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

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

[Out]

17/560*arcsin(a*x)/a^4+3/112*x*(-a^2*x^2+1)^(1/2)/a^3+23/840*x^3*(-a^2*x^2+1)^(1/2)/a-1/42*a*x^5*(-a^2*x^2+1)^
(1/2)-2/35*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^4-1/35*x^2*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^2+8/35*x^4*arctanh(a
*x)*(-a^2*x^2+1)^(1/2)-1/7*a^2*x^6*arctanh(a*x)*(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.57, antiderivative size = 186, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.09, size = 79, normalized size = 0.42 \[ \frac {-48 \left (5 a^2 x^2+2\right ) \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+a x \left (-40 a^4 x^4+46 a^2 x^2+45\right ) \sqrt {1-a^2 x^2}+51 \sin ^{-1}(a x)}{1680 a^4} \]

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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fricas [A]  time = 0.48, size = 106, normalized size = 0.57 \[ -\frac {{\left (40 \, a^{5} x^{5} - 46 \, a^{3} x^{3} - 45 \, a x + 24 \, {\left (5 \, a^{6} x^{6} - 8 \, a^{4} x^{4} + a^{2} x^{2} + 2\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )\right )} \sqrt {-a^{2} x^{2} + 1} + 102 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{1680 \, a^{4}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.38, size = 140, normalized size = 0.75 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (240 \arctanh \left (a x \right ) x^{6} a^{6}+40 x^{5} a^{5}-384 a^{4} x^{4} \arctanh \left (a x \right )-46 x^{3} a^{3}+48 a^{2} x^{2} \arctanh \left (a x \right )-45 a x +96 \arctanh \left (a x \right )\right )}{1680 a^{4}}+\frac {17 i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}+i\right )}{560 a^{4}}-\frac {17 i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-i\right )}{560 a^{4}} \]

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 = 0.41, size = 163, normalized size = 0.88 \[ -\frac {1}{1680} \, a {\left (\frac {5 \, {\left (\frac {8 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} x}{a^{2}} - \frac {2 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{a^{2}} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x}{a^{2}} - \frac {3 \, \arcsin \left (a x\right )}{a^{3}}\right )}}{a^{2}} - \frac {12 \, {\left (2 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x + 3 \, \sqrt {-a^{2} x^{2} + 1} x + \frac {3 \, \arcsin \left (a x\right )}{a}\right )}}{a^{4}}\right )} - \frac {1}{35} \, {\left (\frac {5 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} x^{2}}{a^{2}} + \frac {2 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{a^{4}}\right )} \operatorname {artanh}\left (a x\right ) \]

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*x)/a^3)/a^2 - 12*(2*(-a^2*x^2 + 1)^(3/2)*x + 3*sqrt(-a^2*x^2 + 1)*x + 3*arcsin(a*x)/a)/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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \operatorname {atanh}{\left (a x \right )}\, dx \]

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]

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

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