\(\int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx\) [332]

Optimal result

Integrand size = 12, antiderivative size = 58 \[ \int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx=-\frac {\sqrt {2} \left (1+\frac {1}{x}\right )^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {-\frac {1-x}{x}}}\right )}{\left (\frac {1}{x}\right )^{3/2} (1+x)^{3/2}} \]

[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)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6311, 6316, 95, 209} \[ \int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx=-\frac {\sqrt {2} \left (\frac {1}{x}+1\right )^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {-\frac {1-x}{x}}}\right )}{\left (\frac {1}{x}\right )^{3/2} (x+1)^{3/2}} \]

[In]

Int[E^ArcCoth[x]/(1 + x)^(3/2),x]

[Out]

-((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 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} x^{3/2}} \, dx}{(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 {\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 {\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] (warning: unable to verify)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx=\sqrt {2} \sqrt {\frac {1}{1+x}} \sqrt {1+x} \arctan \left (\frac {\sqrt {\frac {-1+x}{x^2}} x}{\sqrt {2}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.64

method	result	size
default	\(\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {x -1}\, \sqrt {2}}{2}\right ) \sqrt {x -1}}{\sqrt {\frac {x -1}{1+x}}\, \sqrt {1+x}}\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.45 \[ \int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx=\sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1} \sqrt {\frac {x - 1}{x + 1}}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 40.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx=2 \left (\begin {cases} \frac {\sqrt {2} \operatorname {acos}{\left (\frac {\sqrt {2}}{\sqrt {x + 1}} \right )}}{2} & \text {for}\: \sqrt {x + 1} > - \sqrt {2} \wedge \sqrt {x + 1} < \sqrt {2} \end {cases}\right ) \]

[In]

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

[Out]

2*Piecewise((sqrt(2)*acos(sqrt(2)/sqrt(x + 1))/2, (sqrt(x + 1) < sqrt(2)) & (sqrt(x + 1) > -sqrt(2))))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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