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

Optimal. Leaf size=22 \[ \sin ^{-1}(a x)-\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6124, 844, 216, 266, 63, 208} \[ \sin ^{-1}(a x)-\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/x,x]

[Out]

ArcSin[a*x] - 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 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[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 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 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} \, dx &=\int \frac {1+a x}{x \sqrt {1-a^2 x^2}} \, dx\\ &=a \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx+\int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx\\ &=\sin ^{-1}(a x)+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=\sin ^{-1}(a 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^2}\\ &=\sin ^{-1}(a x)-\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 1.18 \[ -\log \left (\sqrt {1-a^2 x^2}+1\right )+\sin ^{-1}(a x)+\log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[a*x]/x,x]

[Out]

ArcSin[a*x] + Log[x] - Log[1 + Sqrt[1 - a^2*x^2]]

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fricas [B]  time = 0.48, size = 44, normalized size = 2.00 \[ -2 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.23, size = 51, normalized size = 2.32 \[ \frac {a \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{{\left | a \right |}} - \frac {a \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 44, normalized size = 2.00 \[ \frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 33, normalized size = 1.50 \[ \arcsin \left (a x\right ) - \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 35, normalized size = 1.59 \[ \frac {a\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}-\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 5.92, size = 70, normalized size = 3.18 \[ a \left (\begin {cases} \sqrt {\frac {1}{a^{2}}} \operatorname {asin}{\left (x \sqrt {a^{2}} \right )} & \text {for}\: a^{2} > 0 \\\sqrt {- \frac {1}{a^{2}}} \operatorname {asinh}{\left (x \sqrt {- a^{2}} \right )} & \text {for}\: a^{2} < 0 \end {cases}\right ) + \begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((sqrt(a**(-2))*asin(x*sqrt(a**2)), a**2 > 0), (sqrt(-1/a**2)*asinh(x*sqrt(-a**2)), a**2 < 0)) + Pi
ecewise((-acosh(1/(a*x)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True))

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