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

Optimal. Leaf size=74 \[ -\frac {\sin ^{-1}(a x)}{a^4}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}+\frac {\tanh ^{-1}(a x)}{a^4 \sqrt {1-a^2 x^2}}-\frac {x}{a^3 \sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 74, 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 = {6028, 5994, 216, 191} \[ -\frac {x}{a^3 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}+\frac {\tanh ^{-1}(a x)}{a^4 \sqrt {1-a^2 x^2}}-\frac {\sin ^{-1}(a x)}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/(a^3*Sqrt[1 - a^2*x^2])) - ArcSin[a*x]/a^4 + ArcTanh[a*x]/(a^4*Sqrt[1 - a^2*x^2]) + (Sqrt[1 - a^2*x^2]*Arc
Tanh[a*x])/a^4

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

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]

Rule 6028

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 76, normalized size = 1.03 \[ \frac {a x \sqrt {1-a^2 x^2}+\left (1-a^2 x^2\right ) \sin ^{-1}(a x)+\sqrt {1-a^2 x^2} \left (a^2 x^2-2\right ) \tanh ^{-1}(a x)}{a^4 \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 94, normalized size = 1.27 \[ \frac {4 \, {\left (a^{2} x^{2} - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (2 \, a x + {\left (a^{2} x^{2} - 2\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )\right )}}{2 \, {\left (a^{6} x^{2} - a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*(a^2*x^2 - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + sqrt(-a^2*x^2 + 1)*(2*a*x + (a^2*x^2 - 2)*log(-(
a*x + 1)/(a*x - 1))))/(a^6*x^2 - 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)^(3/2),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 = 0.42, size = 144, normalized size = 1.95 \[ -\frac {\left (\arctanh \left (a x \right )-1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{4} \left (a x -1\right )}+\frac {\left (\arctanh \left (a x \right )+1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{4} \left (a x +1\right )}+\frac {\arctanh \left (a x \right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{a^{4}}+\frac {i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-i\right )}{a^{4}}-\frac {i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}+i\right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(arctanh(a*x)-1)*(-(a*x-1)*(a*x+1))^(1/2)/a^4/(a*x-1)+1/2*(arctanh(a*x)+1)*(-(a*x-1)*(a*x+1))^(1/2)/a^4/(
a*x+1)+arctanh(a*x)*(-(a*x-1)*(a*x+1))^(1/2)/a^4+I*ln((a*x+1)/(-a^2*x^2+1)^(1/2)-I)/a^4-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 = 96, normalized size = 1.30 \[ a {\left (\frac {\frac {x}{\sqrt {-a^{2} x^{2} + 1} a^{2}} - \frac {\arcsin \left (a x\right )}{a^{3}}}{a^{2}} - \frac {2 \, x}{\sqrt {-a^{2} x^{2} + 1} a^{4}}\right )} - {\left (\frac {x^{2}}{\sqrt {-a^{2} x^{2} + 1} a^{2}} - \frac {2}{\sqrt {-a^{2} x^{2} + 1} a^{4}}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*((x/(sqrt(-a^2*x^2 + 1)*a^2) - arcsin(a*x)/a^3)/a^2 - 2*x/(sqrt(-a^2*x^2 + 1)*a^4)) - (x^2/(sqrt(-a^2*x^2 +
1)*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 \[ \int \frac {x^3\,\mathrm {atanh}\left (a\,x\right )}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]

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 \[ \int \frac {x^{3} \operatorname {atanh}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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