3.39 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{x^2} \, dx\)

Optimal. Leaf size=37 \[ a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{x} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6124, 807, 266, 63, 208} \[ a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 44, normalized size = 1.19 \[ -\frac {\sqrt {1-a^2 x^2}}{x}+a \log \left (\sqrt {1-a^2 x^2}+1\right )-a \log (x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 40, normalized size = 1.08 \[ -\frac {a x \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt {-a^{2} x^{2} + 1}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^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 [B]  time = 0.04, size = 162, normalized size = 4.38 \[ -a \sqrt {-a^{2} x^{2}+1}+a \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{x}-a^{2} x \sqrt {-a^{2} x^{2}+1}-\frac {a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}+a \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 33, normalized size = 0.89 \[ a\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )-\frac {\sqrt {1-a^2\,x^2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*atanh((1 - a^2*x^2)^(1/2)) - (1 - a^2*x^2)^(1/2)/x

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{x^{2} \left (a x + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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