3.459 \(\int (1-a^2 x^2)^{5/2} \tanh ^{-1}(a x) \, dx\)

Optimal. Leaf size=233 \[ -\frac{5 i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{16 a}+\frac{5 i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{16 a}+\frac{\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac{5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac{5 \sqrt{1-a^2 x^2}}{16 a}+\frac{1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+\frac{5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac{5}{16} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{5 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{8 a} \]

[Out]

(5*Sqrt[1 - a^2*x^2])/(16*a) + (5*(1 - a^2*x^2)^(3/2))/(72*a) + (1 - a^2*x^2)^(5/2)/(30*a) + (5*x*Sqrt[1 - a^2
*x^2]*ArcTanh[a*x])/16 + (5*x*(1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/24 + (x*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x])/6 -
(5*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(8*a) - (((5*I)/16)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[
1 + a*x]])/a + (((5*I)/16)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a

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Rubi [A]  time = 0.12105, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {5942, 5950} \[ -\frac{5 i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{16 a}+\frac{5 i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{16 a}+\frac{\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac{5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac{5 \sqrt{1-a^2 x^2}}{16 a}+\frac{1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+\frac{5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac{5}{16} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{5 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)^(5/2)*ArcTanh[a*x],x]

[Out]

(5*Sqrt[1 - a^2*x^2])/(16*a) + (5*(1 - a^2*x^2)^(3/2))/(72*a) + (1 - a^2*x^2)^(5/2)/(30*a) + (5*x*Sqrt[1 - a^2
*x^2]*ArcTanh[a*x])/16 + (5*x*(1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/24 + (x*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x])/6 -
(5*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(8*a) - (((5*I)/16)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[
1 + a*x]])/a + (((5*I)/16)*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 )^{5/2} \tanh ^{-1}(a x) \, dx &=\frac{\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac{1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+\frac{5}{6} \int \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x) \, dx\\ &=\frac{5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac{\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac{5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac{1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+\frac{5}{8} \int \sqrt{1-a^2 x^2} \tanh ^{-1}(a x) \, dx\\ &=\frac{5 \sqrt{1-a^2 x^2}}{16 a}+\frac{5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac{\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac{5}{16} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\frac{5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac{1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+\frac{5}{16} \int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{5 \sqrt{1-a^2 x^2}}{16 a}+\frac{5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac{\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac{5}{16} x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\frac{5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac{1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)-\frac{5 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)}{8 a}-\frac{5 i \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{16 a}+\frac{5 i \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{16 a}\\ \end{align*}

Mathematica [A]  time = 1.09752, size = 224, normalized size = 0.96 \[ \frac{-225 i \text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )+225 i \text{PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )+24 a^4 x^4 \sqrt{1-a^2 x^2}-98 a^2 x^2 \sqrt{1-a^2 x^2}+299 \sqrt{1-a^2 x^2}+120 a^5 x^5 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-390 a^3 x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+495 a x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-225 i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+225 i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )}{720 a} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 - a^2*x^2)^(5/2)*ArcTanh[a*x],x]

[Out]

(299*Sqrt[1 - a^2*x^2] - 98*a^2*x^2*Sqrt[1 - a^2*x^2] + 24*a^4*x^4*Sqrt[1 - a^2*x^2] + 495*a*x*Sqrt[1 - a^2*x^
2]*ArcTanh[a*x] - 390*a^3*x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] + 120*a^5*x^5*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - (2
25*I)*ArcTanh[a*x]*Log[1 - I/E^ArcTanh[a*x]] + (225*I)*ArcTanh[a*x]*Log[1 + I/E^ArcTanh[a*x]] - (225*I)*PolyLo
g[2, (-I)/E^ArcTanh[a*x]] + (225*I)*PolyLog[2, I/E^ArcTanh[a*x]])/(720*a)

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Maple [A]  time = 0.281, size = 193, normalized size = 0.8 \begin{align*}{\frac{120\,{\it Artanh} \left ( ax \right ){x}^{5}{a}^{5}+24\,{x}^{4}{a}^{4}-390\,{a}^{3}{x}^{3}{\it Artanh} \left ( ax \right ) -98\,{a}^{2}{x}^{2}+495\,ax{\it Artanh} \left ( ax \right ) +299}{720\,a}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{\frac{5\,i}{16}}{\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{5\,i}{16}}{\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{5\,i}{16}}}{a}{\it dilog} \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{5\,i}{16}}}{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)^(5/2)*arctanh(a*x),x)

[Out]

1/720*(120*arctanh(a*x)*x^5*a^5+24*x^4*a^4-390*a^3*x^3*arctanh(a*x)-98*a^2*x^2+495*a*x*arctanh(a*x)+299)*(-a^2
*x^2+1)^(1/2)/a-5/16*I/a*arctanh(a*x)*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))+5/16*I/a*arctanh(a*x)*ln(1-I*(a*x+1)/
(-a^2*x^2+1)^(1/2))-5/16*I/a*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))+5/16*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{5}{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)^(5/2)*arctanh(a*x),x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(5/2)*arctanh(a*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{4} x^{4} - 2 \, 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)^(5/2)*arctanh(a*x),x, algorithm="fricas")

[Out]

integral((a^4*x^4 - 2*a^2*x^2 + 1)*sqrt(-a^2*x^2 + 1)*arctanh(a*x), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)**(5/2)*atanh(a*x),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{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)^(5/2)*arctanh(a*x),x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(5/2)*arctanh(a*x), x)