - (I*a*x*ArcTanh[a*x])/Sqrt[1 - a^2*x^2] + ArcTanh[a*x]*Log[1 - I/E^ArcTanh[a*x]] - Ar
cTanh[a*x]*Log[1 + I/E^ArcTanh[a*x]] + PolyLog[2, (-I)/E^ArcTanh[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]))/a^3

________________________________________________________________________________________

Maple [A]
time = 1.11, size = 190, normalized size = 1.39

method	result	size
default	\(-\frac {\left (\arctanh \left (a x \right )-1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{3} \left (a x -1\right )}-\frac {\left (\arctanh \left (a x \right )+1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{3} \left (a x +1\right )}+\frac {i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{a^{3}}-\frac {i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{a^{3}}+\frac {i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3}}-\frac {i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3}}\)	\(190\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctanh(a*x)/(-a^2*x^2+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*(arctanh(a*x)-1)*(-(a*x-1)*(a*x+1))^(1/2)/a^3/(a*x-1)-1/2*(arctanh(a*x)+1)*(-(a*x-1)*(a*x+1))^(1/2)/a^3/(
a*x+1)+I*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^3-I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)
/a^3+I*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3-I*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3

________________________________________________________________________________________

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(x^2*arctanh(a*x)/(-a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^2*arctanh(a*x)/(-a^2*x^2 + 1)^(3/2), 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(x^2*arctanh(a*x)/(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*x^2 + 1)*x^2*arctanh(a*x)/(a^4*x^4 - 2*a^2*x^2 + 1), x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \operatorname {atanh}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atanh(a*x)/(-a**2*x**2+1)**(3/2),x)

[Out]

Integral(x**2*atanh(a*x)/(-(a*x - 1)*(a*x + 1))**(3/2), x)

________________________________________________________________________________________

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(x^2*arctanh(a*x)/(-a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

integrate(x^2*arctanh(a*x)/(-a^2*x^2 + 1)^(3/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\mathrm {atanh}\left (a\,x\right )}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*atanh(a*x))/(1 - a^2*x^2)^(3/2),x)

[Out]

int((x^2*atanh(a*x))/(1 - a^2*x^2)^(3/2), x)

________________________________________________________________________________________