3.368 \(\int \frac {e^{\tanh ^{-1}(x)}}{1+x} \, dx\)

Optimal. Leaf size=2 \[ \sin ^{-1}(x) \]

[Out]

arcsin(x)

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Rubi [A]  time = 0.02, antiderivative size = 2, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6129, 41, 216} \[ \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]/(1 + x),x]

[Out]

ArcSin[x]

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

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 6129

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[(u*(1 + (d*x)/c)
^p*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ
[p] || GtQ[c, 0])

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 2, normalized size = 1.00 \[ \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[x]/(1 + x),x]

[Out]

ArcSin[x]

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fricas [B]  time = 0.45, size = 18, normalized size = 9.00 \[ -2 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*arctan((sqrt(-x^2 + 1) - 1)/x)

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giac [A]  time = 0.18, size = 2, normalized size = 1.00 \[ \arcsin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(x)

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maple [A]  time = 0.03, size = 3, normalized size = 1.50 \[ \arcsin \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(x)

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maxima [A]  time = 0.40, size = 2, normalized size = 1.00 \[ \arcsin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(x)

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mupad [B]  time = 0.01, size = 2, normalized size = 1.00 \[ \mathrm {asin}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1 - x^2)^(1/2),x)

[Out]

asin(x)

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sympy [A]  time = 0.12, size = 2, normalized size = 1.00 \[ \operatorname {asin}{\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

asin(x)

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