3.5.57 \(\int \frac {(1-a^2 x^2)^{3/2} \tanh ^{-1}(a x)}{x^6} \, dx\) [457]

Optimal. Leaf size=94 \[ \frac {3 a^3 \sqrt {1-a^2 x^2}}{40 x^2}-\frac {a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac {\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}-\frac {3}{40} a^5 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

[Out]

-1/20*a*(-a^2*x^2+1)^(3/2)/x^4-1/5*(-a^2*x^2+1)^(5/2)*arctanh(a*x)/x^5-3/40*a^5*arctanh((-a^2*x^2+1)^(1/2))+3/
40*a^3*(-a^2*x^2+1)^(1/2)/x^2

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Rubi [A]
time = 0.07, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6155, 272, 43, 65, 214} \begin {gather*} -\frac {\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}-\frac {a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac {3}{40} a^5 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+\frac {3 a^3 \sqrt {1-a^2 x^2}}{40 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/x^6,x]

[Out]

(3*a^3*Sqrt[1 - a^2*x^2])/(40*x^2) - (a*(1 - a^2*x^2)^(3/2))/(20*x^4) - ((1 - a^2*x^2)^(5/2)*ArcTanh[a*x])/(5*
x^5) - (3*a^5*ArcTanh[Sqrt[1 - a^2*x^2]])/40

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

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 214

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

Rule 272

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 6155

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 {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{x^6} \, dx &=-\frac {\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}+\frac {1}{5} a \int \frac {\left (1-a^2 x^2\right )^{3/2}}{x^5} \, dx\\ &=-\frac {\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}+\frac {1}{10} a \text {Subst}\left (\int \frac {\left (1-a^2 x\right )^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac {\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}-\frac {1}{40} \left (3 a^3\right ) \text {Subst}\left (\int \frac {\sqrt {1-a^2 x}}{x^2} \, dx,x,x^2\right )\\ &=\frac {3 a^3 \sqrt {1-a^2 x^2}}{40 x^2}-\frac {a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac {\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}+\frac {1}{80} \left (3 a^5\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=\frac {3 a^3 \sqrt {1-a^2 x^2}}{40 x^2}-\frac {a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac {\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}-\frac {1}{40} \left (3 a^3\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )\\ &=\frac {3 a^3 \sqrt {1-a^2 x^2}}{40 x^2}-\frac {a \left (1-a^2 x^2\right )^{3/2}}{20 x^4}-\frac {\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}{5 x^5}-\frac {3}{40} a^5 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 104, normalized size = 1.11 \begin {gather*} \left (-\frac {a}{20 x^4}+\frac {a^3}{8 x^2}\right ) \sqrt {1-a^2 x^2}-\frac {\sqrt {1-a^2 x^2} \left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x)}{5 x^5}+\frac {3}{40} a^5 \log (x)-\frac {3}{40} a^5 \log \left (1+\sqrt {1-a^2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/x^6,x]

[Out]

(-1/20*a/x^4 + a^3/(8*x^2))*Sqrt[1 - a^2*x^2] - (Sqrt[1 - a^2*x^2]*(-1 + a^2*x^2)^2*ArcTanh[a*x])/(5*x^5) + (3
*a^5*Log[x])/40 - (3*a^5*Log[1 + Sqrt[1 - a^2*x^2]])/40

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Maple [A]
time = 1.44, size = 116, normalized size = 1.23

method	result	size
default	\(-\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (8 a^{4} x^{4} \arctanh \left (a x \right )-5 a^{3} x^{3}-16 a^{2} x^{2} \arctanh \left (a x \right )+2 a x +8 \arctanh \left (a x \right )\right )}{40 x^{5}}+\frac {3 a^{5} \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-1\right )}{40}-\frac {3 a^{5} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{40}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/40*(-(a*x-1)*(a*x+1))^(1/2)*(8*a^4*x^4*arctanh(a*x)-5*a^3*x^3-16*a^2*x^2*arctanh(a*x)+2*a*x+8*arctanh(a*x))
/x^5+3/40*a^5*ln((a*x+1)/(-a^2*x^2+1)^(1/2)-1)-3/40*a^5*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))

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Maxima [A]
time = 0.47, size = 126, normalized size = 1.34 \begin {gather*} \frac {1}{40} \, {\left ({\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{4} - 3 \, a^{4} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + 3 \, \sqrt {-a^{2} x^{2} + 1} a^{4} + \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} a^{2}}{x^{2}} - \frac {2 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{x^{4}}\right )} a - \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} \operatorname {artanh}\left (a x\right )}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^6,x, algorithm="maxima")

[Out]

1/40*((-a^2*x^2 + 1)^(3/2)*a^4 - 3*a^4*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) + 3*sqrt(-a^2*x^2 + 1)*a^4
+ (-a^2*x^2 + 1)^(5/2)*a^2/x^2 - 2*(-a^2*x^2 + 1)^(5/2)/x^4)*a - 1/5*(-a^2*x^2 + 1)^(5/2)*arctanh(a*x)/x^5

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Fricas [A]
time = 0.40, size = 93, normalized size = 0.99 \begin {gather*} \frac {3 \, a^{5} x^{5} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + {\left (5 \, a^{3} x^{3} - 2 \, a x - 4 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )\right )} \sqrt {-a^{2} x^{2} + 1}}{40 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^6,x, algorithm="fricas")

[Out]

1/40*(3*a^5*x^5*log((sqrt(-a^2*x^2 + 1) - 1)/x) + (5*a^3*x^3 - 2*a*x - 4*(a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x +
1)/(a*x - 1)))*sqrt(-a^2*x^2 + 1))/x^5

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \operatorname {atanh}{\left (a x \right )}}{x^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)**(3/2)*atanh(a*x)/x**6,x)

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)*atanh(a*x)/x**6, x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^6,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{3/2}}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atanh(a*x)*(1 - a^2*x^2)^(3/2))/x^6,x)

[Out]

int((atanh(a*x)*(1 - a^2*x^2)^(3/2))/x^6, x)

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