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

Optimal. Leaf size=281 \[ -\frac{11 i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{60 a^4}+\frac{11 i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{60 a^4}-\frac{\left (1-a^2 x^2\right )^{3/2}}{30 a^4}+\frac{11 \sqrt{1-a^2 x^2}}{60 a^4}+\frac{1}{5} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^2}+\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^4}-\frac{11 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{30 a^4} \]

[Out]

(11*Sqrt[1 - a^2*x^2])/(60*a^4) - (1 - a^2*x^2)^(3/2)/(30*a^4) + (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(12*a^3) +
(x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(10*a) - (11*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(30*a^4)
- (2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(15*a^4) - (x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(15*a^2) + (x^4*Sqrt[
1 - a^2*x^2]*ArcTanh[a*x]^2)/5 - (((11*I)/60)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^4 + (((11*I)/6
0)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^4

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Rubi [A]  time = 1.07307, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {6014, 6016, 261, 5950, 5994, 266, 43} \[ -\frac{11 i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{60 a^4}+\frac{11 i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{60 a^4}-\frac{\left (1-a^2 x^2\right )^{3/2}}{30 a^4}+\frac{11 \sqrt{1-a^2 x^2}}{60 a^4}+\frac{1}{5} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^2}+\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^4}-\frac{11 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{30 a^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2,x]

[Out]

(11*Sqrt[1 - a^2*x^2])/(60*a^4) - (1 - a^2*x^2)^(3/2)/(30*a^4) + (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(12*a^3) +
(x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(10*a) - (11*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(30*a^4)
- (2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(15*a^4) - (x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(15*a^2) + (x^4*Sqrt[
1 - a^2*x^2]*ArcTanh[a*x]^2)/5 - (((11*I)/60)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^4 + (((11*I)/6
0)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^4

Rule 6014

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Dist
[d, Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(c^2*d)/f^2, Int[(f*x)^(m + 2)*(d +
e*x^2)^(q - 1)*(a + b*ArcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[q
, 0] && IGtQ[p, 0] && (RationalQ[m] || (EqQ[p, 1] && IntegerQ[q]))

Rule 6016

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Sim
p[(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcTanh[c*x])^p)/(c^2*d*m), x] + (Dist[(b*f*p)/(c*m), Int[((f*x)^(m
- 1)*(a + b*ArcTanh[c*x])^(p - 1))/Sqrt[d + e*x^2], x], x] + Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m - 2)*(a
+ b*ArcTanh[c*x])^p)/Sqrt[d + e*x^2], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[p,
0] && GtQ[m, 1]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2 \, dx &=-\left (a^2 \int \frac{x^5 \tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx\right )+\int \frac{x^3 \tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{3 a^2}+\frac{1}{5} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac{4}{5} \int \frac{x^3 \tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx+\frac{2 \int \frac{x \tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx}{3 a^2}+\frac{2 \int \frac{x^2 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{3 a}-\frac{1}{5} (2 a) \int \frac{x^4 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^3}+\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{3 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^2}+\frac{1}{5} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac{1}{10} \int \frac{x^3}{\sqrt{1-a^2 x^2}} \, dx+\frac{\int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{3 a^3}+\frac{4 \int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{3 a^3}+\frac{\int \frac{x}{\sqrt{1-a^2 x^2}} \, dx}{3 a^2}-\frac{8 \int \frac{x \tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx}{15 a^2}-\frac{3 \int \frac{x^2 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{10 a}-\frac{8 \int \frac{x^2 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{15 a}\\ &=-\frac{\sqrt{1-a^2 x^2}}{3 a^4}+\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\frac{10 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)}{3 a^4}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^2}+\frac{1}{5} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac{5 i \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{3 a^4}+\frac{5 i \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{3 a^4}-\frac{1}{20} \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-a^2 x}} \, dx,x,x^2\right )-\frac{3 \int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{20 a^3}-\frac{4 \int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{15 a^3}-\frac{16 \int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{15 a^3}-\frac{3 \int \frac{x}{\sqrt{1-a^2 x^2}} \, dx}{20 a^2}-\frac{4 \int \frac{x}{\sqrt{1-a^2 x^2}} \, dx}{15 a^2}\\ &=\frac{\sqrt{1-a^2 x^2}}{12 a^4}+\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\frac{11 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)}{30 a^4}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^2}+\frac{1}{5} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac{11 i \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{60 a^4}+\frac{11 i \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{60 a^4}-\frac{1}{20} \operatorname{Subst}\left (\int \left (\frac{1}{a^2 \sqrt{1-a^2 x}}-\frac{\sqrt{1-a^2 x}}{a^2}\right ) \, dx,x,x^2\right )\\ &=\frac{11 \sqrt{1-a^2 x^2}}{60 a^4}-\frac{\left (1-a^2 x^2\right )^{3/2}}{30 a^4}+\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac{x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\frac{11 \tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)}{30 a^4}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^2}+\frac{1}{5} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac{11 i \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{60 a^4}+\frac{11 i \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{60 a^4}\\ \end{align*}

Mathematica [A]  time = 0.640527, size = 175, normalized size = 0.62 \[ \frac{\sqrt{1-a^2 x^2} \left (-\frac{11 i \left (\text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-\text{PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )+\tanh ^{-1}(a x) \left (\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )\right )}{\sqrt{1-a^2 x^2}}+12 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2+6 a x \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)+2 \left (a^2 x^2-1\right ) \left (10 \tanh ^{-1}(a x)^2+1\right )+11 a x \tanh ^{-1}(a x)+11\right )}{60 a^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2,x]

[Out]

(Sqrt[1 - a^2*x^2]*(11 + 11*a*x*ArcTanh[a*x] + 6*a*x*(-1 + a^2*x^2)*ArcTanh[a*x] + 12*(-1 + a^2*x^2)^2*ArcTanh
[a*x]^2 + 2*(-1 + a^2*x^2)*(1 + 10*ArcTanh[a*x]^2) - ((11*I)*(ArcTanh[a*x]*(Log[1 - I/E^ArcTanh[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]))/(6
0*a^4)

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Maple [A]  time = 0.273, size = 211, normalized size = 0.8 \begin{align*}{\frac{12\,{a}^{4}{x}^{4} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}+6\,{a}^{3}{x}^{3}{\it Artanh} \left ( ax \right ) -4\,{a}^{2}{x}^{2} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}+2\,{a}^{2}{x}^{2}+5\,ax{\it Artanh} \left ( ax \right ) -8\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}+9}{60\,{a}^{4}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}-{\frac{{\frac{11\,i}{60}}{\it Artanh} \left ( ax \right ) }{{a}^{4}}\ln \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{11\,i}{60}}{\it Artanh} \left ( ax \right ) }{{a}^{4}}\ln \left ( 1-{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{11\,i}{60}}}{{a}^{4}}{\it dilog} \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{11\,i}{60}}}{{a}^{4}}{\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(x^3*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2),x)

[Out]

1/60/a^4*(-(a*x-1)*(a*x+1))^(1/2)*(12*a^4*x^4*arctanh(a*x)^2+6*a^3*x^3*arctanh(a*x)-4*a^2*x^2*arctanh(a*x)^2+2
*a^2*x^2+5*a*x*arctanh(a*x)-8*arctanh(a*x)^2+9)-11/60*I*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^4+11
/60*I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^4-11/60*I*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^4+11
/60*I*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{-a^{2} x^{2} + 1} x^{3} \operatorname{artanh}\left (a x\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*x^3*arctanh(a*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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