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

Optimal. Leaf size=175 \[ -\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{3 a^2}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{3 a^2}+\frac {\sqrt {1-a^2 x^2}}{3 a^2}-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)^2}{3 a^2}+\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{3 a^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 175, 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 = {5994, 5942, 5950} \[ -\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{3 a^2}+\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{3 a^2}+\frac {\sqrt {1-a^2 x^2}}{3 a^2}-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)^2}{3 a^2}+\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{3 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5942

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

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

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Mathematica [A]  time = 0.43, size = 135, normalized size = 0.77 \[ \frac {\sqrt {1-a^2 x^2} \left (-\frac {i \left (\text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-\text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )+\tanh ^{-1}(a x) \left (\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )\right )}{\sqrt {1-a^2 x^2}}-\left (\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2\right )+a x \tanh ^{-1}(a x)+1\right )}{3 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-a^{2} x^{2} + 1} x \operatorname {artanh}\left (a x\right )^{2}, x\right ) \]

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, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.35, size = 175, normalized size = 1.00 \[ \frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (a^{2} x^{2} \arctanh \left (a x \right )^{2}+a x \arctanh \left (a x \right )-\arctanh \left (a x \right )^{2}+1\right )}{3 a^{2}}-\frac {i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{3 a^{2}}+\frac {i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{3 a^{2}}-\frac {i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{3 a^{2}}+\frac {i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{3 a^{2}} \]

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]

1/3/a^2*(-(a*x-1)*(a*x+1))^(1/2)*(a^2*x^2*arctanh(a*x)^2+a*x*arctanh(a*x)-arctanh(a*x)^2+1)-1/3*I*ln(1+I*(a*x+
1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^2+1/3*I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^2-1/3*I*dilog(
1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^2+1/3*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 \[ \int \sqrt {-a^{2} x^{2} + 1} x \operatorname {artanh}\left (a x\right )^{2}\,{d x} \]

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

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

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 \[ \int x \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}^{2}{\left (a x \right )}\, dx \]

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

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