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

Optimal. Leaf size=186 \[ -\frac{1}{42} a x^5 \sqrt{1-a^2 x^2}+\frac{23 x^3 \sqrt{1-a^2 x^2}}{840 a}+\frac{3 x \sqrt{1-a^2 x^2}}{112 a^3}-\frac{1}{7} a^2 x^6 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\frac{8}{35} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{35 a^2}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{35 a^4}+\frac{17 \sin ^{-1}(a x)}{560 a^4} \]

[Out]

(3*x*Sqrt[1 - a^2*x^2])/(112*a^3) + (23*x^3*Sqrt[1 - a^2*x^2])/(840*a) - (a*x^5*Sqrt[1 - a^2*x^2])/42 + (17*Ar
cSin[a*x])/(560*a^4) - (2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(35*a^4) - (x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(35*
a^2) + (8*x^4*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/35 - (a^2*x^6*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/7

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Rubi [A]  time = 0.570737, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6014, 6010, 6016, 321, 216, 5994} \[ -\frac{1}{42} a x^5 \sqrt{1-a^2 x^2}+\frac{23 x^3 \sqrt{1-a^2 x^2}}{840 a}+\frac{3 x \sqrt{1-a^2 x^2}}{112 a^3}-\frac{1}{7} a^2 x^6 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\frac{8}{35} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{35 a^2}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{35 a^4}+\frac{17 \sin ^{-1}(a x)}{560 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x*Sqrt[1 - a^2*x^2])/(112*a^3) + (23*x^3*Sqrt[1 - a^2*x^2])/(840*a) - (a*x^5*Sqrt[1 - a^2*x^2])/42 + (17*Ar
cSin[a*x])/(560*a^4) - (2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(35*a^4) - (x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(35*
a^2) + (8*x^4*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/35 - (a^2*x^6*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/7

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 6010

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

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^3 \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x) \, dx &=-\left (a^2 \int x^5 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x) \, dx\right )+\int x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x) \, dx\\ &=\frac{1}{5} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{1}{7} a^2 x^6 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\frac{1}{5} \int \frac{x^3 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx-\frac{1}{5} a \int \frac{x^4}{\sqrt{1-a^2 x^2}} \, dx-\frac{1}{7} a^2 \int \frac{x^5 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx+\frac{1}{7} a^3 \int \frac{x^6}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{x^3 \sqrt{1-a^2 x^2}}{20 a}-\frac{1}{42} a x^5 \sqrt{1-a^2 x^2}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^2}+\frac{8}{35} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{1}{7} a^2 x^6 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{4}{35} \int \frac{x^3 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx+\frac{2 \int \frac{x \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{15 a^2}+\frac{\int \frac{x^2}{\sqrt{1-a^2 x^2}} \, dx}{15 a}-\frac{3 \int \frac{x^2}{\sqrt{1-a^2 x^2}} \, dx}{20 a}-\frac{1}{35} a \int \frac{x^4}{\sqrt{1-a^2 x^2}} \, dx+\frac{1}{42} (5 a) \int \frac{x^4}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{x \sqrt{1-a^2 x^2}}{24 a^3}+\frac{23 x^3 \sqrt{1-a^2 x^2}}{840 a}-\frac{1}{42} a x^5 \sqrt{1-a^2 x^2}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{35 a^2}+\frac{8}{35} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{1}{7} a^2 x^6 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\frac{\int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{30 a^3}-\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{40 a^3}+\frac{2 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{15 a^3}-\frac{8 \int \frac{x \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{105 a^2}-\frac{3 \int \frac{x^2}{\sqrt{1-a^2 x^2}} \, dx}{140 a}-\frac{4 \int \frac{x^2}{\sqrt{1-a^2 x^2}} \, dx}{105 a}+\frac{5 \int \frac{x^2}{\sqrt{1-a^2 x^2}} \, dx}{56 a}\\ &=\frac{3 x \sqrt{1-a^2 x^2}}{112 a^3}+\frac{23 x^3 \sqrt{1-a^2 x^2}}{840 a}-\frac{1}{42} a x^5 \sqrt{1-a^2 x^2}+\frac{11 \sin ^{-1}(a x)}{120 a^4}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{35 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{35 a^2}+\frac{8}{35} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{1}{7} a^2 x^6 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{280 a^3}-\frac{2 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{105 a^3}+\frac{5 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{112 a^3}-\frac{8 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{105 a^3}\\ &=\frac{3 x \sqrt{1-a^2 x^2}}{112 a^3}+\frac{23 x^3 \sqrt{1-a^2 x^2}}{840 a}-\frac{1}{42} a x^5 \sqrt{1-a^2 x^2}+\frac{17 \sin ^{-1}(a x)}{560 a^4}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{35 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{35 a^2}+\frac{8}{35} x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{1}{7} a^2 x^6 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)\\ \end{align*}

Mathematica [A]  time = 0.084368, size = 79, normalized size = 0.42 \[ \frac{a x \left (-40 a^4 x^4+46 a^2 x^2+45\right ) \sqrt{1-a^2 x^2}-48 \left (5 a^2 x^2+2\right ) \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+51 \sin ^{-1}(a x)}{1680 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x*Sqrt[1 - a^2*x^2]*(45 + 46*a^2*x^2 - 40*a^4*x^4) + 51*ArcSin[a*x] - 48*(1 - a^2*x^2)^(5/2)*(2 + 5*a^2*x^2
)*ArcTanh[a*x])/(1680*a^4)

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Maple [C]  time = 0.18, size = 140, normalized size = 0.8 \begin{align*} -{\frac{240\,{\it Artanh} \left ( ax \right ){x}^{6}{a}^{6}+40\,{x}^{5}{a}^{5}-384\,{a}^{4}{x}^{4}{\it Artanh} \left ( ax \right ) -46\,{x}^{3}{a}^{3}+48\,{a}^{2}{x}^{2}{\it Artanh} \left ( ax \right ) -45\,ax+96\,{\it Artanh} \left ( ax \right ) }{1680\,{a}^{4}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{{\frac{17\,i}{560}}}{{a}^{4}}\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+i \right ) }-{\frac{{\frac{17\,i}{560}}}{{a}^{4}}\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1680/a^4*(-(a*x-1)*(a*x+1))^(1/2)*(240*arctanh(a*x)*x^6*a^6+40*x^5*a^5-384*a^4*x^4*arctanh(a*x)-46*x^3*a^3+
48*a^2*x^2*arctanh(a*x)-45*a*x+96*arctanh(a*x))+17/560*I*ln((a*x+1)/(-a^2*x^2+1)^(1/2)+I)/a^4-17/560*I*ln((a*x
+1)/(-a^2*x^2+1)^(1/2)-I)/a^4

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Maxima [A]  time = 1.46478, size = 248, normalized size = 1.33 \begin{align*} -\frac{1}{1680} \, a{\left (\frac{5 \,{\left (\frac{8 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}} x}{a^{2}} - \frac{2 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{a^{2}} - \frac{3 \, \sqrt{-a^{2} x^{2} + 1} x}{a^{2}} - \frac{3 \, \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}} a^{2}}\right )}}{a^{2}} - \frac{12 \,{\left (2 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x + 3 \, \sqrt{-a^{2} x^{2} + 1} x + \frac{3 \, \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}}\right )}}{a^{4}}\right )} - \frac{1}{35} \,{\left (\frac{5 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}} x^{2}}{a^{2}} + \frac{2 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}}}{a^{4}}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1680*a*(5*(8*(-a^2*x^2 + 1)^(5/2)*x/a^2 - 2*(-a^2*x^2 + 1)^(3/2)*x/a^2 - 3*sqrt(-a^2*x^2 + 1)*x/a^2 - 3*arc
sin(a^2*x/sqrt(a^2))/(sqrt(a^2)*a^2))/a^2 - 12*(2*(-a^2*x^2 + 1)^(3/2)*x + 3*sqrt(-a^2*x^2 + 1)*x + 3*arcsin(a
^2*x/sqrt(a^2))/sqrt(a^2))/a^4) - 1/35*(5*(-a^2*x^2 + 1)^(5/2)*x^2/a^2 + 2*(-a^2*x^2 + 1)^(5/2)/a^4)*arctanh(a
*x)

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Fricas [A]  time = 1.42565, size = 247, normalized size = 1.33 \begin{align*} -\frac{{\left (40 \, a^{5} x^{5} - 46 \, a^{3} x^{3} - 45 \, a x + 24 \,{\left (5 \, a^{6} x^{6} - 8 \, a^{4} x^{4} + a^{2} x^{2} + 2\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )\right )} \sqrt{-a^{2} x^{2} + 1} + 102 \, \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right )}{1680 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1680*((40*a^5*x^5 - 46*a^3*x^3 - 45*a*x + 24*(5*a^6*x^6 - 8*a^4*x^4 + a^2*x^2 + 2)*log(-(a*x + 1)/(a*x - 1)
))*sqrt(-a^2*x^2 + 1) + 102*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)))/a^4

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

[Out]

Timed out

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Giac [A]  time = 1.27316, size = 255, normalized size = 1.37 \begin{align*} -\frac{{\left (\frac{15 \,{\left (a^{2} x^{2} - 1\right )}^{3} \sqrt{-a^{2} x^{2} + 1} + 42 \,{\left (a^{2} x^{2} - 1\right )}^{2} \sqrt{-a^{2} x^{2} + 1} - 35 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{2}} - \frac{7 \,{\left (3 \,{\left (a^{2} x^{2} - 1\right )}^{2} \sqrt{-a^{2} x^{2} + 1} - 5 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}\right )}}{a^{2}}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{210 \, a^{2}} - \frac{\sqrt{-a^{2} x^{2} + 1}{\left (2 \,{\left (20 \, a^{4} x^{2} - 23 \, a^{2}\right )} x^{2} - 45\right )} x - \frac{51 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}}}{1680 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/210*((15*(a^2*x^2 - 1)^3*sqrt(-a^2*x^2 + 1) + 42*(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1) - 35*(-a^2*x^2 + 1)^(3/
2))/a^2 - 7*(3*(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1) - 5*(-a^2*x^2 + 1)^(3/2))/a^2)*log(-(a*x + 1)/(a*x - 1))/a^2
- 1/1680*(sqrt(-a^2*x^2 + 1)*(2*(20*a^4*x^2 - 23*a^2)*x^2 - 45)*x - 51*arcsin(a*x)*sgn(a)/abs(a))/a^3