3.451 \(\int (1-a^2 x^2)^{3/2} \tanh ^{-1}(a x) \, dx\)

Optimal. Leaf size=189 \[ -\frac{3 i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a}+\frac{3 i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a}+\frac{\left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac{3 \sqrt{1-a^2 x^2}}{8 a}+\frac{1}{4} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac{3}{8} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{4 a} \]

[Out]

(3*Sqrt[1 - a^2*x^2])/(8*a) + (1 - a^2*x^2)^(3/2)/(12*a) + (3*x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/8 + (x*(1 - a^
2*x^2)^(3/2)*ArcTanh[a*x])/4 - (3*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(4*a) - (((3*I)/8)*PolyLog
[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a + (((3*I)/8)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a

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Rubi [A]  time = 0.0859679, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {5942, 5950} \[ -\frac{3 i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a}+\frac{3 i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a}+\frac{\left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac{3 \sqrt{1-a^2 x^2}}{8 a}+\frac{1}{4} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac{3}{8} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[1 - a^2*x^2])/(8*a) + (1 - a^2*x^2)^(3/2)/(12*a) + (3*x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/8 + (x*(1 - a^
2*x^2)^(3/2)*ArcTanh[a*x])/4 - (3*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(4*a) - (((3*I)/8)*PolyLog
[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a + (((3*I)/8)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a

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]

Rubi steps

\begin{align*} \int \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x) \, dx &=\frac{\left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac{1}{4} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac{3}{4} \int \sqrt{1-a^2 x^2} \tanh ^{-1}(a x) \, dx\\ &=\frac{3 \sqrt{1-a^2 x^2}}{8 a}+\frac{\left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac{3}{8} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\frac{1}{4} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac{3}{8} \int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{3 \sqrt{1-a^2 x^2}}{8 a}+\frac{\left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac{3}{8} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\frac{1}{4} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)-\frac{3 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)}{4 a}-\frac{3 i \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{8 a}+\frac{3 i \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{8 a}\\ \end{align*}

Mathematica [A]  time = 0.584002, size = 176, normalized size = 0.93 \[ \frac{-9 i \text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )+9 i \text{PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )-2 a^2 x^2 \sqrt{1-a^2 x^2}+11 \sqrt{1-a^2 x^2}-6 a^3 x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+15 a x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-9 i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+9 i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )}{24 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(11*Sqrt[1 - a^2*x^2] - 2*a^2*x^2*Sqrt[1 - a^2*x^2] + 15*a*x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - 6*a^3*x^3*Sqrt[1
- a^2*x^2]*ArcTanh[a*x] - (9*I)*ArcTanh[a*x]*Log[1 - I/E^ArcTanh[a*x]] + (9*I)*ArcTanh[a*x]*Log[1 + I/E^ArcTa
nh[a*x]] - (9*I)*PolyLog[2, (-I)/E^ArcTanh[a*x]] + (9*I)*PolyLog[2, I/E^ArcTanh[a*x]])/(24*a)

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Maple [A]  time = 0.253, size = 173, normalized size = 0.9 \begin{align*} -{\frac{6\,{a}^{3}{x}^{3}{\it Artanh} \left ( ax \right ) +2\,{a}^{2}{x}^{2}-15\,ax{\it Artanh} \left ( ax \right ) -11}{24\,a}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{\frac{3\,i}{8}}{\it Artanh} \left ( ax \right ) }{a}\ln \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{3\,i}{8}}{\it Artanh} \left ( ax \right ) }{a}\ln \left ( 1-{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{3\,i}{8}}}{a}{\it dilog} \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{3\,i}{8}}}{a}{\it dilog} \left ( 1-{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(6*a^3*x^3*arctanh(a*x)+2*a^2*x^2-15*a*x*arctanh(a*x)-11)*(-a^2*x^2+1)^(1/2)/a-3/8*I/a*arctanh(a*x)*ln(1
+I*(a*x+1)/(-a^2*x^2+1)^(1/2))+3/8*I/a*arctanh(a*x)*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))-3/8*I/a*dilog(1+I*(a*x+
1)/(-a^2*x^2+1)^(1/2))+3/8*I/a*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \operatorname{artanh}\left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (a^{2} x^{2} - 1\right )} \sqrt{-a^{2} x^{2} + 1} \operatorname{artanh}\left (a x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \operatorname{artanh}\left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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