3.434 \(\int \frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x^4} \, dx\)

Optimal. Leaf size=70 \[ -\frac {a \sqrt {1-a^2 x^2}}{6 x^2}-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}+\frac {1}{6} a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6008, 266, 47, 63, 208} \[ -\frac {a \sqrt {1-a^2 x^2}}{6 x^2}+\frac {1}{6} a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/x^4,x]

[Out]

-(a*Sqrt[1 - a^2*x^2])/(6*x^2) - ((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/(3*x^3) + (a^3*ArcTanh[Sqrt[1 - a^2*x^2]])
/6

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

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 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x^4} \, dx &=-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}+\frac {1}{3} a \int \frac {\sqrt {1-a^2 x^2}}{x^3} \, dx\\ &=-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}+\frac {1}{6} a \operatorname {Subst}\left (\int \frac {\sqrt {1-a^2 x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {a \sqrt {1-a^2 x^2}}{6 x^2}-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}-\frac {1}{12} a^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {a \sqrt {1-a^2 x^2}}{6 x^2}-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}+\frac {1}{6} a \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )\\ &=-\frac {a \sqrt {1-a^2 x^2}}{6 x^2}-\frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{3 x^3}+\frac {1}{6} a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 79, normalized size = 1.13 \[ -\frac {a^3 x^3 \log (x)+a x \sqrt {1-a^2 x^2}+2 \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)-a^3 x^3 \log \left (\sqrt {1-a^2 x^2}+1\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/x^4,x]

[Out]

-1/6*(a*x*Sqrt[1 - a^2*x^2] + 2*(1 - a^2*x^2)^(3/2)*ArcTanh[a*x] + a^3*x^3*Log[x] - a^3*x^3*Log[1 + Sqrt[1 - a
^2*x^2]])/x^3

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fricas [A]  time = 0.44, size = 75, normalized size = 1.07 \[ -\frac {a^{3} x^{3} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (a x - {\left (a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )\right )}}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)*(-a^2*x^2+1)^(1/2)/x^4,x, algorithm="fricas")

[Out]

-1/6*(a^3*x^3*log((sqrt(-a^2*x^2 + 1) - 1)/x) + sqrt(-a^2*x^2 + 1)*(a*x - (a^2*x^2 - 1)*log(-(a*x + 1)/(a*x -
1))))/x^3

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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(arctanh(a*x)*(-a^2*x^2+1)^(1/2)/x^4,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 [A]  time = 0.42, size = 96, normalized size = 1.37 \[ \frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (2 a^{2} x^{2} \arctanh \left (a x \right )-a x -2 \arctanh \left (a x \right )\right )}{6 x^{3}}-\frac {a^{3} \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-1\right )}{6}+\frac {a^{3} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)*(-a^2*x^2+1)^(1/2)/x^4,x)

[Out]

1/6*(-(a*x-1)*(a*x+1))^(1/2)*(2*a^2*x^2*arctanh(a*x)-a*x-2*arctanh(a*x))/x^3-1/6*a^3*ln((a*x+1)/(-a^2*x^2+1)^(
1/2)-1)+1/6*a^3*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))

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maxima [A]  time = 0.41, size = 90, normalized size = 1.29 \[ \frac {1}{6} \, {\left (a^{2} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \sqrt {-a^{2} x^{2} + 1} a^{2} - \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{x^{2}}\right )} a - \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \operatorname {artanh}\left (a x\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)*(-a^2*x^2+1)^(1/2)/x^4,x, algorithm="maxima")

[Out]

1/6*(a^2*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) - sqrt(-a^2*x^2 + 1)*a^2 - (-a^2*x^2 + 1)^(3/2)/x^2)*a -
1/3*(-a^2*x^2 + 1)^(3/2)*arctanh(a*x)/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (a\,x\right )\,\sqrt {1-a^2\,x^2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atanh(a*x)*(1 - a^2*x^2)^(1/2))/x^4,x)

[Out]

int((atanh(a*x)*(1 - a^2*x^2)^(1/2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}{\left (a x \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)*(-a**2*x**2+1)**(1/2)/x**4,x)

[Out]

Integral(sqrt(-(a*x - 1)*(a*x + 1))*atanh(a*x)/x**4, x)

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