\(\int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^{3/2}} \, dx\) [331]

Optimal result

Integrand size = 13, antiderivative size = 93 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^{3/2}} \, dx=\frac {2 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2}{(1+x)^{3/2}}+\frac {\sqrt {2} \left (1+\frac {1}{x}\right )^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {-\frac {1-x}{x}}}\right )}{\left (\frac {1}{x}\right )^{3/2} (1+x)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6311, 6316, 98, 95, 209} \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^{3/2}} \, dx=\frac {\sqrt {2} \left (\frac {1}{x}+1\right )^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {-\frac {1-x}{x}}}\right )}{\left (\frac {1}{x}\right )^{3/2} (x+1)^{3/2}}+\frac {2 \left (\frac {1}{x}+1\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2}{(x+1)^{3/2}} \]

[In]

Int[(E^ArcCoth[x]*x)/(1 + x)^(3/2),x]

[Out]

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

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 6311

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 6316

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> Dist[(-c^p)*x^m*(1/x)^m, S
ubst[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*} \text {integral}& = \frac {\left (\left (1+\frac {1}{x}\right )^{3/2} x^{3/2}\right ) \int \frac {e^{\coth ^{-1}(x)}}{\left (1+\frac {1}{x}\right )^{3/2} \sqrt {x}} \, dx}{(1+x)^{3/2}} \\ & = -\frac {\left (1+\frac {1}{x}\right )^{3/2} \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x^{3/2} (1+x)} \, dx,x,\frac {1}{x}\right )}{\left (\frac {1}{x}\right )^{3/2} (1+x)^{3/2}} \\ & = \frac {2 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2}{(1+x)^{3/2}}+\frac {\left (1+\frac {1}{x}\right )^{3/2} \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {x} (1+x)} \, dx,x,\frac {1}{x}\right )}{\left (\frac {1}{x}\right )^{3/2} (1+x)^{3/2}} \\ & = \frac {2 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2}{(1+x)^{3/2}}+\frac {\left (2 \left (1+\frac {1}{x}\right )^{3/2}\right ) \text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {\sqrt {\frac {1}{x}}}{\sqrt {\frac {-1+x}{x}}}\right )}{\left (\frac {1}{x}\right )^{3/2} (1+x)^{3/2}} \\ & = \frac {2 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2}{(1+x)^{3/2}}+\frac {\sqrt {2} \left (1+\frac {1}{x}\right )^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {-\frac {1-x}{x}}}\right )}{\left (\frac {1}{x}\right )^{3/2} (1+x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^{3/2}} \, dx=\frac {\sqrt {1+\frac {1}{x}} x \left (2 \sqrt {\frac {-1+x}{x}}-\sqrt {2} \sqrt {\frac {1}{x}} \arctan \left (\frac {\sqrt {\frac {-1+x}{x^2}} x}{\sqrt {2}}\right )\right )}{\sqrt {1+x}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.51

method	result	size
default	\(-\frac {\sqrt {x -1}\, \left (\sqrt {2}\, \arctan \left (\frac {\sqrt {x -1}\, \sqrt {2}}{2}\right )-2 \sqrt {x -1}\right )}{\sqrt {\frac {x -1}{1+x}}\, \sqrt {1+x}}\)	\(47\)
risch	\(\frac {2 x -2}{\sqrt {\frac {x -1}{1+x}}\, \sqrt {1+x}}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {x -1}\, \sqrt {2}}{2}\right ) \sqrt {x -1}}{\sqrt {\frac {x -1}{1+x}}\, \sqrt {1+x}}\)	\(60\)

[In]

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

[Out]

-(x-1)^(1/2)*(2^(1/2)*arctan(1/2*(x-1)^(1/2)*2^(1/2))-2*(x-1)^(1/2))/((x-1)/(1+x))^(1/2)/(1+x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.49 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^{3/2}} \, dx=-\sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1} \sqrt {\frac {x - 1}{x + 1}}\right ) + 2 \, \sqrt {x + 1} \sqrt {\frac {x - 1}{x + 1}} \]

[In]

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

[Out]

-sqrt(2)*arctan(1/2*sqrt(2)*sqrt(x + 1)*sqrt((x - 1)/(x + 1))) + 2*sqrt(x + 1)*sqrt((x - 1)/(x + 1))

Sympy [F]

\[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^{3/2}} \, dx=\int \frac {x}{\sqrt {\frac {x - 1}{x + 1}} \left (x + 1\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^{3/2}} \, dx=\int { \frac {x}{{\left (x + 1\right )}^{\frac {3}{2}} \sqrt {\frac {x - 1}{x + 1}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^{3/2}} \, dx=\text {Exception raised: NotImplementedError} \]

[In]

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

[Out]

Exception raised: NotImplementedError >> unable to parse Giac output: 2*(sqrt(sageVARx-1)+(atan(i)-2*i)/sqrt(2
)-atan(sqrt(sageVARx-1)/sqrt(2))/sqrt(2))*sign(sageVARx)/sign(sageVARx+1)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^{3/2}} \, dx=\int \frac {x}{\sqrt {\frac {x-1}{x+1}}\,{\left (x+1\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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