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

Optimal. Leaf size=120 \[ -\frac {2 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a^2}+\frac {2 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{a^2}-\frac {4 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a^2} \]

[Out]

-4*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)/a^2-2*I*polylog(2,-I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^2+2*
I*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^2-arctanh(a*x)^2*(-a^2*x^2+1)^(1/2)/a^2

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Rubi [A]  time = 0.10, antiderivative size = 120, 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 = {5994, 5950} \[ -\frac {2 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a^2}+\frac {2 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{a^2}-\frac {4 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/a^2 - (Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/a^2 - ((2*I)*Po
lyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^2 + ((2*I)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^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 5994

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

Rubi steps

\begin {align*} \int \frac {x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx &=-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{a^2}+\frac {2 \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{a}\\ &=-\frac {4 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{a^2}-\frac {2 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^2}+\frac {2 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^2}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 104, normalized size = 0.87 \[ -\frac {\tanh ^{-1}(a x) \left (\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+2 i \left (\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )\right )+2 i \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-2 i \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )}{a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} x \operatorname {artanh}\left (a x\right )^{2}}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.33, size = 151, normalized size = 1.26 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \arctanh \left (a x \right )^{2}}{a^{2}}-\frac {2 i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{a^{2}}+\frac {2 i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{a^{2}}-\frac {2 i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2}}+\frac {2 i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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