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

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Rubi [A]  time = 0.0750883, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, 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 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]

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]

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*}

Mathematica [A]  time = 0.008271, 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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Maple [A]  time = 0.058, size = 22, normalized size = 0.7 \begin{align*} 2\,{\frac{-1+x}{\sqrt{1+x}}{\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}}} \end{align*}

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 = 1.11657, size = 9, normalized size = 0.27 \begin{align*} 2 \, \sqrt{x - 1} \end{align*}

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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Fricas [A]  time = 1.5053, size = 50, normalized size = 1.52 \begin{align*} 2 \, \sqrt{x + 1} \sqrt{\frac{x - 1}{x + 1}} \end{align*}

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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Sympy [A]  time = 56.8934, size = 19, normalized size = 0.58 \begin{align*} \begin{cases} 2 \sqrt{x - 1} & \text{for}\: \left |{x}\right | > 1 \\2 i \sqrt{1 - x} & \text{otherwise} \end{cases} \end{align*}

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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Giac [C]  time = 1.15067, size = 18, normalized size = 0.55 \begin{align*} -2 i \, \sqrt{2} + 2 \, \sqrt{x - 1} \end{align*}

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)