3.328 \(\int \frac {e^{\coth ^{-1}(x)}}{\sqrt {1+x}} \, dx\)

Optimal. Leaf size=33 \[ \frac {2 \sqrt {\frac {1}{x}+1} \sqrt {-\frac {1-x}{x}} x}{\sqrt {x+1}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6176, 6181, 37} \[ \frac {2 \sqrt {\frac {1}{x}+1} \sqrt {-\frac {1-x}{x}} x}{\sqrt {x+1}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]/Sqrt[1 + x],x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 6176

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

Rule 6181

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.64 \[ \frac {2 \sqrt {1-\frac {1}{x^2}} x}{\sqrt {x+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[x]/Sqrt[1 + x],x]

[Out]

(2*Sqrt[1 - x^(-2)]*x)/Sqrt[1 + x]

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fricas [A]  time = 0.59, size = 18, normalized size = 0.55 \[ 2 \, \sqrt {x + 1} \sqrt {\frac {x - 1}{x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x + 1)*sqrt((x - 1)/(x + 1))

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giac [C]  time = 0.14, size = 13, normalized size = 0.39 \[ -2 i \, \sqrt {2} + 2 \, \sqrt {x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*sqrt(2) + 2*sqrt(x - 1)

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maple [A]  time = 0.03, size = 22, normalized size = 0.67 \[ \frac {-2+2 x}{\sqrt {\frac {-1+x}{1+x}}\, \sqrt {1+x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 7, normalized size = 0.21 \[ 2 \, \sqrt {x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x - 1)

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mupad [B]  time = 1.22, size = 18, normalized size = 0.55 \[ 2\,\sqrt {\frac {x-1}{x+1}}\,\sqrt {x+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 7.49, size = 19, normalized size = 0.58 \[ \begin {cases} 2 \sqrt {x - 1} & \text {for}\: \left |{x}\right | > 1 \\2 i \sqrt {1 - x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*sqrt(x - 1), Abs(x) > 1), (2*I*sqrt(1 - x), True))

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