3.319 \(\int e^{\coth ^{-1}(x)} x (1+x)^{3/2} \, dx\)

Optimal. Leaf size=144 \[ \frac{2 \sqrt{-\frac{1-x}{x}} (x+1)^{3/2} x^2}{7 \left (\frac{1}{x}+1\right )^{3/2}}+\frac{8 \sqrt{-\frac{1-x}{x}} (x+1)^{3/2} x}{7 \left (\frac{1}{x}+1\right )^{3/2}}+\frac{46 \sqrt{-\frac{1-x}{x}} (x+1)^{3/2}}{21 \left (\frac{1}{x}+1\right )^{3/2}}+\frac{92 \sqrt{-\frac{1-x}{x}} (x+1)^{3/2}}{21 \left (\frac{1}{x}+1\right )^{3/2} x} \]

[Out]

(46*Sqrt[-((1 - x)/x)]*(1 + x)^(3/2))/(21*(1 + x^(-1))^(3/2)) + (92*Sqrt[-((1 - x)/x)]*(1 + x)^(3/2))/(21*(1 +
x^(-1))^(3/2)*x) + (8*Sqrt[-((1 - x)/x)]*x*(1 + x)^(3/2))/(7*(1 + x^(-1))^(3/2)) + (2*Sqrt[-((1 - x)/x)]*x^2*
(1 + x)^(3/2))/(7*(1 + x^(-1))^(3/2))

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Rubi [A]  time = 0.127649, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {6176, 6181, 89, 78, 45, 37} \[ \frac{2 \sqrt{-\frac{1-x}{x}} (x+1)^{3/2} x^2}{7 \left (\frac{1}{x}+1\right )^{3/2}}+\frac{8 \sqrt{-\frac{1-x}{x}} (x+1)^{3/2} x}{7 \left (\frac{1}{x}+1\right )^{3/2}}+\frac{46 \sqrt{-\frac{1-x}{x}} (x+1)^{3/2}}{21 \left (\frac{1}{x}+1\right )^{3/2}}+\frac{92 \sqrt{-\frac{1-x}{x}} (x+1)^{3/2}}{21 \left (\frac{1}{x}+1\right )^{3/2} x} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]*x*(1 + x)^(3/2),x]

[Out]

(46*Sqrt[-((1 - x)/x)]*(1 + x)^(3/2))/(21*(1 + x^(-1))^(3/2)) + (92*Sqrt[-((1 - x)/x)]*(1 + x)^(3/2))/(21*(1 +
x^(-1))^(3/2)*x) + (8*Sqrt[-((1 - x)/x)]*x*(1 + x)^(3/2))/(7*(1 + x^(-1))^(3/2)) + (2*Sqrt[-((1 - x)/x)]*x^2*
(1 + x)^(3/2))/(7*(1 + x^(-1))^(3/2))

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 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 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 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 e^{\coth ^{-1}(x)} x (1+x)^{3/2} \, dx &=\frac{(1+x)^{3/2} \int e^{\coth ^{-1}(x)} \left (1+\frac{1}{x}\right )^{3/2} x^{5/2} \, dx}{\left (1+\frac{1}{x}\right )^{3/2} x^{3/2}}\\ &=-\frac{\left (\left (\frac{1}{x}\right )^{3/2} (1+x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{(1+x)^2}{\sqrt{1-x} x^{9/2}} \, dx,x,\frac{1}{x}\right )}{\left (1+\frac{1}{x}\right )^{3/2}}\\ &=\frac{2 \sqrt{-\frac{1-x}{x}} x^2 (1+x)^{3/2}}{7 \left (1+\frac{1}{x}\right )^{3/2}}-\frac{\left (2 \left (\frac{1}{x}\right )^{3/2} (1+x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{10+\frac{7 x}{2}}{\sqrt{1-x} x^{7/2}} \, dx,x,\frac{1}{x}\right )}{7 \left (1+\frac{1}{x}\right )^{3/2}}\\ &=\frac{8 \sqrt{-\frac{1-x}{x}} x (1+x)^{3/2}}{7 \left (1+\frac{1}{x}\right )^{3/2}}+\frac{2 \sqrt{-\frac{1-x}{x}} x^2 (1+x)^{3/2}}{7 \left (1+\frac{1}{x}\right )^{3/2}}-\frac{\left (23 \left (\frac{1}{x}\right )^{3/2} (1+x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} x^{5/2}} \, dx,x,\frac{1}{x}\right )}{7 \left (1+\frac{1}{x}\right )^{3/2}}\\ &=\frac{46 \sqrt{-\frac{1-x}{x}} (1+x)^{3/2}}{21 \left (1+\frac{1}{x}\right )^{3/2}}+\frac{8 \sqrt{-\frac{1-x}{x}} x (1+x)^{3/2}}{7 \left (1+\frac{1}{x}\right )^{3/2}}+\frac{2 \sqrt{-\frac{1-x}{x}} x^2 (1+x)^{3/2}}{7 \left (1+\frac{1}{x}\right )^{3/2}}-\frac{\left (46 \left (\frac{1}{x}\right )^{3/2} (1+x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} x^{3/2}} \, dx,x,\frac{1}{x}\right )}{21 \left (1+\frac{1}{x}\right )^{3/2}}\\ &=\frac{46 \sqrt{-\frac{1-x}{x}} (1+x)^{3/2}}{21 \left (1+\frac{1}{x}\right )^{3/2}}+\frac{92 \sqrt{-\frac{1-x}{x}} (1+x)^{3/2}}{21 \left (1+\frac{1}{x}\right )^{3/2} x}+\frac{8 \sqrt{-\frac{1-x}{x}} x (1+x)^{3/2}}{7 \left (1+\frac{1}{x}\right )^{3/2}}+\frac{2 \sqrt{-\frac{1-x}{x}} x^2 (1+x)^{3/2}}{7 \left (1+\frac{1}{x}\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0211074, size = 46, normalized size = 0.32 \[ \frac{2 \sqrt{\frac{x-1}{x}} \sqrt{x+1} \left (3 x^3+12 x^2+23 x+46\right )}{21 \sqrt{\frac{1}{x}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[x]*x*(1 + x)^(3/2),x]

[Out]

(2*Sqrt[(-1 + x)/x]*Sqrt[1 + x]*(46 + 23*x + 12*x^2 + 3*x^3))/(21*Sqrt[1 + x^(-1)])

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Maple [A]  time = 0.06, size = 37, normalized size = 0.3 \begin{align*}{\frac{ \left ( -2+2\,x \right ) \left ( 3\,{x}^{3}+12\,{x}^{2}+23\,x+46 \right ) }{21}{\frac{1}{\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)*x*(1+x)^(3/2),x)

[Out]

2/21*(-1+x)*(3*x^3+12*x^2+23*x+46)/(1+x)^(1/2)/((-1+x)/(1+x))^(1/2)

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Maxima [A]  time = 1.07647, size = 36, normalized size = 0.25 \begin{align*} \frac{2 \,{\left (3 \, x^{4} + 9 \, x^{3} + 11 \, x^{2} + 23 \, x - 46\right )}}{21 \, \sqrt{x - 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*x*(1+x)^(3/2),x, algorithm="maxima")

[Out]

2/21*(3*x^4 + 9*x^3 + 11*x^2 + 23*x - 46)/sqrt(x - 1)

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Fricas [A]  time = 1.52876, size = 93, normalized size = 0.65 \begin{align*} \frac{2}{21} \,{\left (3 \, x^{3} + 12 \, x^{2} + 23 \, x + 46\right )} \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)*x*(1+x)^(3/2),x, algorithm="fricas")

[Out]

2/21*(3*x^3 + 12*x^2 + 23*x + 46)*sqrt(x + 1)*sqrt((x - 1)/(x + 1))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))**(1/2)*x*(1+x)**(3/2),x)

[Out]

Timed out

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Giac [C]  time = 1.16036, size = 46, normalized size = 0.32 \begin{align*} \frac{2}{7} \,{\left (x - 1\right )}^{\frac{7}{2}} + 2 \,{\left (x - 1\right )}^{\frac{5}{2}} + \frac{16}{3} \,{\left (x - 1\right )}^{\frac{3}{2}} - \frac{64}{21} i \, \sqrt{2} + 8 \, \sqrt{x - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*x*(1+x)^(3/2),x, algorithm="giac")

[Out]

2/7*(x - 1)^(7/2) + 2*(x - 1)^(5/2) + 16/3*(x - 1)^(3/2) - 64/21*I*sqrt(2) + 8*sqrt(x - 1)