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

Optimal. Leaf size=107 \[ \frac {2 \sqrt {-\frac {1-x}{x}} x (x+1)^{3/2}}{5 \left (\frac {1}{x}+1\right )^{3/2}}+\frac {28 \sqrt {-\frac {1-x}{x}} (x+1)^{3/2}}{15 \left (\frac {1}{x}+1\right )^{3/2}}+\frac {86 \sqrt {-\frac {1-x}{x}} (x+1)^{3/2}}{15 \left (\frac {1}{x}+1\right )^{3/2} x} \]

[Out]

28/15*(1+x)^(3/2)*((-1+x)/x)^(1/2)/(1/x+1)^(3/2)+86/15*(1+x)^(3/2)*((-1+x)/x)^(1/2)/(1/x+1)^(3/2)/x+2/5*x*(1+x
)^(3/2)*((-1+x)/x)^(1/2)/(1/x+1)^(3/2)

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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 = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6176, 6181, 89, 78, 37} \[ \frac {2 \sqrt {-\frac {1-x}{x}} x (x+1)^{3/2}}{5 \left (\frac {1}{x}+1\right )^{3/2}}+\frac {28 \sqrt {-\frac {1-x}{x}} (x+1)^{3/2}}{15 \left (\frac {1}{x}+1\right )^{3/2}}+\frac {86 \sqrt {-\frac {1-x}{x}} (x+1)^{3/2}}{15 \left (\frac {1}{x}+1\right )^{3/2} x} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]*(1 + x)^(3/2),x]

[Out]

(28*Sqrt[-((1 - x)/x)]*(1 + x)^(3/2))/(15*(1 + x^(-1))^(3/2)) + (86*Sqrt[-((1 - x)/x)]*(1 + x)^(3/2))/(15*(1 +
x^(-1))^(3/2)*x) + (2*Sqrt[-((1 - x)/x)]*x*(1 + x)^(3/2))/(5*(1 + x^(-1))^(3/2))

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 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 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)} (1+x)^{3/2} \, dx &=\frac {(1+x)^{3/2} \int e^{\coth ^{-1}(x)} \left (1+\frac {1}{x}\right )^{3/2} x^{3/2} \, 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^{7/2}} \, dx,x,\frac {1}{x}\right )}{\left (1+\frac {1}{x}\right )^{3/2}}\\ &=\frac {2 \sqrt {-\frac {1-x}{x}} x (1+x)^{3/2}}{5 \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 {7+\frac {5 x}{2}}{\sqrt {1-x} x^{5/2}} \, dx,x,\frac {1}{x}\right )}{5 \left (1+\frac {1}{x}\right )^{3/2}}\\ &=\frac {28 \sqrt {-\frac {1-x}{x}} (1+x)^{3/2}}{15 \left (1+\frac {1}{x}\right )^{3/2}}+\frac {2 \sqrt {-\frac {1-x}{x}} x (1+x)^{3/2}}{5 \left (1+\frac {1}{x}\right )^{3/2}}-\frac {\left (43 \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 )}{15 \left (1+\frac {1}{x}\right )^{3/2}}\\ &=\frac {28 \sqrt {-\frac {1-x}{x}} (1+x)^{3/2}}{15 \left (1+\frac {1}{x}\right )^{3/2}}+\frac {86 \sqrt {-\frac {1-x}{x}} (1+x)^{3/2}}{15 \left (1+\frac {1}{x}\right )^{3/2} x}+\frac {2 \sqrt {-\frac {1-x}{x}} x (1+x)^{3/2}}{5 \left (1+\frac {1}{x}\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 0.38 \[ \frac {2 \sqrt {\frac {x-1}{x}} \sqrt {x+1} \left (3 x^2+14 x+43\right )}{15 \sqrt {\frac {1}{x}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(-1 + x)/x]*Sqrt[1 + x]*(43 + 14*x + 3*x^2))/(15*Sqrt[1 + x^(-1)])

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fricas [A]  time = 0.54, size = 28, normalized size = 0.26 \[ \frac {2}{15} \, {\left (3 \, x^{2} + 14 \, x + 43\right )} \sqrt {x + 1} \sqrt {\frac {x - 1}{x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*x^2 + 14*x + 43)*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)^(3/2),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 = 32, normalized size = 0.30 \[ \frac {2 \left (-1+x \right ) \left (3 x^{2}+14 x +43\right )}{15 \sqrt {1+x}\, \sqrt {\frac {-1+x}{1+x}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((-1+x)/(1+x))^(1/2)*(1+x)^(3/2),x)

[Out]

2/15*(-1+x)*(3*x^2+14*x+43)/(1+x)^(1/2)/((-1+x)/(1+x))^(1/2)

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maxima [A]  time = 1.03, size = 22, normalized size = 0.21 \[ \frac {2 \, {\left (3 \, x^{3} + 11 \, x^{2} + 29 \, x - 43\right )}}{15 \, \sqrt {x - 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*x^3 + 11*x^2 + 29*x - 43)/sqrt(x - 1)

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mupad [B]  time = 1.26, size = 38, normalized size = 0.36 \[ \sqrt {\frac {x-1}{x+1}}\,\left (\frac {28\,x\,\sqrt {x+1}}{15}+\frac {86\,\sqrt {x+1}}{15}+\frac {2\,x^2\,\sqrt {x+1}}{5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)^(3/2)/((x - 1)/(x + 1))^(1/2),x)

[Out]

((x - 1)/(x + 1))^(1/2)*((28*x*(x + 1)^(1/2))/15 + (86*(x + 1)^(1/2))/15 + (2*x^2*(x + 1)^(1/2))/5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x + 1\right )^{\frac {3}{2}}}{\sqrt {\frac {x - 1}{x + 1}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x + 1)**(3/2)/sqrt((x - 1)/(x + 1)), x)

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