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

Optimal. Leaf size=146 \[ -\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a^3}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a^3}-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{2 a^2}-\frac {\sqrt {1-a^2 x^2}}{2 a^3} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 146, 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 = {6016, 261, 5950} \[ -\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a^3}+\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a^3}-\frac {\sqrt {1-a^2 x^2}}{2 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{2 a^2}-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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 6016

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 125, normalized size = 0.86 \[ -\frac {\sqrt {1-a^2 x^2}+a x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+i \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-i \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )+i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )}{2 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} x^{2} \operatorname {artanh}\left (a x\right )}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.41, size = 154, normalized size = 1.05 \[ -\frac {\left (a x \arctanh \left (a x \right )+1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{3}}-\frac {i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{2 a^{3}}+\frac {i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{2 a^{3}}-\frac {i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{3}}+\frac {i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \operatorname {atanh}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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