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

Optimal. Leaf size=38 \[ -\frac {\sqrt {1-a^2 x^2}}{x}-a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

[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 = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6124, 807, 266, 63, 208} \[ -\frac {\sqrt {1-a^2 x^2}}{x}-a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[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 = 1.16 \[ -\frac {\sqrt {1-a^2 x^2}}{x}-a \log \left (\sqrt {1-a^2 x^2}+1\right )+a \log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[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 = 0.52, size = 41, 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((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 [B]  time = 0.42, size = 96, normalized size = 2.53 \[ \frac {a^{4} x}{2 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left | a \right |}} - \frac {a^{2} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{2 \, x {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 35, normalized size = 0.92 \[ -a \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\frac {\sqrt {-a^{2} x^{2}+1}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 47, normalized size = 1.24 \[ -a \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {\sqrt {-a^{2} x^{2} + 1}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 34, 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((a*x + 1)/(x^2*(1 - a^2*x^2)^(1/2)),x)

[Out]

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

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sympy [C]  time = 2.37, size = 65, normalized size = 1.71 \[ a \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right ) + \begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((-acosh(1/(a*x)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True)) + Piecewise((-I*sqrt(a**2*x**2 -
1)/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True))

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