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

Optimal. Leaf size=87 \[ \frac {5 \sin ^{-1}(a x)}{6 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^2}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^4}-\frac {x \sqrt {1-a^2 x^2}}{6 a^3} \]

[Out]

5/6*arcsin(a*x)/a^4-1/6*x*(-a^2*x^2+1)^(1/2)/a^3-2/3*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^4-1/3*x^2*arctanh(a*x)*
(-a^2*x^2+1)^(1/2)/a^2

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Rubi [A]  time = 0.12, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6016, 321, 216, 5994} \[ -\frac {x \sqrt {1-a^2 x^2}}{6 a^3}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^2}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^4}+\frac {5 \sin ^{-1}(a x)}{6 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*Sqrt[1 - a^2*x^2])/(6*a^3) + (5*ArcSin[a*x])/(6*a^4) - (2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(3*a^4) - (x^2*S
qrt[1 - a^2*x^2]*ArcTanh[a*x])/(3*a^2)

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 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 \frac {x^3 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx &=-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^2}+\frac {2 \int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{3 a^2}+\frac {\int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx}{3 a}\\ &=-\frac {x \sqrt {1-a^2 x^2}}{6 a^3}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^2}+\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{6 a^3}+\frac {2 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{3 a^3}\\ &=-\frac {x \sqrt {1-a^2 x^2}}{6 a^3}+\frac {5 \sin ^{-1}(a x)}{6 a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 60, normalized size = 0.69 \[ -\frac {a x \sqrt {1-a^2 x^2}+2 \sqrt {1-a^2 x^2} \left (a^2 x^2+2\right ) \tanh ^{-1}(a x)-5 \sin ^{-1}(a x)}{6 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(a*x*Sqrt[1 - a^2*x^2] - 5*ArcSin[a*x] + 2*Sqrt[1 - a^2*x^2]*(2 + a^2*x^2)*ArcTanh[a*x])/a^4

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fricas [A]  time = 0.49, size = 72, normalized size = 0.83 \[ -\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x + {\left (a^{2} x^{2} + 2\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )\right )} + 10 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{6 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(sqrt(-a^2*x^2 + 1)*(a*x + (a^2*x^2 + 2)*log(-(a*x + 1)/(a*x - 1))) + 10*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*arctanh(a*x)/(-a^2*x^2+1)^(1/2),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 = 99, normalized size = 1.14 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (2 a^{2} x^{2} \arctanh \left (a x \right )+a x +4 \arctanh \left (a x \right )\right )}{6 a^{4}}+\frac {5 i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}+i\right )}{6 a^{4}}-\frac {5 i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-i\right )}{6 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/a^4*(-(a*x-1)*(a*x+1))^(1/2)*(2*a^2*x^2*arctanh(a*x)+a*x+4*arctanh(a*x))+5/6*I*ln((a*x+1)/(-a^2*x^2+1)^(1
/2)+I)/a^4-5/6*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 = 88, normalized size = 1.01 \[ -\frac {1}{6} \, a {\left (\frac {\frac {\sqrt {-a^{2} x^{2} + 1} x}{a^{2}} - \frac {\arcsin \left (a x\right )}{a^{3}}}{a^{2}} - \frac {4 \, \arcsin \left (a x\right )}{a^{5}}\right )} - \frac {1}{3} \, {\left (\frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{a^{2}} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{a^{4}}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a*((sqrt(-a^2*x^2 + 1)*x/a^2 - arcsin(a*x)/a^3)/a^2 - 4*arcsin(a*x)/a^5) - 1/3*(sqrt(-a^2*x^2 + 1)*x^2/a^
2 + 2*sqrt(-a^2*x^2 + 1)/a^4)*arctanh(a*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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