3.5.41 \(\int x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2 \, dx\) [441]

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

[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 = 0.09, 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 = {6141, 6089, 6097} \begin {gather*} -\frac {2 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{3 a^2}-\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} \end {gather*}

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 6089

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 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 6141

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 = 0.28, size = 135, normalized size = 0.77 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (1+a x \tanh ^{-1}(a x)-\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {i \left (\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 )+\text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-\text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}\right )}{3 a^2} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [A]
time = 0.70, size = 175, normalized size = 1.00

method	result	size
default	\(\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (a^{2} x^{2} \arctanh \left (a x \right )^{2}+a x \arctanh \left (a x \right )-\arctanh \left (a x \right )^{2}+1\right )}{3 a^{2}}-\frac {i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{3 a^{2}}+\frac {i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{3 a^{2}}-\frac {i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{3 a^{2}}+\frac {i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{3 a^{2}}\)	\(175\)

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,method=_RETURNVERBOSE)

[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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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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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(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,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

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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