3.40 \(\int \frac{e^{-\tanh ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=63 \[ \frac{a \sqrt{1-a^2 x^2}}{x}-\frac{\sqrt{1-a^2 x^2}}{2 x^2}-\frac{1}{2} a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0604718, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6124, 835, 807, 266, 63, 208} \[ \frac{a \sqrt{1-a^2 x^2}}{x}-\frac{\sqrt{1-a^2 x^2}}{2 x^2}-\frac{1}{2} a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6124

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/
2)*Sqrt[1 - a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(n - 1)/2]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0419927, size = 57, normalized size = 0.9 \[ \frac{1}{2} \left (\frac{(2 a x-1) \sqrt{1-a^2 x^2}}{x^2}-a^2 \log \left (\sqrt{1-a^2 x^2}+1\right )+a^2 \log (x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.043, size = 186, normalized size = 3. \begin{align*}{\frac{a}{x} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{a}^{3}x\sqrt{-{a}^{2}{x}^{2}+1}+{{a}^{3}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{{a}^{2}}{2}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{a}^{2}}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }-{a}^{2}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }-{{a}^{3}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{\frac{1}{2\,{x}^{2}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.93545, size = 113, normalized size = 1.79 \begin{align*} \frac{a^{2} x^{2} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt{-a^{2} x^{2} + 1}{\left (2 \, a x - 1\right )}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- \left (a x - 1\right ) \left (a x + 1\right )}}{x^{3} \left (a x + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.21876, size = 215, normalized size = 3.41 \begin{align*} \frac{{\left (a^{3} - \frac{4 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a}{x}\right )} a^{4} x^{2}}{8 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}{\left | a \right |}} - \frac{a^{3} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{2 \,{\left | a \right |}} + \frac{\frac{4 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a{\left | a \right |}}{x} - \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}{\left | a \right |}}{a x^{2}}}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^3 - 4*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a/x)*a^4*x^2/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*abs(a)) - 1/2*a^3
*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + 1/8*(4*(sqrt(-a^2*x^2 + 1)*abs(a) + a)
*a*abs(a)/x - (sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*abs(a)/(a*x^2))/a^2