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

Optimal. Leaf size=70 \[ \frac {2 \sqrt {-\frac {1-x}{x}} \sqrt {x+1} x}{3 \sqrt {\frac {1}{x}+1}}+\frac {10 \sqrt {-\frac {1-x}{x}} \sqrt {x+1}}{3 \sqrt {\frac {1}{x}+1}} \]

[Out]

10/3*((-1+x)/x)^(1/2)*(1+x)^(1/2)/(1/x+1)^(1/2)+2/3*x*((-1+x)/x)^(1/2)*(1+x)^(1/2)/(1/x+1)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6176, 6181, 78, 37} \[ \frac {2 \sqrt {-\frac {1-x}{x}} \sqrt {x+1} x}{3 \sqrt {\frac {1}{x}+1}}+\frac {10 \sqrt {-\frac {1-x}{x}} \sqrt {x+1}}{3 \sqrt {\frac {1}{x}+1}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]*Sqrt[1 + x],x]

[Out]

(10*Sqrt[-((1 - x)/x)]*Sqrt[1 + x])/(3*Sqrt[1 + x^(-1)]) + (2*Sqrt[-((1 - x)/x)]*x*Sqrt[1 + x])/(3*Sqrt[1 + x^
(-1)])

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 e^{\coth ^{-1}(x)} \sqrt {1+x} \, dx &=\frac {\sqrt {1+x} \int e^{\coth ^{-1}(x)} \sqrt {1+\frac {1}{x}} \sqrt {x} \, dx}{\sqrt {1+\frac {1}{x}} \sqrt {x}}\\ &=-\frac {\left (\sqrt {\frac {1}{x}} \sqrt {1+x}\right ) \operatorname {Subst}\left (\int \frac {1+x}{\sqrt {1-x} x^{5/2}} \, dx,x,\frac {1}{x}\right )}{\sqrt {1+\frac {1}{x}}}\\ &=\frac {2 \sqrt {-\frac {1-x}{x}} x \sqrt {1+x}}{3 \sqrt {1+\frac {1}{x}}}-\frac {\left (5 \sqrt {\frac {1}{x}} \sqrt {1+x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x^{3/2}} \, dx,x,\frac {1}{x}\right )}{3 \sqrt {1+\frac {1}{x}}}\\ &=\frac {10 \sqrt {-\frac {1-x}{x}} \sqrt {1+x}}{3 \sqrt {1+\frac {1}{x}}}+\frac {2 \sqrt {-\frac {1-x}{x}} x \sqrt {1+x}}{3 \sqrt {1+\frac {1}{x}}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 0.49 \[ \frac {2 \sqrt {\frac {x-1}{x}} \sqrt {x+1} (x+5)}{3 \sqrt {\frac {1}{x}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[x]*Sqrt[1 + x],x]

[Out]

(2*Sqrt[(-1 + x)/x]*Sqrt[1 + x]*(5 + x))/(3*Sqrt[1 + x^(-1)])

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fricas [A]  time = 0.52, size = 21, normalized size = 0.30 \[ \frac {2}{3} \, {\left (x + 5\right )} \sqrt {x + 1} \sqrt {\frac {x - 1}{x + 1}} \]

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]

2/3*(x + 5)*sqrt(x + 1)*sqrt((x - 1)/(x + 1))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.04, size = 25, normalized size = 0.36 \[ \frac {2 \left (-1+x \right ) \left (x +5\right )}{3 \sqrt {\frac {-1+x}{1+x}}\, \sqrt {1+x}} \]

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]

2/3*(-1+x)*(x+5)/((-1+x)/(1+x))^(1/2)/(1+x)^(1/2)

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maxima [A]  time = 0.89, size = 15, normalized size = 0.21 \[ \frac {2 \, {\left (x^{2} + 4 \, x - 5\right )}}{3 \, \sqrt {x - 1}} \]

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]

2/3*(x^2 + 4*x - 5)/sqrt(x - 1)

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mupad [B]  time = 1.24, size = 21, normalized size = 0.30 \[ \frac {2\,\sqrt {\frac {x-1}{x+1}}\,\sqrt {x+1}\,\left (x+5\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*((x - 1)/(x + 1))^(1/2)*(x + 1)^(1/2)*(x + 5))/3

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sympy [A]  time = 4.83, size = 39, normalized size = 0.56 \[ 2 \left (\begin {cases} 2 \sqrt {2} \left (\frac {\sqrt {2} \left (x - 1\right )^{\frac {3}{2}}}{12} + \frac {\sqrt {2} \sqrt {x - 1}}{2}\right ) & \text {for}\: x \geq -1 \wedge x < 1 \end {cases}\right ) \]

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]

2*Piecewise((2*sqrt(2)*(sqrt(2)*(x - 1)**(3/2)/12 + sqrt(2)*sqrt(x - 1)/2), (x >= -1) & (x < 1)))

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