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

Optimal. Leaf size=107 \[ \frac {2 \sqrt {-\frac {1-x}{x}} \sqrt {x+1} x^2}{5 \sqrt {\frac {1}{x}+1}}+\frac {6 \sqrt {-\frac {1-x}{x}} \sqrt {x+1} x}{5 \sqrt {\frac {1}{x}+1}}+\frac {12 \sqrt {-\frac {1-x}{x}} \sqrt {x+1}}{5 \sqrt {\frac {1}{x}+1}} \]

[Out]

12/5*((-1+x)/x)^(1/2)*(1+x)^(1/2)/(1/x+1)^(1/2)+6/5*x*((-1+x)/x)^(1/2)*(1+x)^(1/2)/(1/x+1)^(1/2)+2/5*x^2*((-1+
x)/x)^(1/2)*(1+x)^(1/2)/(1/x+1)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6176, 6181, 78, 45, 37} \[ \frac {2 \sqrt {-\frac {1-x}{x}} \sqrt {x+1} x^2}{5 \sqrt {\frac {1}{x}+1}}+\frac {6 \sqrt {-\frac {1-x}{x}} \sqrt {x+1} x}{5 \sqrt {\frac {1}{x}+1}}+\frac {12 \sqrt {-\frac {1-x}{x}} \sqrt {x+1}}{5 \sqrt {\frac {1}{x}+1}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]*x*Sqrt[1 + x],x]

[Out]

(12*Sqrt[-((1 - x)/x)]*Sqrt[1 + x])/(5*Sqrt[1 + x^(-1)]) + (6*Sqrt[-((1 - x)/x)]*x*Sqrt[1 + x])/(5*Sqrt[1 + x^
(-1)]) + (2*Sqrt[-((1 - x)/x)]*x^2*Sqrt[1 + x])/(5*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 45

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] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 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)} x \sqrt {1+x} \, dx &=\frac {\sqrt {1+x} \int e^{\coth ^{-1}(x)} \sqrt {1+\frac {1}{x}} x^{3/2} \, 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^{7/2}} \, dx,x,\frac {1}{x}\right )}{\sqrt {1+\frac {1}{x}}}\\ &=\frac {2 \sqrt {-\frac {1-x}{x}} x^2 \sqrt {1+x}}{5 \sqrt {1+\frac {1}{x}}}-\frac {\left (9 \sqrt {\frac {1}{x}} \sqrt {1+x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x^{5/2}} \, dx,x,\frac {1}{x}\right )}{5 \sqrt {1+\frac {1}{x}}}\\ &=\frac {6 \sqrt {-\frac {1-x}{x}} x \sqrt {1+x}}{5 \sqrt {1+\frac {1}{x}}}+\frac {2 \sqrt {-\frac {1-x}{x}} x^2 \sqrt {1+x}}{5 \sqrt {1+\frac {1}{x}}}-\frac {\left (6 \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 )}{5 \sqrt {1+\frac {1}{x}}}\\ &=\frac {12 \sqrt {-\frac {1-x}{x}} \sqrt {1+x}}{5 \sqrt {1+\frac {1}{x}}}+\frac {6 \sqrt {-\frac {1-x}{x}} x \sqrt {1+x}}{5 \sqrt {1+\frac {1}{x}}}+\frac {2 \sqrt {-\frac {1-x}{x}} x^2 \sqrt {1+x}}{5 \sqrt {1+\frac {1}{x}}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 39, normalized size = 0.36 \[ \frac {2 \sqrt {\frac {x-1}{x}} \sqrt {x+1} \left (x^2+3 x+6\right )}{5 \sqrt {\frac {1}{x}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[x]*x*Sqrt[1 + x],x]

[Out]

(2*Sqrt[(-1 + x)/x]*Sqrt[1 + x]*(6 + 3*x + x^2))/(5*Sqrt[1 + x^(-1)])

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fricas [A]  time = 0.41, size = 26, normalized size = 0.24 \[ \frac {2}{5} \, {\left (x^{2} + 3 \, x + 6\right )} \sqrt {x + 1} \sqrt {\frac {x - 1}{x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(x^2 + 3*x + 6)*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)*x*(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 = 30, normalized size = 0.28 \[ \frac {2 \left (-1+x \right ) \left (x^{2}+3 x +6\right )}{5 \sqrt {\frac {-1+x}{1+x}}\, \sqrt {1+x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(-1+x)*(x^2+3*x+6)/((-1+x)/(1+x))^(1/2)/(1+x)^(1/2)

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maxima [A]  time = 0.46, size = 20, normalized size = 0.19 \[ \frac {2 \, {\left (x^{3} + 2 \, x^{2} + 3 \, x - 6\right )}}{5 \, \sqrt {x - 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(x^3 + 2*x^2 + 3*x - 6)/sqrt(x - 1)

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mupad [B]  time = 1.25, size = 38, normalized size = 0.36 \[ \sqrt {\frac {x-1}{x+1}}\,\left (\frac {6\,x\,\sqrt {x+1}}{5}+\frac {12\,\sqrt {x+1}}{5}+\frac {2\,x^2\,\sqrt {x+1}}{5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x - 1)/(x + 1))^(1/2)*((6*x*(x + 1)^(1/2))/5 + (12*(x + 1)^(1/2))/5 + (2*x^2*(x + 1)^(1/2))/5)

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sympy [C]  time = 7.93, size = 133, normalized size = 1.24 \[ - 2 \left (\begin {cases} \frac {x \sqrt {x - 1}}{3} + \frac {5 \sqrt {x - 1}}{3} & \text {for}\: \frac {\left |{x + 1}\right |}{2} > 1 \\\frac {i x \sqrt {1 - x}}{3} + \frac {5 i \sqrt {1 - x}}{3} & \text {otherwise} \end {cases}\right ) + 2 \left (\begin {cases} \frac {8 x \sqrt {x - 1}}{15} + \frac {\sqrt {x - 1} \left (x + 1\right )^{2}}{5} + \frac {8 \sqrt {x - 1}}{3} & \text {for}\: \frac {\left |{x + 1}\right |}{2} > 1 \\\frac {8 i x \sqrt {1 - x}}{15} + \frac {i \sqrt {1 - x} \left (x + 1\right )^{2}}{5} + \frac {8 i \sqrt {1 - x}}{3} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*Piecewise((x*sqrt(x - 1)/3 + 5*sqrt(x - 1)/3, Abs(x + 1)/2 > 1), (I*x*sqrt(1 - x)/3 + 5*I*sqrt(1 - x)/3, Tr
ue)) + 2*Piecewise((8*x*sqrt(x - 1)/15 + sqrt(x - 1)*(x + 1)**2/5 + 8*sqrt(x - 1)/3, Abs(x + 1)/2 > 1), (8*I*x
*sqrt(1 - x)/15 + I*sqrt(1 - x)*(x + 1)**2/5 + 8*I*sqrt(1 - x)/3, True))

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